Light Shadows and Reflection – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Sources Of Light :

Any object that gives out light is called a source of light. Luminous objects are also called sources of light. Sources of light can be natural or artificial (man-made) (Fig. 13.1).
Fig. 13.1 Sources of light

Examples of natural sources of light are ‘he sun and other stairs and insects like the firefly. Some artificial sources of light are candle, electric bulb, and laser.

Transparent, Translucent, And Opaque Materials :

Different types of materials transmit light differently. Based on the way they transmit light, materials can be divided into transparent, translucent, and opaque materials.
Materials that allow light to pass through significant scattering or absorption [Fig. 13.2(a)] are called transparent materials. We will be able to see through these materials very clearly. Examples of transparent materials are clear air, clear glass, clean water, some kinds of plastic, and cellophane paper.
Materials that allow light to pass through them, but scatter or diffuse the light as it passes through, i.e., a parallel beam of light comes through in all directions are called translucent materials [Fig. 13.2(b)], That is why an object cannot be seen clearly through a translucent material. Examples of translucent materials are butter paper, a frosted glass, paper smeared with oil, and smoked glass.
Materials that completely block light are called opaque materials [Fig. 13.2(c)]. We will not be able to see through these materials at all. Examples of an opaque materials are metal, mud, cement, coal, and wood. A mirror is a very good example of opaque material. An ideal mirror does not let any light pass through it.

Propagation Of Light :

Usually light travels in a straight line. When we want to represent the propagation of light with a diagram, we represent it with the help of rays and beams.
Ray A ray is a line with an arrow that shows the direction of propagation of light, and such a diagram is called a ray diagram.
Beam A group of light rays moving in an organized manner is called a beam of light.

The property of light to travel in straight lines explains many interesting phenomena related to light, like formation of shadows by opaque objects and formation of images in a pin-hole camera.

Shadows :

An opaque object blocks the light falling on it. This creates an area of darkness on the side of the object away from the source of light. A translucent object also creates a faint area of darkness. An area of darkness formed by an opaque object obstructing light is called a shadow. The following three things are required for a shadow to form (Fig. 13.4):
• a source of light
• an opaque object
• a screen or irface behind the object.

A shadow will not form if any of these is absent. This explains why we cannot see a shadow in the dark. It is only when light rays are obstructed by an opaque object that we get a shadow of the object.
Let us perform an activity to learn about the characteristics of a shadow.
Characteristics of a Shadow:
A shadow has the following three characteristics:
1. It is always black, regardless of the colour of the object used to make the shadow
2. It only shows the shape or outline of the object and not the details.
3. The size of a shadow varies depending on the distance between the object and the source of light, and the distance between the object and the screen.

Reflection Surfaces :

We say light is reflected when it bounces off a surface. Reflection of light helps us to see most of the things around us.
Reflection of light by a surface depends on the nature of the surface. A rough and bumpy surface (also called an irregular surface) reflects a parallel beam of light incident upon it in different directions (Fig.
13.5). A good example of a rough surface is bark of a tree and blanket. This kind of reflection is called diffused reflection.
A smooth surface (a highly polished surface) reflects a parallel beam of light incident upon it in one direction. (Fig. 13.6). A good example of a smooth surface is a mirror. When you stand in front of a mirror, you can see yourself in the mirror. This is called your image.

A very interesting phenomenon occurs when an object forms an image by reflection. This is something all of us must have noticed while seeing ourselves in the mirror. When we lift our right hand, the image in the mirror appears to lift its left hand. This seeming left-right reversal is called lateral inversion.
An image is different from a shadow. Some of the differences between an image and a shadow are given in Table 13.1.

A Pin-hole Camera :
A pin-hole camera is just a box (Fig. 13.7) with a very tiny hole on one of its sides. Light falls on the hole, and an inverted image is formed on the side opposite to the hole. The human eye acts very much like a pin-hole camera.

Key Words :

Source of light: An object that gives out light (luminous object) is called a source of light.
Transparent material: A material that transmits all the light is called a transparent material.
Translucent material: A material that transmits some amount of light is called a translucent material.
Opaque material: A material that completely blocks the light is called an opaque material.
Shadow: An area of darkness formed by an opaque object obstructing light is called a shadow.

Summary :

• Objects can be transparent, translucent, or opaque, depending on how much light can pass through them.
• A shadow is formed when an opaque object blocks the light falling on it.
• A shadow is always black regardless of the colour of the object.
• We say light is reflected when it bounces off a surface.
• A rough and bumpy surface reflects light in different directions.
• A smooth surface reflects light in only one direction.
• An image shows the colour, outline, and details of the object.

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NCERT Exemplar Problems Class 11 Mathematics Chapter 1 Sets

Short Answer Type Questions
Q1. Write the following sets in the roaster from

Q2. Write the following sets in the roaster form:

Q3. If Y = {x\x is a positive factor of the number 2^P(2^P – 1), where 2^P – 1 is a prime number}. Write Y in the roaster form.

Sol: Y- {x | x is a positive factor of the number 2^P-1 (2^P – 1), where 2^P – 1 is a prime number}.
So, the factors of 2^P-1 are 1,2,2²,2³,…, 2^{P- 1}.
Y= {1,2,2²,2³, …,2^p-1,2^p-1}

Q4. State which of the following statements are true and which are false. Justify your answer.
(i) 35 ∈ {x | x has exactly four positive factors}.
(ii) 128 e {y | the sum of all the positive factorsofy is 2y}
(iii) 3∉{x|x⁴-5x³ + 2jc²-112x + 6 = 0}
(iv) 496 ∉{y | the sum of all the positive factors of y is 2y}.
Sol: (i) The factors of 35 are 1, 5, 7 and 35. So, 35 is an element of the set. Hence, statement is true.

(ii) The factors of 128 hre 1,2,4, 8, 16, 32, 64 and 128.
Sum of factors = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255 * 2 x 128 Hence, statement is false.

(iii) We have, x⁴ – 5x³ + 2x² – 1 12jc + 6 = 0 Forx = 3, we have
(3)⁴ – 5(3)³ + 2(3)² – 112(3) + 6 = 0
=> 81 – 135 + 18-336 + 6 = 0
=>    -346 = 0, which is not true.
So 3 is not an element of the set
Hence, statement is true.

(iv) 496 = 2⁴ x 31
So, the factors of 496 are 1,2,4, 8, 16,31,62, 124,248 and 496.
Sum of factors = 1 +2 + 4 + 8+ 16 + 31+62+124 + 248 + 496 = 992 = 2(496)
So, 496 is the element of the set Hence, statement is false

Q5. Given L, = {1,2, 3,4},M= {3,4, 5, 6} and N= {1,3,5}
Verify that L-(M⋃N) = (L-M)⋂(L-N)
Sol: Given L,= {1,2, 3,4}, M= {3,4,5,6} and N= {1,3,5}
M⋃N= {1,3,4, 5,6}
L – (M⋃N) = {2}
Now, L-M= {1, 2} and L-N= {2,4}
{L-M) ⋂{L-N)= {2}
Hence, L-{M⋃N) = {L-M) ⋂ (L-N).

Q6. If A and B are subsets of the universal set U, then show that
(i) A⊂ A∪ B
(ii) A⊂B⟺A∪B = B
(iii) (A∩B)⊂ A



Q7. Given that N= {1,2,3, …, 100}. Then write
(i) the subset of N whose elements are even numbers.
(ii) the subset of N whose element are perfect square numbers.
Sol: We have, N= {1,2, 3,4,…, 100}
(i) subset of N whose elements are even numbers = {2,4, 6, 8,…, 100}
(ii) subset of N whose elements are perfect square = {1,4, 9, 16, 25, 36,49, 64, 81, 100}

Q8. If X= {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by
(i) 4n                     
(ii) n + 6                 
(iii) n/2
(iv) n-1 

Q9.If Y= {1,2,3,…, 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.

Q10. A, B and C are subsets of Universal Set If A = {2, 4, 6, 8, 12, 20}, B= {3,6,9,12,15}, C= {5,10,15,20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Q11. Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationship among sets U, G, B and S.

Q12. For all sets A, B and C, show that (A – B) ∩ (C – B) = A – (B ∪ C)

Instruction for Exercises 13-17: Determine whether each of the statements in these exercises is true or false. Justify your answer.

Q13. For all sets A and B, (A – B)∪ (A∩ B) = A
Sol: True
L.H.S. = (A-B) ∪ (A∩B) = [(A-B) ∪A] ∩ [(A – B) ∪B]
= A∩ (A-B) = A= R.H.S.
Hence, given statement is true.

Q14. For all sets A, B and C, A – (B-C) = (A- B)-C
Sol: False 

Q15. For all sets A, B and C, if A ⊂ B, then  A ∩C<⊂B ∩C

Sol: True
Let          x ∈A∩C
=> x ∈ A and x∈ C

Q16. For all sets A, B and C, if A⊂ B, then A ∪ C⊂ B ∪ C
Sol: True

Q17. For all sets A, B and C, if A⊂ C and B ⊂ C,then A∪ B ⊂ C

Instruction for Exercises 18-22: Using properties of sets prove the statements given in these exercises.

Q18. For all sets A and B, A ∪ (B -A) = A ∪ B

Q19. For all sets A and B, A – (A – B) = A ∩ B

Q20. For all sets A and B, A – (A ∩ B) = A – B

Q21. For all sets A and B,(A ∪ B)- B = A-B

Long Answer Type Questions

Q23. Let A, B and C be sets. Then show that A ∩ (B ∪ C) = (A ∩ B)∪ (A ∩ C).

From (i) and (ii), we get .                              .
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Q24. Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science,6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed
(i) in English and Mathematics but not in Science
(ii) in Mathematics and Science but not in English
(iii) in Mathematics only
(iv) in more than one subject only
Sol. Let M be the set of students who passed in Mathematics, E be the set of students who passed in English and S be the set of students who passed in Science.

Given n (U) = 100,
n(E) = 15, n(M) = 12, n(S) = 8,
n(E ∩ M) = 6, n(M ∩S) = 7, n(E ∩ S) — 4, and n(E ∩M ∩ S) = 4,

Number of students passed in English and Mathematics but not in Science = b = 2
(ii) Number of students passed in Mathematics and Science but not in English = d = 3
(iii) Number of students passed in Mathematics only = e = 3
(iv) Number of students passed in more than one subject = a + b + c + d =4+2+0+3=9

Q25. In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither.
Sol: Let C be the set of students who play cricket and T be the set of students who play tennis.
n(U) = 60, n(C) = 25, n(T) = 20, and n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C n T) = 25 + 20 – 10 = 35

Q26. In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.
Sol: Let M be the set of students who study Mathematics, P be the set of students who study E Physics and C be the set of students who study Chemistry

Q27. In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find
(a) The number of families which buy newspaper A only.
(b) The number of families which buy none of A, B and C.
Sol: Let A be the set of families which buy newspaper A, B be the set of families which buy newspaper B and C be the set of families which buy newspaper C. The

Number of families which buy none of A, B and C = 10000 x (40/100)

Q28. In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows: French = 17, English = 13, Sanskrit = 15, French and English = 09, English and Sanskrit = 4,French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study
(i) French only 
(ii) English only 
(iii) Sanskrit only 
(iv) English and Sanskrit 
(v) French and  Sanskrit but not English
(vi) French and  English but not Sanskrit
(vii)     at least one  of the three languages
(viii) none of the  three languages but not French

Sol: Let F be the set of students who study French, E be the set of students who study English and S be the set of students who study Sanskrit.
Then, n{U) = 50, n(F) =17, n{E) = 13, and n{S) = 15,
n(F ∩ E) = 9, n(E ∩ S) = 4, n(F ∩ S) = 5, n(F ∩ E ∩ S) = 3

(i) Number of students studying French only = e = 6
(ii) Number of students studying English only = g = 3
(iii) Number of students studying Sanskrit only =f= 9
(iv) Number of students studying English and Sanskrit but not French = c = 1
(v) Number of students studying French and Sanskrit but not English = d = 2
(vi) Number of students studying French and English but not Sanskrit = b = 6
(vii) Number of students studying at least one of the three languages = a + b + c + d + e+f+g = 30
(viii) Number of students studying none of the three languages but not French = 50-30 = 20

Q30. Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n are, respectively, (a) 4,7 (b) 7,4 (c) 4,4 (d) 7, 7

Q32. Let F₁ be the set of parallelograms, F₂ the set of rectangles, F₃ the set of rhombuses, F₄ the set of squares and F₅ the set of trapeziums in a plane. Then F₁may be equal to
(a) F₂ ∩F₃                                              
(b) F₃ ∩F₄
(c) F₂ u F_s
(d) F₂ ∪ F₃ ∪ F₄ ∪ F₁
Sol: (d) Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram.
F₁ = F₂ ∪ F₃ ∪ F₄ ∪ F₁

Q33. Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square, then

Q34. Let R be set of points inside a rectangle of sides a and b (a, b> 1) with two sides along the positive direction of x-axis andy-axis. Then

Q35. In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is
(a) 0 (b) 25 (c) 35 (d) 45
Sol: Let C be the set of students who play cricket and T be the set of students who play tennis.
n(U) = 60, n(C) = 25, n(T) = 20, and n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C n T) = 25 + 20 – 10 = 35

Q36. In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
(a) 210 (b) 290 (c) 180 (d) 260
Sol: (b) Let H be the set of persons who read Hindi and E be the set of persons who read English.

NCERT Exemplar Problems Class 11 Mathematics

The post NCERT Exemplar Problems Class 11 Mathematics Chapter 1 Sets appeared first on Learn CBSE.

NCERT Exemplar Problems Class 11 Mathematics Chapter 2 Relations and Functions 

Short Answer Type Questions
Q1. If A = {-1, 2, 3 } and B = {1, 3}, then determine
(i) AxB (ii) BxC (c) BxB (iv) AxA
Sol: We have A = {-1,2,3} and B = {1,3}
(i) A x B = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3,1), (3, 3)}
(ii) BxA = {( 1, -1), (1, 2), (1,3), (3,-1), (3,2), (3, 3)}
(iii) BxB= {(1,1), (1,3), (3,1), (3, 3)}
(iv) A xA = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3,3)}

Q2. If P = {x : x < 3, x e N}, Q= {x : x≤2,x ∈ W}. Find (P∪ Q) x (P∩ Q), where W is the set of whole numbers.
Sol: We have, P={x: x<3,x ∈ N} = {1,2}
And Q = {x :x≤ 2,x∈ W] = {0,1,2}
P∪Q= {0, 1,2} and P ∩ Q= {1,2}
(P ∪ Q) x (P ∩ Q) = {0,1, 2} x {1,2}
= {(0,1), (0, 2), (1,1), (1,2), (2,1), (2, 2)}

Q3. lfA={x:x∈ W,x < 2}, 5 = {x : x∈N, 1 <.x < 5}, C= {3, 5}. Find
(i) Ax(B∩Q) (ii) Ax(B∪C)
Sol: We have, A = {x :x∈ W,x< 2} = {0, 1};
B = {x : x ∈ N, 1 <x< 5} = {2, 3,4}; and C= {3, 5}

(i) B∩ C = {3}
A x (B ∩ C) = {0, 1} x {3} = {(0, 3), (1, 3)}

(ii) (B ∪ C) ={2,3,4, 5}
A x (B ∪ C) = {0, 1} x {2, 3,4, 5}
= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1, 5)}

Q4. In each of the following cases, find a and b. (2a + b, a – b) = (8, 3) (ii) {a/4, a – 2b) = (0, 6 + b)
Sol: (i) We have, (2a + b,a-b) = (8,3)
=> 2a + b = 8 and a – b = 3
On solving, we get a = 11/3 and b = 2/3

Q5. Given A = {1,2,3,4, 5}, S= {(x,y) :x∈ A,y∈ A}.Find the ordered pairs which satisfy the conditions given below
x+y = 5 (ii) x+y<5 (iii) x+y>8
Sol: We have, A = {1,2, 3,4, 5}, S= {(x,y) : x ∈ A,y∈ A}
(i) The set of ordered pairs satisfying x + y= 5 is {(1,4), (2,3), (3,2), (4,1)}
(ii) The set of ordered pairs satisfying x+y < 5 is {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}
(iii) The set of ordered pairs satisfying x +y > 8 is {(4, 5), (5,4), (5, 5)}.

Q6. Given R = {(x,y) : x,y ∈ W, x² + y² = 25}. Find the domain and range of R
Sol: We have, R = {(x,y):x,y∈ W, x² + y² = 25}
= {(0,5), (3,4), (4, 3), (5,0)}
Domain of R = Set of first element of ordered pairs in R = {0,3,4, 5}
Range of R = Set of second element of ordered pairs in R = {5,4, 3, 0}

Q7. If R₁ = {(x, y)| y = 2x + 7, where x∈ R and -5 ≤ x ≤ 5} is a relation. Then find the domain and range of R₁.
Sol: We have, R₁ = {(x, y)|y = 2x + 7, where x∈ R and -5 ≤x ≤ 5}
Domain of R₁ = {-5 ≤ x ≤ 5, x ∈ R} = [-5, 5]
x ∈ [-5, 5]
=> 2x ∈ [-10,10]
=>2x + 7∈ [-3, 17]
Range is [-3, 17]

Q8. If R₂ = {(x, y) | x and y are integers and x² +y² = 64} is a relation. Then find R₂
Sol: We have, R₂ = {(x, y) | x and y are integers and x² + y² – 64}
Clearly, x² = 0 and y² = 64 or x² = 64 andy² = 0
x = 0 and y = ±8
or x = ±8 and y = 0
R₂ = {(0, 8), (0, -8), (8,0), (-8,0)}

Q9. If R₃ = {(x, |x|) | x is a real number} is a relation. Then find domain and range
Sol: We have, R₃ = {(x, |x)) | x is real number}
Clearly, domain of R₃ = R
Now, x ∈ R and |x| ≥ 0 .
Range of R₃ is [0,∞)

Q10. Is the given relation a function? Give reasons for your answer.
(i) h={(4,6), (3,9), (-11,6), (3,11)}
(ii) f = {(x, x) | x is a real number}
(iii) g = {(n, 1 In)| nis a positive integer}
(iv) s= {(n, n²) | n is a positive integer}
(v) t= {(x, 3) | x is a real number}
Sol: (i) We have, h = {(4,6),(3,9), (-11,6), (3,11)}.
Since pre-image 3 has two images 9 and 11, it is not a function.
(ii) We have, f = {(x, x) | x is a real number}
Since every element in the domain has unique image, it is a function.
(iii) We have, g= {(n, 1/n) | nis a positive integer}
For n, it is a positive integer and 1/n is unique and distinct. Therefore,every element in the domain has unique image. So, it is a function.
(iii) We have, s = {(n, n²) | n is a positive integer}
Since the square of any positive integer is unique, every element in the domain has unique image. Hence, ibis a function.
(iv) We have, t = {(x, 3)| x is a real number}.
Since every element in the domain has the image 3, it is a constant function.

Q11. If f and g are real functions defined byf( x) = x² + 7 and g(x) = 3x + 5, find each of the following

Q12. Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7.
(i) For what real numbers x,f(x)= g(x)?
(ii) For what real numbers x,f (x) < g(x)?
Sol: We have,f(x) = 2x + 1 and g(x) = 4x-7
(i) Now f (x) = g(x)
=> 2x+l=4x-7
=> 2x = 8 =>x = 4
(ii) f (x) < g(x)
=> 2x + 1 < 4x – 7
=> 8 < 2x
=> x > 4

Q13. If f and g are two real valued ftmctions defined as f(x) = 2x + 1, g(x) = x² + 1, then find.

Q14. Express the following functions as set of ordered pairs and determine their range.
f:X->R,f{x) =  x³ + 1, where X= {-1,0, 3, 9, 7}
Sol: We have, f:X→ R,flx) = x³ + 1.
Where X = {-1, 0, 3, 9, 7}
Now f (-l) = (-l)³+1 =-l + 1 =0
f(0) = (0)³+l=0+l = l
f(3) = (3)³ + 1 = 27 + 1 = 28
f(9) = (9)³ + 1 = 729 + 1 = 730
f(7) = (7)³ + 1 = 343 + 1 = 344
f= {(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}
Range of f= {0, 1, 28, 730, 344}

Q15. Find the values of x for which the functions f(x) = 3x² -1 and g(x) = 3+ x are equal.
Sol: f(x) = g(x)
=> 3x²-l=3+x => 3x²-x-4 = 0 => (3x – 4)(x+ 1) – 0
x= -1,4/3

Q16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g(x) = x +, then what values should be assigned to and ?
Sol:We have, g = {(1, 1), (2, 3), (3, 5), (4,7)}
Since, every element has unique image under g. So, g is a function.
Now, g(x) = x + For (1,1), g(l) = a(l) + P
=>             l = +            (i)
For (2, 3), g(2) = (2) +
=>             3 = 2 +         (ii)
On solving Eqs. (i) and (ii), we get = 2, = -l
f(x) = 2x-1
Also, (3, 5) and (4, 7) satisfy the above function.

Q17. Find the domain of each of the following functions given by

Q18. Find the range of the following functions given by

Q19. Redefine the function f(x) = |x-2| + |2+x| , -3 ≤x ≤3

Q21. Let f (x) = √x and g(x) = xbe two functions defined in the domain R⁺ ∪ {0}. Find
(i) (f+g)(x)        
(ii) (f-g)(x)
(iii) (fg)(x)
(iv) f/g(x)

Q23. If f( x)= y = ax-b/ cx-a then prove that f (y) = x

Objective Type Questions
Q24. Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) mⁿ                     
(b) n^m– 1                  
(c) mn – 1               
(d) 2^mn– 1

Sol: (d) We have, n(A) = m and n(B) = n
n(A xB) = n(A). n(B) = mn
Total number of relation from A to B = Number of subsets of AxB = 2^mn
So, total number of non-empty relations = 2^mn – 1

Q25. If [x]² – 5[x] + 6 = 0, where [. ] denote the greatest integer function, then
(a) x ∈ [3,4]
(b) x∈ (2, 3]            
(c) x∈ [2, 3]           
(d) x ∈ [2, 4)
Sol: (d) We have [x]² – 5[x] + 6 = 0 => [(x – 3)([x] – 2) = 0
=> [x] = 2,3 .
For [x] = 2, x ∈ [2, 3)
For [x] = 3, x ∈ [3,4)
x ∈ [2, 3) u [3,4)
Or x ∈ [2,4)

Q29. If fx) ax+ b, where a and b are integers,f(-1) = -5 and f(3) – 3, then a and b are equal to
(a) a = -3, b =-1
(b) a = 2, b =-3
(c) a = 0, b = 2
(d) a = 2, b = 3

Fill in the Blanks Type Questions

Q36. Let f and g be two real functions given by  f= {(0, 1), (2,0), (3,.-4), (4,2), (5, 1)}
g= {(1,0), (2,2), (3,-1), (4,4), (5, 3)}  then the domain of f  x g is given by________ .
Sol: We have, f = {(0, 1), (2, 0), (3, -4), (4, 2), (5,1)} and g= {(1, 0), (2, 2), (3, 1), (4,4), (5, 3)}
Domain of  f = {0,2, 3, 4, 5}
And Domain of g= {1, 2, 3,4, 5}
Domain of (f x g) = (Domain of f) ∩ (Domain of g) = {2, 3,4, 5}

Matching Column Type Questions

Q37. Let f= {(2,4), (5,6), (8, -1), (10, -3)} andg = {(2, 5), (7,1), (8,4), (10,13), (11, 5)} be two real functions. Then match the following:

True/False Type Questions

Q38. The ordered pair (5,2) belongs to the relation R ={(x,y): y = x – 5, x,y∈Z}
Sol: False
We have, R = {(x, y): y = x – 5, x, y ∈ Z}
When x = 5, then y  = 5-5=0 Hence, (5, 2) does not belong to R.

Q39. If P = {1, 2}, then P x P x P = {(1, 1,1), (2,2, 2), (1, 2,2), (2,1, 1)}
Sol:False
We have, P = {1, 2} and n(P) = 2
n(P xPxP) = n(P) x n(P) x n(P) = 2 x 2 x 2
= 8 But given P x P x P has 4 elements.

Q40. If A= {1,2, 3}, 5= {3,4} and C= {4, 5, 6}, then (A x B) ∪ (A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3, 5), (3,6)}.
Sol: True
We have.4 = {1,2, 3}, 5= {3,4} andC= {4,5,6}
AxB= {(1, 3), (1,4), (2, 3), (2,4), (3, 3), (3,4)}
And A x C =  {(1,4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6)}
(A x B)∪(A xC)= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3,3), (3,4), (3, 5), (3,6)}

Q42. If Ax B= {(a, x), (a, y), (b, x), (b, y)}, thenM = {a, b},B= {x, y}.
Sol: True
We have, AxB= {{a, x), {a, y), (b, x), {b, y)}
A = Set of first element of ordered pairs in A x B = {a, b}
B = Set of second element of ordered pairs in A x B = {x, y}

NCERT Exemplar Problems Class 11 Mathematics

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Electricity and Circuits – CBSE Notes for Class 6 Science

CBSE NotesCBSE Notes Class 6 ScienceNCERT Solutions Science

Life without electricity is difficult to imagine. Many devices and machines in our day-to-day life run on electricity.
Look at the pictures given below. Which of these run on electricity? Write their names in the space provided.

Things that run on electricity have electric current passing through them. In this, chapter, you will learn about electric current, what is needed to produce it, the conditions required to make an electric current flow, and the materials through which current can flow. You will learn some very interesting things like how to make a small bulb glow and how to make an electric switch. Answers: Refrigerator, Fan.

Electric Current :

Most of the devices and machines we use like an electric iron, oven, room heater, refrigerator, ceiling fan or an electric bulb work when an electric current flows through them.
With help from an adult, look at what is inside a transparent electric bulb (Fig. 14.1). Among other things, you will see that it has a thin filament (a very thin metal wire). The filament heats up when an electric current is passed through it. It heats up so much that it begins to glow and give out light.
Now, we will learn what produces an electric current.

Source Of Electric Current :

A device that can be used to produce an electric current is called a source of electric current. Common sources of electric current are cells and batteries (collection of cells) which comes in various shapes and sizes (Fig. 14.2), and electric current that we get from plug points in houses. A very useful kind of cell which we use very often is the dry cell. Due to a chemical reaction that takes place in cells and batteries, electric current is produced.
For large-scale production of electricity, flowing water or steam is used.

The Dry Cell:
A dry cell is a very convenient source of electric current. The dry cell, as its name suggests, contains dry or semi-solid ingredients.
Let us take a look inside a dry cell [Fig. 14.3(a and b)].

The dry cell contains a paste of ammonium chloride inside a zinc container. Inside the paste, a cardboard container containing powdered manganese dioxide and carbon is placed. The cardboard container has microscopic ‘holes’ in it (such materials are called porous materials) through which a chemical reaction takes place between ammonium chloride paste and powdered manganese dioxide. A rod, usually carbon, with a metal cap is dipped into the manganese dioxide. The whole thing is then sealed (with only the metal cap sticking out), so that the contents do not spill out. The zinc can is also wrapped so that only the base is exposed. Every source of electric current has two ends or terminals where conducting wires are connected to draw electric current. The tip of the metal cap and the base of the zinc can are called the positive and negative terminals of the dry cell, respectively. Electric current can be thought of as ‘flowing in’ from one terminal and ‘flowing out’ from the other. If the tip of the metal cap and the base of the zinc can are connected by a metal wire, current will flow through it.
Different Types of Electric Cells:
Apart from the simple primary cells like dry cell, there are different types of electric cells. Different cells use different methods for producing an electric current. Primary cells can be used only once, and have to be thrown away once they have been used up.

There are cells that can be recharged once they are drained. These are called secondary cells. They are used in mobile phones, laptops, and car batteries.
Nowadays, solar cells are being used in many applications. Solar cells use sunlight to produce electric current. Fig. 14.4 Some devices that work on dry cell.
Many calculators are powered with solar cells. Solar panels made up of solar cells are used to light up streets and many homes.

Flow Of Electric Current :

Three basic conditions (Fig. 14.5) are required for an electric current to flow.
• A device used to produce an electric current like cell, battery, or a plug point acting as a source.
• A wire made of a metal like copper, silver, or aluminium, which will allow electric current to flow through easily.
• An unbroken loop (of the wire) running from one terminal of the source, through various appliances, back to the other terminal of the source.

Making a Simple Electric Circuit:
When we connect the terminals of a pencil cell (name given to the cell due to its shape) to a bulb using two wires, the bulb glows. This happens because we provide a path for the current to flow. A path for an electric current to flow is called an electric circuit.
In Figure 14.6(a), one wire from the pencil cell is connected to the torch bulb, while the other wire is not. The electric circuit is not complete here. In Figure 14.6(b), both the wires from the cell are connected to the torch bulb. The electric circuit is complete in this case. Electric current flows only if there is an unbroken path or closed circuit starting from one terminal of the source, through the torch bulb, to the other terminal of the source. Thus, the bulb glows in Figure 14.6(b) but not in Figure 14.6(a). The circuit in Figure 14.6(a) is not complete. Hence, current cannot flow through the circuit and the bulb does not glow. Such a circuit is called an open circuit. The circuit in Figure 14.6(b) is complete. Electric current flows through the circuit and, as a result, the bulb glows. Such a circuit is called a closed circuit.


Electric current flows in a particular direction. In an electric circuit, the electric current flows from the positive terminal to the negative terminal of the electric cell. Figure 14.7 shows the direction of flow of electric current in a circuit.

Electric Switch:

We use electric switches (Fig. 14.8) to put on or off the electrical devices and machines. But do you know how it works?

An electric switch is a device that is used to open or close an electric circuit. When we open an electric circuit, the flow of electric current in the circuit stops [Fig. 14.9(a)], and when we close an electric circuit, an electric current flows through it [Fig. 14.9(b)]. In an electrical circuit, a switch is sometimes.

Electric Torch :

A schematic diagram of an electric torch is shown in Fig. 14.10. An electric torch has one or more dry cells inside it, which act as the ‘source’. These cells are connected through a switch to a small bulb. When the switch is pushed to the ‘on’ position, the circuit is complete and the bulb glows. When the switch is pushed to the ‘off’ position, the circuit is incomplete (broken). Now the current cannot flow through the circuit, and the light goes out.

Conductors And Insulators :

Look at Figures 14.11(a)-(d). Each shows a complete electric circuit. Then why is it that the bulb glows only in circuits (b) and (d)? It is because not all materials allow electric current to pass through them.
A material that allows electric current to pass through it is called a conductor of electricity like the key and the safety pin [Figs. 14.11(b) and (d)]. A material that does not allow electric current to pass through it is called an insulator of electricity like the rubber band and the plastic pen [Figs. 14.11(a) and (c)].

All metals are conductors of electricity while some are better conductors than others. A few non-metals like graphite (pencil lead is made of graphite) are also conductors of electricity.
Examples of insulators are glass, wood, rubber, pure water, and dry air. Flowever, the smallest impurity in water (impurities are substances like salts, dissolved in water) makes it a conductor.
The handles of screwdrivers and testers used by electricians are usually made of wood or hard plastic. They also wear rubber gloves while repairing an electric switch to avoid electric shock.

Electrical Safety :

Electricity can be very dangerous, if you do not handle electrical devices carefully. One should never play with electrical wires and sockets. Electricity from cells is safe and you can experiment with it, but you have to be careful not to connect the two terminals of a cell directly through a wire/conductor. Electricity generated by portable generators is dangerous and should not be used for experiments.

Key Words :

Source of electric current: A device that can be used to produce an electric current is called a source of electric current.
Electric circuit: A path for an electric current to flow is called an electric circuit.
Closed circuit: A circuit which has an ‘unbroken path’ through which an electric current can flow is called a closed circuit.
Open circuit: A circuit with a break in it is called an open circuit.
Electric switch: A device that is used to open or close a circuit is called an electric switch.
Conductor (in this chapter): A material that allows electric current to pass through it easily is called conductor.
Insulator (in this chapter): A material that does not allow electric current to pass through it easily is called an insulator.

Summary :

• In a dry cell, a chemical reaction takes place to produce an electric current.
• A dry cell contains solid or semisolid ingredients.
• All cells have two terminals: the positive and the negative terminal.
• Electric current flows only if there is an unbroken or complete path, starting from one terminal of the source, through various devices back to the other terminal of the source.
• An electric switch is a device that is used to open or close a circuit.

The post Electricity and Circuits – CBSE Notes for Class 6 Science appeared first on Learn CBSE.

Water – CBSE Notes for Class 6 Science

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Water Available For Use:

About three-fourth of the Earth’s surface is covered with water. That is why it is also called the water planet. But do you know how much water is readily available for use? Most of the water (about 97%) is in the seas and oceans as salt water. This water is too salty to be used for drinking and irrigation. Thus, only a tiny fraction (about 3%) of the Earth’s water is available to us as freshwater. Out of this, 2.997% is locked up in the mountains or glaciers or is buried so deep under that it costs too much to extract.
So, only about 0.003% of the fresh water is easily available to us in the form of groundwater, river, lake, stream, soil moisture, and water vapour. See Figure 15.1.

Uses Of Water :

Besides being essential for life, water is used for many other purposes. In India, about 70% of the total water available is used for agriculture, 20-22% by industries, and only 8% is used for personal or domestic needs. Figure 15.2 shows a pie chart that gives the percentage use of water.
Let us learn more about the various uses of water.

Agricultural needs Our country depends a lot on agriculture. Farmers rely on water to sustain their agricultural crops, e.g., wheat, paddy, etc. Many a times, rainfall is not sufficient to water these crops, and farmers have to use artificial watering systems, referred to as irrigation.

Industrial needs Factories use a large amount of water every day—as raw material, for cleaning, heating, cooling, generating electricity (e.g., water turbines), etc. The amount of required depends on the kind and size of the factory, and water.
Personal/domestic needs We need water to drink. Water that is suitable for drinking is called potable water. We also need water to bathe, wash clothes and dishes, clean our house, and to water plants.
Apart from these uses, water is also used for transportation and recreation (Fig. 15.6). It also regulates the climate of a place and provides homes to many animals.

Different States Of Water :

In nature, water exists in three states. It could be in the form of liquid (e.g., rain, river, sea), solid (e.g., ice, snow, hail), or gas (e.g., water vapour) (Fig. 15.7).
You can heat water over a stove to convert it into vapour. What happens if you leave water in an uncovered vessel on a summer afternoon outside your house? After a few hours, you will find that the level of water in the vessel has decreased. This is because a lot of it would have escaped into the atmosphere in the form of water vapour. The process by which a liquid is converted to its vapour is called evaporation.
What about the reverse process? The process by which the vapour of a substance is converted to its liquidform is called condensation. Water vapour is also added into the air by the leaves of plants, through the process of transpiration.
Evaporation and condensation of water take place on a very large scale on the surface of the Earth and its atmosphere. These processes play a key role in cloud formation and rain.

Cloud Formation :

When the temperature of air increases, it expands (i.e., its particles move away from one another). This makes the air lighter and it rises in the atmosphere, taking water vapour with it. As the air rises, it begins to cool. The water vapour condenses on dust particles present in the atmosphere to form millions of tiny droplets. Tiny ice crystals will be formed instead if it is very cold. This cluster of tiny water droplets floating in air is what we call a cloud.

Water Cycle :

Water droplets in the clouds keep bumping against one another, and sometimes stick to form bigger drops. When these drops become too heavy to float in the air, they drop down back to the Earth as rain. The water that comes down as rain, in time, evaporates and goes up to form clouds again. This leads to form a cycle, known as the water cycle.
Water cycle is the cyclic movement of water from the atmosphere to the Earth and back to the atmosphere through various processes.
Figure 15.9 shows how the water cycle works.

Evaporation Sun’s heat changes the water in oceans, rivers, ponds, and other water bodies into water vapour. Transpiration Plants give off water vapour through their leaves.
Condensation Water vapour rises up and condenses on dust particles to form clouds.
Precipitation Water stored in clouds reaches Earth in the form of rain, snow, etc.
Collection Some rainwater seeps into the ground, forming ground water. Rainwater also flows into streams and rivers, and then finds its way into seas and oceans.

Drought :

Sometimes it does not rain for a long time—for an entire month, two months, the whole season, two years, etc. The abnormally long period of insufficient or no rainfall at all is called drought. During drought, rivers run dry, water level in lakes goes down, and even the water in the soil dries up (Fig. 15.10).

There is very little or no water available during a drought. Most plants die leading to lack of food in the region. The lack of food in a region for a long period is called famine. When there are no plants, animals that survive on plants also die. Animals and humans also die due to thirst and dehydration (excessive loss of water from the body).

Flood :

Just as too little rain is bad, too much rain is bad as well. It leads to water being everywhere, sometimes the entire area remains under water or submerged. A condition when ground becomes submerged under water, due to heavy rain and overflowing of rivers is called flood.
During a flood, plants and crops die either due to suffocation by excess water or due to soil being washed away, robbing their roots of support.

In such waterlogged conditions, many disease-causing germs start multiplying and
cause water-borne diseases. Sometimes water-borne diseases affect thousands of people at the same time. A disease affecting thousands of people at the same time is called an epidemic.
A flood can thus lead to many conditions—disease, starvation, loss of life, and property. A flood also can have the same effect on the food chain as a drought does.
Both drought and flood are natural disasters—unfortunate events brought about by nature—that can have a very strong effect on the lives of people. The economy of a country can crumble as a result of loss of property and life and many more things. But we can and must try to reduce the bad effects of these disasters as much as possible. We must understand that such things can happen, and so we must be prepared to face them as and when they happen.

Conservation Of Water :

Since only a small percentage of water on our planet is usable, it is very important that we use water carefully. Conservation of water can be done by building dams, avoiding wastage of water at homes, both indoors and outdoors, harvesting rainwater, and preventing pollution of water.
Building Dam :
A dam is a structure built to hold back water in order to prevent floods, and to provide water for irrigation and storage.
Building a dam is a solution to both drought and flood (Fig. 15.12).

Dams are also used in producing electricity. Built on rivers, a dam has high walls and has many openings or gates to both let in and hold back water.
During heavy rainfall, when the rivers fill up, water enters the dam. When water is needed later, the gates are opened to let out water.
Avoiding Wastage at Homes:
Some ways to avoid wastage of water at homes are given below.

Rainwater Harvesting:
The process of collecting and storing rainwater from roofs or a surface catchment is called rainwater harvesting (fig. 15.14). Storing rainwater that collects on roofs instead of letting it go down the drain, is a practical solution in case of droughts. This technique is known as rooftop rainwater harvesting. This involves collecting rainwater from rooftops in dugout ponds, vessels, or underground tanks to store water for long periods. Another option is to allow water to go into the ground directly from the roadside drains that collect rainwater. The stored rainwater is treated before use because it may contain bird faeces, chemicals, and other pollutants, which need to be removed before use.

Preventing Water Pollution:
Another method to conserve water is to safeguard our fresh water bodies from pollution. Garbage and harmful chemicals pollute the water and make it unfit for use. Polluted water (Fig. 15.15) is also very bad for aquatic life. Plants and animals in and around polluted water may die or get infected. And when human beings consume the contaminated fish, etc., they are also put at risk of diseases.

Key Words :

Salt water: Water of the seas and oceans that has a high salt content, making it unfit for drinking and irrigation, is called salt water.
Fresh water: Water found in rivers, lakes, and ponds used for domestic and commercial purposes is called fresh water.
Irrigation: Watering crops by artificial means is called irrigation.
Potable water: Water fit for human consumption is called potable water.
Transpiration: The release of water vapour into the atmosphere through the leaves of plants is called transpiration.
Water cycle: The cyclic movement of water from the atmosphere to the Earth and back to the atmosphere through various processes is called water cycle.
Drought: Abnormally long period of insufficient or no rainfall is called drought.
Famine: Lack of food in a region for a long period is called famine.
Flood: A condition when ground becomes submerged under water, due to heavy rain and overflowing of rivers is called flood.
Epidemic: A disease affecting thousands of people at the same time is called epidemic.
Dam: A structure built on a river to store and hold back water is called a dam.
Rainwater harvesting: The process of collecting and storing rainwater from roofs or a surface catchment is called rainwater harvesting.

Summary :

Only a tiny fraction of the Earth’s water is available as fresh water. All living things have a lot of water in their body. Almost 70% of our body weight contains to water.
• We need water for many purposes—drinking, personal needs, agricultural needs, industrial needs, for transportation and recreation, and regulating the climate. It is home to various plants and animals.
• Water exists in solid, iiquid, and gaseous states.
• Evaporation and condensation play an important role in cloud formation.
• Water cycle is the cyclic movement of water from the atmosphere to the Earth and back to the atmosphere through various processes.
• Abnormally long period of insufficient or no rainfall is known as drought.
• When there is too much rainfall in an area, rivers overflow and water covers all the area around. This is called a flood. A flood can cause great destruction.
• Rainwater harvesting is one of the ways to conserve water.

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Air Around Us – CBSE Notes for Class 6 Science

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Air Is All Around Us :

A thick blanket of air, called the atmosphere, surrounds our Earth. Air is also present in things which seem to be empty. Let us find this out by doing an activity.

Composition Of Air :
It contains mainly nitrogen, and oxygen. It also contains carbon dioxide, noble gases, water vapour, dust particles, and traces of other gases. The composition of air is shown in Figure 16.1. Let us now verify the presence of some constituents of air.

Nitrogen and Oxygen :
Air contains about 78% nitrogen and 21% oxygen. Oxygen in air supports burning whereas nitrogen does not. Let us prove this by doing a simple activity.
Carbon Dioxide:
Air contains about 0.03% of carbon dioxide. Plants and animals take in oxygen and give out carbon dioxide during respiration. When you burn something, carbon dioxide is also produced.
Water Vapour:
Air contains varying amounts of water vapour depending on the weather of a place. You have learnt about the water cycle. The sun heats up the water in seas and oceans. This water evaporates and forms water vapour. You can verify the presence of water vapour in air by observing wet clothes drying on a clothesline (Fig. 16.2). Where does the water from these wet clothes disappear? The water from the wet clothes forms water vapour and mixes with the air.

Dust and Smoke:
Have you seen sun rays entering a dark room? Have you noticed tiny particles in the rays? These are dust particles. Air contains dust. Air also contains smoke released from factories and vehicles (Fig. 16.3).

Air Supports Life :

We all need air to survive. Air contains oxygen and carbon dioxide useful to plants and animals. Plants use carbon dioxide of the air to make their own food by a process called photosynthesis. Let us see how air supports life in plants and animals.

In Plants:
Plants have tiny pores called stomata, found on the underside of a leaf (Fig. 16.4). Air containing carbon dioxide and oxygen enters the plant through these openings where it gets used in photosynthesis and respiration.
In Animals:
All animals need to respire, be it a cockroach, a fish, or an elephant. It is just that they use different organs and mechanisms for respiration.
Sometimes, we wrongly use the terms breathing and respiration interchangeably. Breathing is a physical act of taking in oxygen and giving out carbon dioxide, whereas respiration is a chemical process by which glucose in the body breaks down to give energy.

In Aquatic Animals and Plants:
Most aquatic animals like fish, tadpole, crab, and shrimp have special organs for respiration called gills. Gills help to take in oxygen and give out carbon dioxide. Some aquatic animals like dolphin and whale come to the surface of the water regularly to take in air, since they breathe with the help of lungs.

Aquatic plants like Hydrilla also breathe in oxygen dissolved in water through their stomata.
In Amphibians:
Amphibians like frog, newt, and salamander need breathing systems for both air and water. Crocodile and alligator swim through water with part of their snout above the water surface to breathe easily through nostrils.

In Birds:
Birds have an efficient respiratory system as they need high levels of oxygen during flight. Birds have a pair of lungs with air sacs that remain open all the time, so that air can easily pass through them.

In Mammals:
Most mammals breathe with the help of lungs.
They take in oxygen and give out carbon dioxide.

Balance Of Oxygen And Carbon Dioxide In The Air :

The balance of oxygen and carbon dioxide in the atmosphere is maintained through respiration in plants and animals and by photosynthesis in plants. Plants produce oxygen during photosynthesis and utilize oxygen during respiration. They produce much more oxygen during photosynthesis than they consume, during respiration This is how the oxygen consumed by plants and to a large extent by animals is replenished in the air through photosynthesis.

Air Pollution :

The addition of substances in the environment in quantities that are harmful to living beings is called pollution. Air is getting polluted day by day because of various human activities. Burning of fuels like coal and petroleum, excessive burning of fuels like wood, smoke and harmful gases released from industries (Fig. 16.5), smoke released by vehicles (Fig. 16.6), and machines releasing gases are the major causes of air pollution. These gases spread and mix in the air and spoil the quality of air, thereby making it impure.

Air pollution has major adverse effects on plants, animals as well as human beings. Harmful gases present in the polluted air make breathing difficult. Air pollution also leads to a lot of lung disease like asthma and lung cancer. Air pollution also damages crops.
If we do not start looking after the quality of air around us, the oxygen and carbon dioxide levels will no longer be balanced and living beings will be the ones getting affected.
There are a number of ways by which we can reduce air pollution. Some of them are: planting more and more trees, recycling plastics, regular checking of vehicles for emission of harmful gases, etc.

Key Words :

Atmosphere:  A thick blanket of air Pollution The addition of substances in the surrounding the Earth’s surface is called the environment in quantities that are harmful atmosphere. to living beings is called pollution.

Summary :

• All living organisms need air to survive. Air cannot be seen but can be felt when it moves.
• Air is a mixture of several gases.
• Oxygen is needed for respiration. Carbon dioxide is given out as a by-product after respiration.
• Insects take in air with the help of tiny holes in their bodies called spiracles; earthworm breathes through their skin, which is kept moist with the help of a substance called mucous.
• Some aquatic animals like whale and dolphin as well as mammals breathe with the help of lungs.
• Amphibians like frog breathe with the help of lungs, when on land. In water, these animals breathe with the help of their moist skin.
• Birds breathe through lungs and air sacs that are open all the time.
• There are several causes of air pollution: excessive burning of fuels like wood, coal, and petroleum, machines releasing gases, vehicles releasing smoke, and several types of harmful gases released by industries.

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Garbage In Garbage Out – CBSE Notes for Class 6 Science

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Segregation Of Wastes :

Garbage or waste may be in the form of fruit or vegetable peels, discarded objects, wrapping materials, wasted food as household garbage, or discarded chemicals and fertilizers washed into rivers, domestic sewage, etc. These wastes can be segregated into biodegradable and non-biodegradable. Wastes that rot (undergo degradation) by the action of decomposers (tiny organisms found in the soil) are called biodegradable wastes. Dead plants and animals and their products (e.g., fruit and vegetable peels, paper, and leaves) decay very easily These wastes mix with the soil and produce manure. Wastes that do not rot by the action of decomposers are called non-biodegradable wastes. For example, glass, plastic, and metals. Many of them can be recycled to produce new things.
Depending on the type of wastes, two garbage bins—one for biodegradable wastes and other for non-biodegradable wastes should be used. This will help in easy sorting and recycling of wastes to make beneficial products.

Best out of waste :
The Rock Garden in Chandigarh is an excellent example of how solid wastes can be utilized. Every item in this garden is made from waste materials like tyres, plastic bottles, eggshells, and tube lights. This innovative idea of utilizing solid waste has made the Rock Garden very popular tourist attraction.

Management Of Biodegradable Wastes :

Some of the ways to manage biodegradable wastes are as follows:
Composting:
Since biodegradable or organic wastes like vegetable peels, waste food, leaves, dead flowers, and egg shells can be recycled, they are converted into manure by burying them in compost pits. Recycling of organic wastes like vegetable peels, waste food, leaves, etc., by burying them in compost pits is called composting. Composting is a simple and almost effortless process of recycling. The biodegradable wastes are degraded by the action of small organisms like bacteria and fungi. There is also a different kind of composting where a kind of earthworm called red worms (or red wrigglers) act on wastes and degrade them.
This type of composting with the help of a type of earthworm called red worms, is called vermicomposting (Fig. 17.1). Red worms break down the organic matter into nutrient-rich manure which increases soil fertility.
Vermicompost can be made in 3-4 weeks and it appears as loose soil-like material. One should not put animal product or oily substance in the pit as it could lead to the growth of disease-causing organisms.

Landfills:
Large areas used for waste disposal are called landfills. Landfill is another method to manage huge amount, of biodegradable waste. In a landfill, garbage is buried in such a way that it does not damage the environment (Fig. 17.2). Garbage buried inside landfills stay here for a long time as it decomposes very slowly. After a landfill is full, it can be converted into a park. For example, Indraprastha Park in New Delhi is built on a landfill site.

Management Of Non-Biodegradable Wastes :

As non-biodegradable wastes like plastic bags, glass bottles, etc., cannot be broken down by decomposers, their disposal poses a big problem.
Non-biodegradable wastes can be managed by practicing the concept of 3 Rs—Reduce, Reuse, and Recycle.
Reduce:
We need to reduce the amount of waste generated by consuming more and throwing away less. We often buy more things than we really need. Nowadays, disposable items have become popular, for example, ballpoint pen, plastic bag, disposable napkins etc. We are using them frequently without giving a thought to their hazardous effects.
Here are some tips for reducing wastes:
• Use fountain pen in place of a ballpoint pen,
• Use old newspapers for packaging, and
• Use cloth napkins in place of disposable ones.
Reuse:
We can reuse certain things for more than one purpose. If we reuse them for other purposes, we can help in reducing the waste. Here are some of the tips for reusing things:
• Small jars and bottles can be cleaned to keep some other kitchen stuffs.
• One should prefer glass bottle to metal can when buying juice or soft drink. The bottles can be refilled.
• Old clothes can be made into other usable items like cushion cover, handkerchief, etc.
• Old mobile phones can be donated to friends or family.
Recycle:
The process by which waste materials are used to make new products is called recycling. Materials like glass, metal, plastic, and paper are collected, separated, and recycled to make new things.
Recycling of Plastics :
Bucket, bottle, toy, shoe, bag, pen, and comb are a few things made of plastic. Use of plastics has become a major concern nowadays because they are non-biodegradable and release harmful gases upon heating or burning. They can also contaminate foodstuffs. If eaten by animals, plastics can choke and kill them. Therefore, one should reduce and reuse plastic items as far as possible. When plastic items are to be discarded, they should be recycled to make new things.
Not all the plastics generated are recycled, hence causing much damage to life on Earth. In many places in India, plastics are totally banned seeing its adverse effects on the environment.
Some of the ways to reduce, reuse, and recycle plastics are as follows:
• Paper, cotton, and jute bag should be preferred over plastic bags (Fig. 17.4).
• Some disposable plastic containers and jars can be used to grow plants (Fig. 17.5).
• Empty bottles can be refilled for storage of water or any other liquid stuff. Think about the number of times you buy a bottle of water when you are out. Instead you can fill the used bottle and carry it whenever you are out.
• Zip foils can be reused after cleaning thoroughly each time after storing foodstuffs.

Recycling of paper:
Paper is made from trees and trees are essential for our survival on Earth. So, even if paper is biodegradable, depletion of trees at a fast pace is a big concern. Therefore, to save trees we must use paper carefully. Some of the ways to save paper are:
• We should always write on both sides of paper sheets.
• Unused pages from old notebooks can be torn off and made into a new notebook for doing rough work and other miscellaneous work.
• We can also reuse envelopes and covers by using stickers to write new addresses.
• We can also make fresh paper from old newspapers.
By now, we know how harmful garbage accumulation can be. But, by following the 3Rs concept, we can manage the garbage generated in our surroundings so that it is beneficial to the environment.

Key Words :

Biodegradable wastes: Wastes that rot by the action of decomposers are called biodegradable wastes.
Non-biodegradable wastes: Wastes that do not rot by the action of decomposers are called non-biodegradable wastes.
Composting: Recycling of organic wastes like vegetable peels, waste food, leaves, etc., by burying them in compost pits is called composting.
Vermicomposting: Composting with the help of a type of earthworm, called red worms, is called vermicomposting.
Landfills: Large areas used for waste disposal are called landfills.
Recycling: The process by which waste materials are used to make new products is called recycling.

Summary :

• There are mainly two types of wastes based on their ability to rot over a period of time: biodegradable and non-biodegradable wastes.
• Dead plants and animals and their products decay while objects like plastic and glass do not.
• Vermicomposting and landfills are two ways of managing solid wastes.
• By practicing the concept of 3Rs (Reduce, Reuse, and Recycle), we can manage our wastes well.

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NCERT Exemplar Problems Class 11 Mathematics Chapter 3  Trigonometric Functions

Q4. If cos (α + ) =4/5 and sin (α- )=5/13 , where α lie between 0 and π/4, then find the value of tan 2α.

Q6. Prove that cos cos /2- cos 3 cos 9/2 = sin 7/2 sin 4 .

Q7. If a cos θ + b sin θ =m and a sin θ -b cosθ = n, then show that a² + b²-m² + n²

Sol: We have, a cos θ + b sin θ = m (i)
and a sin θ -bcos θ = n (ii)

Q8. Find the value of tan 22°30′

Q9. Prove that sin 4A = 4 sin A cos³A – 4 cos A sin³ A.

Sol: L.H.S. = sin 4A
= 2 sin 2A- cos 2A = 2(2 sin A cosA)(cos² A – sin² A)
= 4 sin A • cos³ A – 4 cos A sin³ A = R.H.S.

Q10. If tan + sin = m and tan – sin = n, then prove that m²-n² = 4 sin tan

Sol:We have, tan + sin = m   (i)
And tan -sin =n  (ii)
Now,         m + n = 2 tan
And          m – n = 2 sin.
(m + n)(m -n) = 4 sin 6
tan m² -n² = 4 sin -tan

Q11. If tan (A + B) =p and tan (A – B) = q, then show that tan 2A = p+q / 1 – pq

Sol: We have tan (A + B) =p and tan (A – B) = q
tan2A = tan [(A + B) + (A-B)]

Q12. If cos + cos = 0 = sin + sin β, then prove that cos 2 + cos 2β = -2 cos (α + ).

Q15.  If sin θ+ cos θ =1, then find the general value of θ

Q16. Find the most general value of θ satisfying the equation tan θ = -1 and cos θ = 1/√2 .
Sol: We have tan θ = -1 and cos θ =1/√2 .
So, θ lies in IV quadrant.
θ = 7/4
So, general solution is θ = 7π/4 + 2 n π, n∈ Z

Q17. If cot θ + tan θ = 2 cosec θ, then find the general value of θ
Sol: Given that, cot θ + tan θ = 2 cosec θ

Q18. If 2 sin² θ =3 cos θ, where O≤θ≤2, then find the value of θ

Q19. If sec x cos 5x + 1 = 0, where 0 < x <π/2 , then find the value of x.

Long Answer Type Questions

Q20. If sin(θ + α) = a and sin(θ + β) = b , then prove that cos2(α – β) – 4abcos(α – β) = 1-2a² -2b²

Sol: We have sin(θ + α) = a —(i)
sin(θ + β) = b ——-(ii)

Q22. Find the value of the expression

Q23. If a cos 2+b sin 2 = c has α and β as its roots, then prove that tan α +tan β = 2b/a+c

Q24. If x = sec ϕ-tanϕandy = cosec ϕ + cot ϕ then show that xy + x -y +1=0.

Q25. If lies in the first quadrant and cos =8/17 , then find the value of cos (30° + ) + cos (45° – ) + cos (120° – ).

Q26. Find the value of the expression cos⁴ π/8 + cos⁴ 3π/8  + cos⁴ 5π/8  + cos⁴7π/8

Q27. Find the general solution of the equation 5 cos² +7 sin² -6 = 0.

Q28. Find the general solution of‘the equation sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x.
Sol: We have, (sin x + sin 3x) – 3 sin 2x = (cos x + cos 3x) – 3 cos 2x
=> 2 sin 2x cos x – 3 sin 2x = 2 cos 2x.cos x – 3 cos 2x
=> sin 2x(2 cos x – 3) = cos 2x(2 cos x – 3)
=> sin 2x = cos 2x (As cos x ≠ 3/2)
=>              tan 2x = 1    => tan 2x = tan π/4
=>              2x = nπ + π/4 , n∈Z
x = nπ/2 +π/8 , n∈Z

Q29. Find the general solution of the equation (√3- l)cos + (√3+ 1)sin = 2.

Objective Type Questions

Q30. If sin + cosec =2, then sin² + cosec² is equal to
(a) 1
(b) 4                          
(c) 2                         
(d) None of these

Q31. If f(x) = cos² x + sec² x, then ‘
(a) f(x) <1             
(b) f(x) = 1              
(c) 2 <f(x) < 1      
(d) fx) ≥ 2

Q32. If tan θ = 1/2 and tan ϕ = 1/3, then the value of θ + ϕ is

Q33. Which of the following is not correct?

(a) sin θ = – 1/5 (b) cos θ = 1                 (c) sec θ = -1/2         (d) tan θ = 20
Sol: (c)
We know that, the range of sec θ is R – (-1, 1).
Hence, sec θ cannot be equal to -1/2

Q34. The value of tan 1° tan 2° tan 3° … tan 89° is
(a) 0
(b) 1
(c) 1/2
(d) Not defined

Sol: (b)
tan 1° tan 2° tan 3° … tan 89°
= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]
= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°]
= 1-1… 1-1 = 1

Q36. The value of cos 1° cos 2° cos 3° … cos 179° is
(a) 1/√2
(b) 0
(c) 1
(d) -1

Sol: (b)
Since cos 90° = 0, we have
cos 1° cos 2° cos 3° …cos 90°… cos 179° = 0

Q37. If tan θ = 3 and θ lies in the third quadrant, then the value of sin θ is

Q38. The value of tan 75° – cot 75° is equal to

Q39. Which of the following is correct?
(a) sin 1° > sin 1                                     
(b) sin 1° < sin 1
(c) sin l° = sin l
(d) sin l° = π/18° sin 1

Sol: We know that, in first quadrant if θ is increasing, then sin θ is also increasing.
∴sin 1° < sin 1 [∵ 1 radian = 57◦30′]

Q41. The minimum value of 3 cos x + 4 sin x + 8 is
(a) 5
(b) 9
(c) 7
(d) 3
Sol: (d)
3 cos x + 4sin x + 8 = 5 (3/5 cos x + 4/5sin x) + 8
= 5(sin α cos x + cos α sin x) + 8
= 5 sin(α + x) + 8, where tan α = 3/4

Q42. The value of tan 3A – tan 2A – tan A is
(a) tan 3A . tan 2A . tan A
(b) -tan 3A .tan 2A . tan A
(c) tan A . tan 2A – tan 2A . tan 3A – tan 3A . tan A
(d) None of these
Sol: (a)
3A= A+ 2A
=> tan 3A = tan (A + 2A)
=> tan 3 A = tanA + tan2A/ 1 – tan A . tan 2A
=> tan A + tan 2A = tan 3A – tan 3A• tan 2A . tan A
=> tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Q43. The value of sin (45° + )- cos (45° – ) is
(a) 2 cos              
(b) 2 sin              
(c) 1                         
(d) 0
Sol: (d)
sin (45° + ) – cos (45° – ) = sin (45° + ) – sin (90° – (45° – ))
= sin (45° + ) – sin (45°+ ) = 0

Q44. The value of (π/4+ ) cot (π/4- ) is
(a) -1                       
(b)  0  
(c)  1                     
(d)   Not defined

Q46. The value of cos 12° + cos 84° + cos 156° + cos 132° is
(a) 1/2            
(b) 1                       
(c) -1/2            
(d) 1/8

Q47. If tan A = 1/2 and tan B = 1/3 then tan (2A + B) is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Q49. The value of sin 50° – sin 70° + sin 10° is equal to
(a) 1                       
(b) 0                       
(c) 1
(d) 2

Q50. If sin + cos =1, then the value of sin 2 is
(a) 1                      
(b) 1   
(c) 0                        
(d) -1

NCERT Exemplar Problems Class 11 Mathematics

The post NCERT Exemplar Problems Class 11 Mathematics Chapter 3 Trigonometric Functions appeared first on Learn CBSE.

NCERT Exemplar Problems Class 11 Mathematics Chapter 4 Principle of Mathematical Induction

Short Answer Type Questions

Q1. Give an example of a statement P(n) which is true for all n≥ 4 but P(l), P(2) and P(3) are not true. Justify your answer.

Sol. Consider the statement P(n): 3n < n!

For n = 1, 3 x 1 < 1!, which is not true
For n = 2, 3 x 2 < 2!, which is not true
For n = 3, 3 x 3 < 3!, which is not true
For n = 4, 3 x 4 < 4!, which is true
For n = 5, 3 x 5 < 5!, which is true

Q2. Give an example of a statement P(n) which is true for all Justify your answer.

Instruction for Exercises 3-16: Prove each of the statements in these Exercises by the Principle of Mathematical Induction.

Q3. 4ⁿ – 1 is divisible by 3, for each natural number
Sol: Let P(n): 4ⁿ – 1 is divisible by 3 for each natural number n.
Now, P(l): 4¹ – 1 = 3, which is divisible by 3 Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 4^k – 1 is divisible by 3
or               4^k – 1 = 3m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 4^k+1 – 1
= 4^k-4-l
= 4(3m + 1) – 1  [Using (i)]
= 12 m + 3
= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q4. 2³ⁿ – 1 is divisible by 7, for all natural numbers
Sol: Let P(n): 2³ⁿ – 1 is divisible by 7
Now, P( 1): 2³ — 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 2^3k – 1 is divisible by 7.
or               2^3k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 2^3(k+1)– 1
= 2^3k.2³– 1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q5. n³ – 7n + 3 is divisible by 3, for all natural numbers
Sol: Let P(n): n³ – 7n + 3 is divisible by 3, for all natural numbers n.
Now P(l): (l)³ – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) = K³ – 7k + 3 is divisible by 3
or K³ – 7k + 3 = 3m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1 ):(k + l)³ – 7(k + 1) + 3
= k³ + 1 + 3k(k + 1) – 7k— 7 + 3 = k³ -7k + 3 + 3k(k + l)-6
= 3m + 3[k(k+l)-2]  [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q6. 3²ⁿ – 1 is divisible by 8, for all natural numbers
Sol: Let P(n): 3²ⁿ – 1 is divisible by 8, for all natural numbers n.
Now, P(l): 3² – 1 = 8, which is divisible by 8.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): 3^2k – 1 is divisible by 8
or               3^2k -1 = 8m, m ∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 3^2(k+1)– l
= 3^2k • 3² — 1
= 9(8m + 1) – 1     (using (i))
= 72m + 9 – 1
= 72m + 8
= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q7. For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Sol: Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n.
Now, P(l) = 7¹-2¹ = 5, which is divisible by 5.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.

.’.  P(k) = 7^k -2^k is divisible by 5
or  7^k – 2^k = 5m, m∈ N                                                                           (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 7^k+1 -2^k+1
= 7^k-7-2^k-2
= (5 + 2)7^k -2^k-2
= 5.7^k + 2.7^k-2-2^k
= 5.7^k + 2(7^k – 2^k)
= 5 • 7^k + 2(5 m)     (using (i))
= 5(7^k + 2m), which divisible by 5.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q8. For any natural number n, xⁿ -yⁿ is divisible by x -y, where x and y are any integers with x ≠y
Sol: Let P(n) : xⁿ – yⁿ is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x¹ -y¹ = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): x^k -y^k is divisible by (x – y)
or   x^k-y^k = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):x^k+l-y^k+l
= x^k-x-x^k-y + x^k-y-y^ky
= x^k(x-y) +y(x^k-y^k)
= x^k(x – y) + ym(x – y)  (using (i))
= (x -y) [x^k+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q9. n³ -n is divisible by 6, for each natural number n≥
Sol: Let P(n): n³ – n is divisible by 6, for each natural number n> 2.
Now, P(2): (2)³ -2 = 6, which is divisible by 6.
Hence, P(2) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): k³ – k is divisible by 6
or    k³ -k= 6m, m∈ N       (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): (k+ l)³-(k+ 1)
= k³+ 1 +3k(k+ l)-(k+ 1)
= k³+ 1 +3k² + 3k-k- 1 = (k³-k) + 3k(k+ 1)
= 6m + 3 k(k +1)  (using (i))
Above is divisible by 6.   (∴ k(k + 1) is even)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.

Q10. n(n² + 5) is divisible by 6, for each natural number
Sol: Let P(n): n(n² + 5) is divisible by 6, for each natural number.
Now P(l): 1 (l² + 5) = 6, which is divisible by 6.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): k( k² + 5) is divisible by 6.
or K (k²+ 5) = 6m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(K+l):(K+l)[(K+l)² + 5]
= (K + l)[K² + 2K+6]
= K³ + 3 K² + 8K + 6
= (K² + 5K) + 3 K² + 3K + 6 =K(K² + 5) + 3(K² + K + 2)
= (6m) + 3(K² + K + 2)        (using (i))
Now, K² + K + 2 is always even if A is odd or even.
So, 3(K² + K + 2) is divisible by 6 and hence, (6m) + 3(K² + K + 2) is divisible by 6.
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q11. n² < 2ⁿ, for all natural numbers n ≥
Sol: Let P(n): n² < 2ⁿ for all natural numbers n≥ 5.
Now P(5): 5² < 2⁵ or 25 < 32, which is true.
Hence, P(5) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): k² < 2^k  (i)
Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)² <2^k+1
Using (i), we get
(k + l)² = k² + 2k + 1 < 2^k + 2k + 1         (ii)
Now let, 2^k + 2k + 1 < 2^k+1     (iii)
∴ 2^k + 2k + 1 < 2 • 2^k
2k + 1 < 2^k, which is true for all k > 5 Using (ii) and (iii), we get (k + l)² < 2^k+1 Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 5.

Q12. 2n<(n + 2)! for all natural numbers
Sol: Let P(n): 2n < (n + 2)! for all natural numbers n.
P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence,P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) :2k<(k + 2)!  (i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k+ 1) < (k + 1 + 2)!
or 2(k+ 1) < (k + 3)!
Using (i), we get
2(k + 1) = 2k + 2<(k+2) !  +2  (ii)
Now let, (k + 2)! + 2 < (k + 3)!  (iii)
=>  2 < (k+ 3)! – (k+2) !
=> 2 < (k + 2) ! [k+ 3-1]
=>2<(k+ 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii), we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q14. 2 + 4 + 6+… + 2n = n² + n, for all natural numbers
Sol: Let P(n) :2 + 4 + 6+ …+2 n = n² + n
P(l): 2 = l² + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k² + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k² + k + 2(k+ 1)  [Using (i)]
= k² + k + 2k + 2
= k² + 2k+1+k+1
= (k + 1)² + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q15. 1 + 2 + 2² + … + 2ⁿ = 2^{n +1} – 1 for all natural numbers
Sol: Let P(n): 1 + 2 + 2² + … + 2ⁿ = 2^{n +1} – 1, for all natural numbers n
P(1): 1 =2^{0 + 1} — 1 = 2 — 1 = 1, which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 2²+…+2^k = 2^k+1-l              (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 2²+ …+2^k + 2^k+1
= 2^{k +1} – 1 + 2^k+1  [Using (i)]
= 2.2^k+l– 1 = 1
₌ 2^(k+1)+1-1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q16. 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers
Sol: Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k):l+5 + 9 +…+(4k-3) = k(2k-1)  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … +  (4k- 3) + [4(k+ 1) – 3]
= 2k² -k+4k+ 4-3
= 2k² + 3k + 1
= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

Long Answer Type Questions
Q17. A sequence a_x, a₂, a₃, … is defined by letting a₁=3 and a_k = 7a_k–₁ for all natural numbers k≥ Show that a_n = 3 • 7 ^n-1 for all natural numbers.
Sol: We have a sequence a_x, a₂, a₃… defined by letting a, = 3 and a_k = 7a_k–₁, for all natural numbers k≥2.

Q18. A sequence b₀, b₁, b₂, … is defined by letting b₀ = 5 and b_k = 4 + b_k–₁, for all natural numbers Show that b_n = 5 + 4n, for all natural number n using mathematical induction.
Sol. We have a sequence b₀, b₁, b₂,… defined by letting b₀ = 5 and b_k = 4 + b_k–₁,, for all natural numbers k.

So, by the principle of mathematical induction P(n) is true for any natural number rt,n> 1.

Q25. Prove that number of subsets of a set containing n distinct elements is 2″, for all n ∈
Sol: Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.
For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 2¹ elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2^k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2^k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2^k = 2^k+1
So, P(k + 1) is true. Hence, P(n) is true.

 NCERT Exemplar Problems Class 11 Mathematics  

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NCERT Exemplar Problems Class 11 Mathematics Chapter 5 Complex Numbers and Quadratic Equations

Short Answer Type Questions

Q6. If a = cos θ + i sin θ, then find the value of (1+a/1-a) 
Sol:  a = cos θ + i sin θ


Q10. Show that the complex number z, satisfying the condition arg lies on arg (z-1/z+1) = π/4 lies on a  circle.

Sol: Let z = x + iy

Q11. Solve the equation |z| = z + 1 + 2i.
Sol: We have |z| = z + 1 + 2i
Putting z = x + iy, we get
|x + iy| = x + iy + 1+2i

Long Answer Type Questions
Q12. If |z + 1| = z + 2( 1 + i), then find the value of z.
Sol: We have |z + 11 = z + 2(1+ i)
Putting z = x + iy, we get
Then, |x + iy + 11 = x + iy + 2(1 + i)
⟹|x + iy + l|=x + iy + 2(1 +i)

Q13. If arg (z – 1) = arg (z + 3i), then find (x – 1) : y, where z = x + iy.
Sol: We have arg (z – 1) = arg (z + 3i), where z = x + iy
=>  arg (x + iy – 1) = arg (x + iy + 3i)
=> arg (x – 1 + iy) = arg [x + i(y + 3)]

Q14. Show that | z-2/z-3| = 2 represents a circle . Find its center and radius .
Sol: We have | z-2/z-3| = 2
Puttingz=x + iy, we get

Q15. If z-1/z+1 is a purely imaginary number (z ≠1), then find the value of |z|.

Sol: Let   z = x + iy

Q17. If |z₁ | = 1 (z₁≠ -1) and z₂ = z₁ – 1/ z₁ + 1 , then show that real part of z₂ is zero .

Q18. If Z₁, Z₂ and Z₃, Z₄ are two pairs of conjugate complex numbers, then find arg (Z_1/ Z₄) + arg (Z_2/ Z₃)
Sol. It is given that z₁ and z₂ are conjugate complex numbers.

Q20. If for complex number z₁ and z₂, arg (z₁) – arg (z₂) = 0, then show that |z₁ – z₂| = | z₁|- |z₂ |

Q21. Solve the system of equations Re (z²) = 0, |z| = 2.

Sol: Given that, Re(z²) = 0, |z| = 2

Q22. Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.

Fill in the blanks 

True/False Type Questions

Q26. State true or false for the following.
(i) The order relation is defined on the set of complex numbers.
(ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z, the minimum value of |z| + |z – 11 is 1.
(iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1).
(v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0.
(vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Sol:(i) False
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.

(ii) False
Let z = x + iy, where x, y > 0
i.e., z or point A(x, y) lies in first quadrant. Now, —iz = -i(x + iy)
= -ix – i²y = y – ix
Now, point B(y, – x) lies in fourth quadrant. Also, ∠AOB = 90°
Thus, B is obtained by rotating A in clockwise direction about origin.

Matching Column Type Questions
Q24. Match the statements of Column A and Column B.

Q28. What is the conjugate of 2-i / (1 – 2i)²

Q29. If |Z₁| = |Z₂|, is it necessary that Z₁ = Z₂?
Sol: If |Z₁| = |Z₂| then z₁ and z₂ are at the same distance from origin.
But if arg(Z₁) ≠arg(z₂), then z₁ and z₂ are different.
So, if (z₁| = |z₂|, then it is not necessary that z₁ = z₂.
Consider Z₁ = 3 + 4i and Z₂ = 4 + 3i

Q30.If  (a²+1)² / 2a –i = x + iy, then what is the value of x² + y²?
Sol: (a²+1)² / 2a –i = x + iy

Q31. Find the value of z, if |z| = 4 and arg (z) = 5π/6

Q34. Where does z lies, if | z – 5i / z + 5i  |  = 1?
Sol: We have | z – 5i / z + 5i  |

Instruction for Exercises 35-40: Choose the correct answer from the given four options indicated against each of the Exercises.

Q35. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for

Q41. Which of the following is correct for any two complex numbers z₁ and z₂?

NCERT Exemplar Problems Class 11 Mathematics

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ऐसे ऐसे – CBSE Notes for Class 6 Hindi

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टिकट अलबम – CBSE Notes for Class 6 Hindi

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झाँसी की रानी – CBSE Notes for Class 6 Hindi

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जो देखकर भी नहीं देखते – CBSE Notes for Class 6 Hindi

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संसार पुस्तक है – CBSE Notes for Class 6 Hindi

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CBSE English Communicative Class 9 Syllabus 

Exam Structure for English Communicative Class IX (2017-18)

The annual examination will be of 80 marks, with a duration of three hours.

Section A: Reading  [20 Marks]

This section will have two reading passages as per the details below :
Q1: A Factual passage 300-350 words with eight Very Short Answer Type Questions. [8 marks]
Q2: A Discursive passage of 350-400 words with four Short Answer Type Questions to test inference, evaluation and analysis and four Very Short Answer Type Questions to test vocabulary. [12 marks]

Section B: Writing and Grammar [30 Marks]

Q3: Writing a diary/article in about 100-120 words using visual or verbal cue/s. The questions will be thematically based on MCB. [8 marks]
Q4: Writing a short story based on a given outline or cue/s in about 200 – 250 words. [12 marks]
The Grammar syllabus will include the following areas in class IX :

1. Tenses
2. Modals
3. Use of passive voice
4. Subject – verb concord
5. Reporting
   1. Commands and requests
   2. Statements
   3. Questions
6. Clauses:
   1. Noun clauses
   2. Adverb clauses of condition and time
   3. Relative clauses
7. Determiners
8. Prepositions

The above items may be tested through test types as given below:
Q5: Gap filling with one or two words to test Prepositions, Articles, Conjunctions and Tenses. [3 marks]
Q6: Editing or Omission. [4 marks]
Q7: Sentences Reordering or Sentence Transformation in context. [3 marks]

Section C: Literature Textbook & Extended Reading Text [30 Marks]

Q8: One out of two extracts from prose/poetry/play for reference to the context.
Four Very Short Answer Questions: two questions of one mark each for global comprehension and two questions of one mark each for interpretation. [4 marks]
Q9: Four Short Answer type Questions from the Literature Reader to test local and global comprehension of theme and ideas (30-40 words each). [4 × 2 = 08 Marks]
Q10: One out of two Long Answer type Questions to assess how the values inherent in the text have been brought out. Creativity, imagination and extrapolation beyond the text and across the texts will be assessed. (100-120 words). [8 marks]
Q11: One out of two very Long Answer Questions on theme, plot involving interpretation, inference and character sketch, in about 150-200 words based on the prescribed extended reading text. [10 Marks]

Prescribed Books: Published by CBSE, New Delhi

Interact in English Series

• Main Course Book (Revised Edition)
• Workbook (Revised Edition)
• Literature Reader (Revised Edition)

Extended Reading Texts (either one)

• Gulliver’s Travels (unabridged) by Jonathan Swift
• Three Men in a Boat (unabridged) by Jerome K. Jerome

NCERT Solutions

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CBSE English Language And Literature Class 9 Syllabus

Exam Structure for English Language And Literature Class IX (2017-18)

The annual examination will be of 80 marks, with a duration of three hours.

Section A: Reading  [20 Marks]

This section will have two reading passages as per the details below :
Q1: A Factual passage 300-350 words with eight Very Short Answer Type Questions. [8 marks]
Q2: A Discursive passage of 350-400 words with four Short Answer Type Questions to test inference, evaluation and analysis and four Very Short Answer Type Questions to test vocabulary. [12 marks]

Section B: Writing and Grammar [30 Marks]

Q3: Writing a diary/article in about 100-120 words using visual or verbal cue/s. The questions will be thematically based on the prescribed books. [8 marks]
Q4: Writing a short story based on a given outline or cue/s in about 200 – 250 words. [12 marks]
The Grammar syllabus will include the following areas in class IX :

1. Tenses
2. Modals
3. Use of passive voice
4. Subject – verb concord
5. Reporting
   1. Commands and requests
   2. Statements
   3. Questions
6. Clauses:
   1. Noun clauses
   2. Adverb clauses of condition and time
   3. Relative clauses
7. Determiners
8. Prepositions

The above items may be tested through test types as given below:
Q5: Gap filling with one or two words to test Prepositions, Articles, Conjunctions and Tenses. [3 marks]
Q6: Editing or Omission. [4 marks]
Q7: Sentences Reordering or Sentence Transformation in context. [3 marks]

Section C: Literature Textbooks [30 Marks]

Q8: One out of two extracts from prose/poetry/play for reference to the context.
Four Very Short Answer Questions: two questions of one mark each for global comprehension and two questions of one mark each for interpretation. [4 marks]
Q9: Four Short Answer type Questions from the Literature Reader to test local and global comprehension of theme and ideas (30-40 words each). [4 × 2 = 08 Marks]
Q10: One out of two Long Answer type Questions to assess how the values inherent in the text have been brought out. Creativity, imagination and extrapolation beyond the text and across the texts will be assessed. (100-120 words). [8 marks]
Q11: One out of two very Long Answer Questions on theme, plot involving interpretation, inference and character sketch, in about 150-200 words based on the prescribed extended reading text. [10 Marks]

Prescribed Books:

Published by NCERT, New Delhi

• BEEHIVE – Textbook for class IX
• MOMENTS Supplementary Reader for Class IX

Extended Reading Texts (either one)

• Gulliver’s Travels (unabridged) by Jonathan Swift
• Three Men in a Boat (unabridged) by Jerome K. Jerome

NCERT Solutions

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CBSE Hindi A Class 9 Syllabus

Exam Structure for Hindi A Class IX (2017-18)

NCERT Solutions

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CBSE Hindi B Class 9 Syllabus

Exam Structure for Hindi B Class IX (2017-18)

NCERT Solutions

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NCERT Exemplar Problems Class 11 Mathematics Chapter 7 Permutations and Combinations

Short Answer Type Questions
Q1. Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
Sol. First the women choose the chairs from amongst the chairs numbered 1 to 4.

Q2. If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary, then what is the rank of the word RACHIT?

Sol: The alphabetical order of the letters of the word RACHIT is: A, C, H, I, R, T. Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with 1 = 5!
Clearly, the first word beginning with R is RACHIT.
.•. Rank of the word RACHIT in dictionary = 4×5! + 1= 4 x120+1= 481

Q3. A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.

Sol: Since the candidate cannot attempt more than 5 questions from either group, he is able to attempt minimum two questions from either group.
The possible number of questions attempted from each group will be as given in the following table:

Q4. Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.
Sol: There are 18 point in a plane, of which 5 points are collinear.

Q5. We wish to select 6 persons from 8 but, if the person A is chosen, then B must be chosen. In how many ways can selections be made?
Sol: Total number of persons = 8
Number of person to be selected = 6
It is given that, if A is chosen then, B must be chosen. Therefore, following cases arise:

Q6. How many committee of five persons with a chairperson can be selected form 12 persons?
Sol: Total number of persons =12
Number of persons to be selected = 5

Q7. How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
Sol: There are 26 English alphabets and 10 digits (0 to 9).
It is given that each plate contains two different letters followed by three different digits.

Q8. A bag contains 5 black and 6 red balls. Determine the number of ways in • which 2 black and 3 red balls can be selected from the lot.
Sol: The bag contains 5 black and 6 red balls.

Q9. Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
Sol: Total number of things. = n
We have to arrange r things out of n in which three particular things must occur together.

Q10. Find the number of different words that can be formed from the letters of the word TRIANGLE, so that no vowels are together.
Sol: Given word is: TRIANGLE Consonants are: T, R, N, G, L Vowels are: I, A, E
Since we have to form words in such a way that no two vowels are together, we first arrange consonants.
Five consonants can be arranged in 5! ways.

Q11. Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Sol: We have to form 4-digit numbers which are greater than 6000 and less than 7000.
We know that a number is divisible by 5, if at the unit place of the number there is 0 or 5.
So, unit digit can be filled in 2 ways.
The thousandth place can be filled by ‘6’ only.
The hundredth place and tenth place can be filled together in 8 x 7 = 56 ways. So, total number of ways = 56 x 2 = 112

Q12. Thereare 10persons named P₁,P₂,P₃,…,P_!0.Outof 10 persons, 5 persons are to be arranged in a line such that in each arrangement P, must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.
Sol. Given that, P₁, P₂, …, P₁₀, are 10 persons, out of which 5 persons are to be arranged but P, must occur whereas P₄ and P₅ never occurs.
As P, is already occurring w’e have to select now 4 out of 7 persons.
.•. Number of selections = ⁷C₄ = 35 Number of arrangements of 5 persons = 35 x 5! = 35 x 120 = 4200

Q13. There are 10 lamps in a hall each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
Sol: There are 10 lamps in a hall.
The hall can be illuminated if at least one lamp is switched.
.•. Total number of ways =        ¹⁰C₁+ ¹⁰C₂ + ^l0C₃… + ¹⁰C_]0
= 2¹⁰– 1 = 1024- 1 = 1023

Q14. A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Sol: There are two white, three black and four red balls.
We have to draw 3 balls, out of these 9 balls in which at least one black ball is included.
So we have following possibilities:

.’. Number of selections = ³C₁ x ⁶C₂ + ³C₂ x ⁶C, + ³C₃ x ⁶C₀
= 3×15+ 3×6+1= 45+ 18 + 1= 64

Q15. If ⁿC_r-1=  36 ⁿC_r = 84 and ⁿC_r+1= 126, then find the value of ^rC₂.

Q16. Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Sol: We have to find the number of integers greater than 7000 with the digits 3,5, 7, 8 and 9.
So, with these digits, we can make maximum five-digit numbers because repetition is not allowed.
Since all the five-digit numbers are greater than 7000, we have Number of five-digit integers = 5x4x3x2x1 = 120 A four-digit integer is greater than 7000 if thousandth place has any one of 7, 8 and 9.
Thus, thousandth place can be filled in 3 ways. The remaining three places can be filled from remaining four digits in ⁴P₃ ways.
So, total number of four-digit integers = 3x ⁴P₃ = 3x4x3x2 = 72 Total number of integers = 120 + 72 = 192

Q17. If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
Sol: It is given that no two lines are parallel which means that all the lines are intersecting and no three lines are concurrent.
One point of intersection is created by two straight lines.
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time

Q18. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?
Sol: If first two digit is 41, the remaining 4 digits can be arranged in ⁸P₄ = 8 x 7 x 6×5 = 1680 ways.
Similarly, if first two digit is 42, 46, 62, or 64, the remaining 4 digits can be arranged in ⁸P₄ ways i.e., 1680 ways.
.’. Total number of telephone numbers having all six digits distinct = 5x 1680 = 8400

Q19. In an examination, a student has to answer 4 questions out of 5 questions, questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Sol: It is given that 2 questions are compulsory out of 5 questions.
So, the other 2 questions can be selected from the remaining 3 questions in ³C₂ = 3 ways.

Q20. A convex polygon has 44 diagonals. Find the number of its sides.
[Hint: Polygon of n sides has (ⁿC₂ – n) number of diagonals.]
Sol: Let the convex polygon has n sides.
Number of diagonals=Number of ways of selecting two vertices – Number of sides = ⁿC₂ – n
It is given that polygon has 44 diagonals.

Long Answer Type Questions
Q21. 18 mice were placed in two experimental groups and one control group with all groups equally large. In how many ways can the mice be placed into three groups?
Sol: It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
Each group is of 6 mice.

Q22. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if (i) they can be of any colour (ii) two must be white and two red and (iii) they must all be of the same colour.
Sol:Total number of marbles = 6 white +- 5 red = 11 marbles

(a) If they can be of any colour means we have to select 4 marbles out of 11
∴ Required number of ways = ¹¹C₄
(b) Two white marbles can be selected in ⁶C₂
Two red marbles can be selected in ⁵C₂ ways.
∴ Total number of ways = ⁶C₂ x ⁵C₂ = 15 x 10 = 150
(c) If they all must be of same colour,
Four white marbles out of 6 can be selected in ⁶C₄ ways.
And 4 red marbles out of 5 can be selected in ⁵C₄ ways.
∴ Required number of ways = ⁶C₄ + ⁵C₄ = 15 + 5 = 20

Q23. In how many ways can a football team of 11 players be selected from 16 players? How many of them will

• include 2 particular players?
• exclude 2 particular players?

Sol: Total number of players = 16
We have to select a team of 11 players
So, number of ways = ¹⁶C₁₁
(i) If two particular players are included then more 9 players can be selected from remaining 14 players in ¹⁴C₉
(ii) If two particular players are excluded then all 11 players can be selected from remaining 14 players in ¹⁴C₁₁
Q24.  sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Sol: Total number of students in each class = 20
We have to select at least 5 students from each class.
So we can select either 5 students from class XI and 6 students from class XII or 6 students from class XI and 5 students from class XII.
∴ Total number of ways of selecting a team of 11 players = ²⁰C₅ x ²⁰C₆ + ²⁰C₆ x ²⁰C₅ = 2 x ²⁰C₅ x ²⁰C₆

Q25. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls
Sol: Number of girls = 4;
Number of boys = 7
We have to select a team of 5 members provided that

Objective Type Questions
Q26. If ⁿC₁₂ = ⁿC₈, then n is equal to 
(a) 20     
(b) 12
(c) 6
(d) 30

Q27. The number of possible outcomes when a coin is tossed 6 times is
(a) 36                    
(b) 64                       
(c) 12                       
(d) 32
Sol: (b) Number of outcomes when a coin tossed = 2 (Head or Tail)
∴Total possible outcomes when a coin tossed 6 times = 2x2x2x2x2x 2 = 64

Q28. The number of different four-digit numbers that can be formed with the digits
2, 3, 4, 7 and using each digit only once is
(a) 120                  
(b) 96                       
(c) 24                       
(d) 100
Sol: (c) Given digits 2,3,4 and 7, we have to form four-digit numbers using these digits.
∴Required number of ways = ⁴P₄ = 4!=4x3x2x1 = 24

Q29. The sum of the digits in unit place of all the numbers formed with the help of 3,4, 5 and 6 taken all at a time is
(a)    432               
(b)    108                  
(c)      36                   
(c)    18
Sol: (b) If the unit place is ‘3’ then remaining three places can be filled in 3! ways.
Thus ‘3’ appears in unit place in 3! times.
Similarly each digit appear in unit place 3! times.
So,    sum of digits in unit place = 3!(3 + 4 + 5 + 6) = 18 x 6 = 108

Q30. The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is
(a) 60                 
(b) 120                  
(c) 7200              
(d)  720
Sol: (c) Given number of vowels = 4 and number of consonants = 5 We have to form words by 2 vowels and 3 consonants.
So, lets first select 2 vowels and 3 consonants.
Number of ways of selection = ⁴C₂ x ⁵C₃ = 6 x 10 = 60 Now, these letters can be arranged in 5! ways.
So, total number of words = 60 x 5! = 60 x 120 = 7200

Q31. A five-digit number divisible by 3 is to be formed using the numbers 0, 1,2,4, and 5 without repetitions. The total number of ways this can be done is
(a)  216               
(b)  600                  
(c)  240                 
(d) 3125
[Hint: 5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.]

Sol:(a) We know that a number is divisible by 3 if the sum of its digits is divisible by 3.
Now sum of the given six digits is 15 which is divisible by 3. So to form a number of five-digit which is divisible by 3 we can remove either ‘O’ or ‘3’. If digits 1, 2, 3,4, 5 are used then number of required numbers = 5!
If digits 0, 1,2,4, 5 are used then first place from left can be filled in 4 ways and remaining 4 places can be filled in 4! ways. So in this case required numbers are 4 x 4! ways.
So, total number of numbers = 120 + 96 = 216

Q32. Everybody in a room shakes hands with everybody else. If the total number of hand shakes is 66, then the total number of persons in the room is
(a)    11                 
(b)     12                    
(c)     13                   
(d)   14
Sol: (b) Between any two person there is one hand shake.

Q33. The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is
(a)    105                (b)     15                     (c)     175                  (d)  185
Sol: (d) Number of ways of selecting 3 points from given 12 points = ¹²C₃
But any three points selected from given seven collinear points does not form triangle.
Number of ways of selecting three points from seven collinear points = ⁷C₃Required number of triangles = ¹²C₃ – ⁷C₃ = 220 -35 = 185

Q34. The number of parallelograms that can be formed form a set of four parallel lines intersecting another set of three parallel lines is
(a)    6                     
(b)   18                    
(c)   12                   
(d)   9
Sol: (b) To form parallelogram we required a pair of line from a set of 4 lines and another pair of line from another set of 3 lines.
Required number of parallelograms = ⁴C₂ x ³C₂ = 6×3 = 18

Q35. The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
(a)   ¹⁶C₁₁               
(b)    ¹⁶C₅                    
(c)     ¹⁶C₉                  
(d)   ²⁰C₉

Sol: (c) Total number of players = 22
We have to select a team of 11 players.
We have to exclude 4 particular of them, so only 18 players are now available. Also from these 2 particular players are always included. Therefore we have to select 9 more players from the remaining 16 players.
So, required number of ways = ¹⁶C₉

Q36. The number of 5-digit telephone numbers having at least one of their digits repeated is
(a) 900000
(b) 10000                 
(c) 30240                
(d) 69760
Sol: (d) Total number of telephone numbers when there is no restriction = 10⁵ Also number of telephone numbers having all digits different = ^l0P₅ Required number of ways = 10⁵ – ^l0P₅ = 1000000 -10x9x8x7x6 = 1000000-30240 = 69760

Q37. The number of ways in which we.can choose a committee from four men and six women, so that the committee includes at least two men and exactly twice as many women as men is
(a) 94                    
(b) 126                     
(c) 128                    
(d) none of these
Sol: (a) Number of men = 4; Number of women = 6
It is given that committee includes at least two men and exactly twice as many women as men.
So, we can select either 2 men and 4 women or 3 men and 6 women.
∴ Required number of committee formed = ⁴C₂ x ⁶C₄ + ⁴C₃ x ⁶C₆
= 6×15 + 4×1=94

Q38. The total number of 9-digit numbers which have all different digits is
(a) 10!                   
(b) 9!                        
(c) 9×9!               
(d) 10×10!
Sol: (c) We have to form 9-digit number which has all different digit.
First digit from the left can be filled in 9 ways (excluding ‘0’).
Now nine digits are left including ‘O’.
So remaining eight places can be filled with these nine digits in ⁹P_S ways.
So, total number of numbers = 9 x ⁹P₈ = 9×9!

Q39. The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is
(a) 1440               
(b) 144                     
(c) 7!                        
(d) ⁴C₄ x ³C₃
Sol: (b) We have word ARTICLE.
Vowels are A, I, E and consonants are R, T, C, L.
Now vowels occupy three even places (2^nd, 4^th and 6^th) in 3! ways.
In remaining four places four consonants can be arranged in 4! ways.
So, total number of words = 3! x4! = 6×24= 144

Q40. Given five different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
(a) 3600               
(b) 3720                   
(c) 3800                  
(d) 3600
[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes and 3 red dyes are 2⁵, 2⁴ and 2³, respectively.]

Fill in the Blanks Type Questions

True/False Type Questions

Q51. There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ¹²C₂ – ⁵C₂.
Sol: False
Required number of lines = ¹²C₂ – ⁵C₂.+ 1

Q52. Three letters can be posted in fiv.e letter boxes in 3⁵
Sol: False
Each letter can be posted in any one of the five letter boxes.
So, total number of ways of posting three letters = 5x5x5 = 125

Q53. In the permutations of n things r taken together, the number of permutations in which m particular things occur together is

Q54. In a steamer there are stalls for 12 animals and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in 3¹² ways.
Sol:True
In each stall any one of the three animals can be shipped.
So total number of ways of loading = 3x3x3x…xl2 times = 3¹²

Q55. If some or all of n objects are taken at a time, then the number of combinations is 2ⁿ– 1.
Sol: True
If some or all objects taken at a time, then number of combinations would be ⁿC₁ + ⁿC₂ + ⁿC₃ + … + ⁿC_n = 2ⁿ – 1

Q56. There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.
Sol: False
Number of ways of selecting any number of objects from given n identical objects is 1.
Now selecting zero or more red ball from 4 identical red balls = 1 + 1 + 1 + 1 + 1=5
Selecting at least 1 black ball from 5 identical black balls =1 + T+1 + 1 + 1= 5 So, total number of ways = 5 x 5 = 25

Q58. A candidate is required to answer 7 questions, out of 12 questions which are divided into two groups, each containing 6 He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.
Sol: False
A candidate can attempt questions in following maimer

Q59. To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ⁵C₃ x ²²C₉.
Sol:True
We can select 3 scheduled caste candidate out of 5 in ⁵C₃ ways.
And we can select 9 other candidates out of 22 in ²²C₉ways.
.’. Total number of selections = ⁵C₃ x ²²C₉

 NCERT Exemplar Problems Class 11 Mathematics

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