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Since a point is in the xz-plane, then it is of the form (x,0, z). So its y-coordinate is zero.

Answer is (b) 5π/6

Answer is (a) π/4

We have, the coordinates of a point dividing the line joining the points A (x₁, y₁, z₁) and B (x₂, y₂, z₂) in the ratio k : 1 are,

P = ((kx₂ + x₁)/(k + 1), (ky₂ + y₁)/(k + 1), (kz₂ + z₁)/(k + 1))

(i) If xy – plane divides a line segment joining the given points, A, B, then z – coordinate of P is zero.

i.e., (kz₂ + z₁)/(k + 1) = 0 ⇒ kz₂ + z₁ = 0 ⇒ k = - z₁/z₂

∴ xy-plane divides the line segment AB in the ratio -z₁ : z₂ Similarly, 

(ii) The yz-plane in the ratio -x₁: x₂ 

(iii) The zx-plane in the ratio -y₁ : y₂

Since a point is on the x-axis then it is of the form (x,0,0) 

∴ y-coordinate = 0 and z-coordinate = 0.

Answer is (b) - k

Answer is (b) e^x + e^-y = c

Answer is (A) log(e - 1)

Answer is (b) y = kx

Answer is (b) 1/(x log x) x (log(log x))

Answer is (D) (1 + x)/x

Answer is (C) i

Answer is (C) 1/e

Larkin uses the “fresh-peeled voice” of the thrush as an adjective to beautify the evening setting of the new season, the spring. The phrase describes the freshness and sharpness of the thrush's humming. This freshness is symbolic of the freshness that has dissolved in the air with the advent of the new season that the poet celebrates. The thrush sings, sitting in a “laurel-surrounded in the deep bare garden”. It hums repeatedly that “it will be spring soon”. Its singing marks an “astonishing” effect on the brickwork of the houses. The song of the thrush also acts a catalyst in the transcending of Philip to his childhood flashing the “forgotten-boredom” right in front of his eyes. He is transformed into a child. The poet probably tries to draw a parallelism between the freshness of the bird's song and the innocence of the child. The “fresh-peeled voice” is symbolic of the spring succeeding the winter. The transformation of the season. It appears as the winter gave birth to the spring and now that it is coming, the whole universe and humanity dance to the tune of thrush to join in the celebration.

The relationship of Mrs. Crost's with that of her daughter did not seem to be that of love and affection. They shared a very practical life based on strong reasons and logic. She seemed to be a matter of fact person who didn't believe in getting  emotional on stuff. She carried on her responsibilities for her mother as a mere schedule. Care and empathy was what lacked in her. She knew what were her mother's weaknesses. This is clear when she says: ‘She slips sometimes.’ The gap in between their generations was visible when they differed on women wearing miniskirts on the streets. Helen was quite indifferently calm of the fact that Mrs. Croft was three years older than a century and needed constant assistance, supervision and above all, love.

It is difficult to judge whether the poet is trying to compare or is drawing a relation. Philip Larkin, in his poem Coming, celebrates the advent of the new season, spring, with the “fresh-peeled voice” of the thrush. He creates the imagery of the spring peeled out of the winter. The old season giving birth to the new season. The nature had been sleeping in the cold and gloomy winter and now the freshness of the new season sparked a new life in it. The birds, houses, gardens, the whole nature has joined the party to welcome the spring. Seeing this transformation the poet is so happy that he himself transcends into childhood. Here Larkin highlights the difference between innocence and experience. He presents an innocent watching the adults, laughing and reconciling, probably after a fight or reconciling with the life. How he begins to feel happy though he understands nothing. This is the innocence of the child that his happiness lies in others happiness, which is juxtaposed with the experienced adults, who engage themselves in trivial issues creating troubles for themselves and others. The poet has tried to bring out the difference between two seasons and stages of human life. This mystique is beyond Larkin's comprehension and he is only left wondering about it all.

As Larkin is absorbed in the resonant humming of the thrush, he transcends present to his boring childhood, which he feels is best forgotten. The poet is transformed into a child. He feels happy like a child who feels happy just by watching elders reconciling with each other. The child comprehends nothing yet smiles just because the adults are happy. This might appear surprising however, if one may look more closely, the scene reflects the innocence of a child. Probably the poet has tried to make a point that our happiness lies in other's happiness. 

The whole thought makes Larkin happy and he wonders about the mystiques of universe and human life.

The autobiographical element makes a reader curious to know about Larkin's childhood. His parents were very loving and affectionate. However, he recalls his childhood as a dejected one. He talks of himself in depreciating terms. It appears that the poet had a very poor concept of himself. It is also known that Philip suffered slight stammer in childhood that endured for the rest of his life, though reduced. If the reader pays attention, argues John Woley, it is not difficult to note the contradiction in the term “forgotten boredom”. If, like Larkin says, he has forgotten his childhood, the question is, how can he comment so confidently that it was a bored one? However, it may be concluded that Larkin's childhood did not have any memories that he was fond of.He even remarked once that his biography could begin when he was 21, which implies that nothing spectacular happened before that. Thus, Philip Larkin recalls his childhood as “a forgotten boredom”.

The following details depict a very uncomfortable picture of life, faced by financial crisis: The author travelled in a third class cabin. The author talks about 'struggling' to earn a livelihood and establishing oneself abroad. The author were three or four friends residing together and sharing a wash-room which was icy cold. The author, along with his friends, took turns cooking pots of curry which they ate with their hands on a table covered with newspapers. His rugged life could well be understood by instances such as watching cricket at Lord's or listening to Mukesh. Habits such as walking barefoot, smoking, drinking tea and strolling around on weekends show that he wasn't all set in his life. The word 'penniless' quite describes the financial crunch which he faced while he was studying at LSE.

Indian life and American life have a very obvious line of difference. The author found it quite difficult to adjust to the American lifestyle.

There was a time when the narrator went to buy tea bags and a 'flask'. This is when he came to know that the thing which he was looking for was actually known as a 'thermos' and 'flask' was something which was used to store Whisky. The reason for the narrator having limited knowledge on the subject was the fact that he hadn't consumed whisky till then.

The poem 'Coming' by Philip Larkin is a celebration ofthe advent of the spring. To express the happiness the poet sets the plot of house fronts clothedin chilly and yellow light. Amidst all this, a thrush sings a welcoming song. It seems the whole nature is dancing with joy at the arrival of the new season. The thrush, sitting in a garden shrub,laurel”, in the deep bare garden, is humming repeatedly in” its fresh-peeled voice “that” it will be spring soon”. This joy full singing of the thrush imparts an “astonishing” effect on the brickwork of the houses. The poet feels happy as well to see the beauty that nature encompasses. In fact it is through the voice of the thrush that the poet has tried to express that how overwhelmed he is on the “coming” of the spring.

The narrator has seen various shifts in his life. Firstly, during his bachelorhood, he adjusted a great deal surviving under very absurd and unfavourable conditions. He started working in a library to meet his expenses while he was attending lectures at the LSE. He adjusted his accommodation with other people like him, sharing food, toilet, etc. He led a reckless life eating with hands, soaking dirty dishes in the tub, lounging barefoot, doing things only expected from aimless and lethargic people. Next came a phase in his life when he got married and found a permanent job in America. He shifted because he was given a full-time job in the processing department of the library at MIT. This, however, did not mean that he had started living lavishly. His budget remained just the same it was when he was a student. The aspect that changed drastically was the pace of life in America. 

Every person was in a hurry to get to the top. Even in his new accommodation, he adjusted to various noises such as flashing sirens, fleet of buses rumbling along all night, distracting and suffocating him. Life wasn't as peaceful and calm as he had expected with no “glittering ocean to thrill” him or no “breeze to cool” his face. His change in schedule can be reflected at various incidents such as buying milk, etc. His life at the YMCA building turned to be stifling and intolerable with the inconvenient living conditions and unbearable noise because of which he had disturbed sleep.

(a) Compass is considered as a symbol of eternity. However, in the poem, it represents the change that a relationship goes through as it grows, old and less intense. How a man's love for his young beloved grows less as he gave value to beauty over the spiritual love. Though the love never dies, the physical beauty of a lover may fade and withers as it falls to time's compass' sickle. 

(b) Shakespeare says that love is not at the mercy of time. True love is ageless. No matter how old a human grows, with it love grows; though the physical beauty falls prey to time's sickle. It is the constancy that holds two people in a relationship together.

It is not surprising that Shakespeare has chosen love as the subject and has not involved human agents. The reason for his choice is obvious and it the underlying theme of the poem, the fickle nature of humans. However, the love, and eternal being, is constant. Hence, the poet when talked of all the qualities that love embodies and should be sans of, he chose love itself and not a human, whose emotions and sentiments undergo frequent changes. Shakespeare should be appreciated for his understanding of human nature and thus a wise decision made by him in his composition.

It implies that if the poet is proved wrong about these thoughts on love, then he will recant all that he has written, and no man has ever loved.

"Negatives" are an effective tool to prove one's point. It highlights the other side of the coin to bring home the positive points of the statement very effectively. In this case the poet puts forward all the negative aspects that love is taken for, and then argues that love is something permanent and beyond physical beauty.

Answer is (A) sin x

Let e^x = z

∴ e^x.dx = dz

∴ I = ∫e^x.cos(e^x) dx = ∫cos z.dz = sin z + k = sin e^x + k

Answer is (c) e - 1

Answer is (c) e^3x/3 + k

Answer is (a) 3x²

Answer is (d) k + 3log|x|

We have provided Class 11 Maths MCQ Questions with Answers to help students understand the concept very well. Class 11 Maths MCQ Questions of Principle of Mathematical Induction with Answers were prepared based on the latest exam pattern.

MCQ Questions with Answers provide here with detailed solutions so that Students can easily understand the logic behind each answer. It will help students to attempt the exam with confidence. You also get ideas about the type of questions and methods to answer in your Class 11th examination. Practice Class 11 Maths MCQ Questions of Principle of Mathematical Induction given below.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The sum of the series 1³ + 2³+ 3³ + ………..n³ is

(a) {(n + 1)/2}²
(b) {n/2}²
(c) n(n + 1)/2
(d) {n(n + 1)/2}²

2. if n is an odd positive integer, then an + bn is divisible by :

(a) a² + b²
(b) a + b
(c) a – b
(d) none of these

3. 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)}

(a) n(n + 1)
(b) n/(n + 1)
(c) 2n/(n + 1)
(d) 3n/(n + 1)

4. The sum of the series 1² + 2² + 3² + ………..n² is

(a) n(n + 1)(2n + 1)
(b) n(n + 1)(2n + 1)/2
(c) n(n + 1)(2n + 1)/3
(d) n(n + 1)(2n + 1)/6

5. {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =

(a) 1/(n + 1) for all n ∈ N
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N
(d) n/(n + 1) for all n ∈ R

6. For any natural number n, 7ⁿ – 2ⁿ is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

7. The nth terms of the series 3 + 7 + 13 + 21 +………. is

(a) 4n – 1
(b) n² + n + 1
(c) none of these
(d) n + 2

8. Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.

(a) n(n+1)(n+2)/3
(b) n(n+1)(n+2)/6
(c) n(n+2)/6
(d) (n+1)(n+2)/6

9. For any natural number n, 7ⁿ – 2ⁿ is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

10. For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by

(a) 19
(b) 17
(c) 23
(d) 25

11. For all n ∈ N, 7²ⁿ − 48n −1 is divisible by :

(a) 25
(b) 2304
(c) 1234
(d) 26

12. Let P(n) be the statement 2ⁿ <n! where n is a natural number, then P(n) is true for:

(a) all n
(b) all n>2
(c) all n>3
(d) all n<3

13. Let T(k) be the statement 1+3+5+....+(2k−1) = k²+10. Which of the following is correct?

(a) T(1) is true
(b) T(k) is true ⇒ T(k + 1) is true
(c) T(n) is true for all n ∈ N
(d) All above are correct

14. Let P(n)  :n²+n+1 is an even integer. If P(k) is assumed true ⇒ P(k+1) is true. Therefore, P(n) is

(a) true for n>1,n∈N
(b) true for all n∈N
(c) true for n>2,n∈N
(d) none of these

15. If x ⁿ⁻¹ is divisible by x−k, then the least positive integral value of k is

(a) 1
(b) 2
(c) 3
(d) 4

16. if n is a +ve integer, then 2.4²ⁿ⁺¹ + 3³ⁿ⁺¹is divisible by

(a) 2
(b) 7
(c) 11
(d) 27

17. A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P (5) is true. On the basis of this he could conclude that P(n) is true

(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5

18.The greatest positive integer, which divides n(n +1)(n + 2)(n + 3) for all n ∈ N , is

(a) 2
(b) 6
(c) 24
(d) 120

19. n(n + 1) (n + 5) is a multiple of

(a) 3
(b) 8
(c) 5
(d) 7

20. For every positive integer n, 7n – 3n is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

Answer:

1. Answer: (d) {n(n + 1)/2}²

Explanation: Given, series is 1³ + 2³ + 3³ + ……….. n³

Sum = {n(n + 1)/2}²

2. Answer: (b) a + b

Explanation: Given number = an + bn

Let n = 1, 3, 5, ……..

a_{n +} b_n = a + b

a_n + b_n = a³ + b³ = (a + b) × (a²+ b² + ab) and so on.

Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..

So, the given number is divisible by (a + b)

3. Answer: (b) n/(n + 1)

Explanation: Let the given statement be P(n). Then,

P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)} = n/(n + 1).

Putting n = 1 in the given statement, we get

LHS = 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2.

LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} = k/(k + 1) ..…(i)

Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)}

[1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)}

= k/(k + 1)+1/{ (k + 1)(k + 2)}.

= {k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)

= {k(k + 2) + 1}/{(k + 1)(k + 2}

= {(k + 1)² }/{(k + 1)(k + 2)}

= (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1)

⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)}

= (k + 1)/(k + 1 + 1)

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

Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

4. Answer: (d) n(n + 1)(2n + 1)/6

Explanation: Given, series is 1² + 2²+ 3² + ………..n²

Sum = n(n + 1)(2n + 1)/6

5. Answer: (a) 1/(n + 1) for all n ∈ N.

Explanation: Let the given statement be P(n). Then,

P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).

When n = 1, LHS = {1 – (1/2)} = 1/2 and RHS = 1/(1 + 1) = 1/2.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)

Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]

= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]

= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]

= 1/(k + 2)

Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)

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

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

6. Answer: (c) 5

Explanation: Given, 7ⁿ – 2ⁿ

Let n = 1

7ⁿ – 2ⁿ = 7¹ – 2¹ 

= 7 – 2 

= 5

which is divisible by 5

Let n = 2

7ⁿ – 2ⁿ = 7² – 2² 

= 49 – 4

 = 45

which is divisible by 5

Let n = 3

7ⁿ – 2ⁿ = 7³ – 2³ 

= 343 – 8 

= 335

which is divisible by 5

Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5.

7. Answer: (b) n² + n + 1

Explanation: Let S = 3 + 7 + 13 + 21 +……….a_n-1 + an …………1

and S = 3 + 7 + 13 + 21 +……….a_n-1 + an …………2

Subtract equation 1 and 2, we get

S – S = 3 + (7 + 13 + 21 +……….a_n-1 + a_n) – (3 + 7 + 13 + 21 +……….a_n-1 + a_n)

⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_n – a_n-1) – a_n

⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – an

⇒ a_n = 3 + {4 + 6 + 8 + ……(n-1)terms}

⇒ a_n = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}

⇒ a_n = 3 + (n – 1)/2 × {8 + (n – 2)2}

⇒ a_n = 3 + (n – 1) × {4 + n – 2}

⇒ a_n = 3 + (n – 1) × (n + 2)

⇒ a_n = 3 + n²+ n – 2

⇒ an = n² + n + 1

So, the nth term is n² + n + 1

8. Answer: (d) (n+1)(n+2)/6

Explanation: Let each side of the base contains n shots,

then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1

= n(n + 1)/2

= (n  + n)/2

Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers

So, Total shots = ∑(n² + n)/2

= (1/2)×{∑n² + ∑n}

= (1/2)×{n(n+1)(2n+1)/6 + n(n+1)/2}

= n(n+1)(n+2)/6

9. Answer: (c) 5

Explanation: Given, 7ⁿ – 2ⁿ

Let n = 1

7n – 2n 

= 7¹ – 2¹ 

= 7 – 2 

  = 5

which is divisible by 5

Let n = 2

7ⁿ – 2ⁿ = 7² – 2² 

= 49 – 4 

= 45

which is divisible by 5

Let n = 3

7ⁿ – 2ⁿ = 7³ – 2³ 

= 343 – 8 

= 335

which is divisible by 5

Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

10. Answer: (b) 17

Explanation: Given, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹

Let n = 1,

3 × 5^2×1+1 + 2^3×1+1 

= 3 × 5²⁺¹ + 2³⁺¹ 

= 3 × 5³ + 2⁴ 

= 3 × 125 + 16 

= 375 + 16 

= 391

Which is divisible by 17

Let n = 2,

3 × 5^2×2+1 + 2^3×2+1 

= 3 × 5⁴⁺¹ + 2⁶⁺¹ 

= 3 × 5⁵ + 2⁷ 

= 3 × 3125 + 128 

= 9375 + 128

= 9503

Which is divisible by 17

Hence, For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17

11. Answer: (b) 2304

Explanation: Given number = 72n − 48n − 1

Let n = 1, 2 ,3, 4, ……..

7²ⁿ − 48n − 1 = 72 − 48 − 1 

= 49 – 48 – 1 

= 49 – 49 

= 0

7²ⁿ − 48n − 1 

= 7⁴ − 48 × 2 − 1 

= 2401 – 96 – 1 

= 2401 – 97 

= 2304

7²ⁿ − 48n − 1 

= 7⁶ − 48 × 3 − 1

= 117649 – 144 – 1 

= 117649 – 145 

= 117504 

= 2304 × 51

Since, all these numbers are divisible by 2304 for n = 1, 2, 3,…..

So, the given number is divisible by 2304

12. Answer: (c) all n>3

Explanation: We have,P(n) be the statement 2ⁿ <n!

Where n is a natural number

Then,

Put n=1,2,3....

So, P(1) be the statement of 2¹<1! = 2<1 (It is wrong)

P(2) be the statement of 2² < 2! = 4<2 (It is wrong)

P(3)  be the statement of  <3!=8<6(It is wrong)

But,

P(4)  be the statement of 2⁴ <4! =16<24(Itisright)

Similarly,

P(5), P(6)....... So, all number n is a natural number.

13. Answer: (b) T(k) is true ⇒ T(k + 1) is true

Explanation: When k = 1, LHS = 1 but RHS = 1 + 10 = 11

∴ T(1) is not true

Let T(k) is true.

i.e., 1+3+5+.....+(2k−1) = k²+10

Now, 1+ 3+ 5 + ..... + (2k -1) + (2k +1)

= k²+10+2k+1 = (k+1)²+10

∴ T(k +1) is true.

i.e., T(k) is true ⇒ T (k +1) is true.

But T(n) is not true for all n ∈ N , as T(1) is not true.

14. Answer: (d) none of these    

Explanation: Given, P(n)  :n²+n+1 is an even integer.

For n=1,P(1)=3, which is not an even integer.

∴P(1) is not true.

Also, n²+n+1 = n(n+1)+1 is always an odd integer. (principle of induction is not applicable).

15. Answer: (a) 1

Explanation: x⁻¹ is always a factor of xⁿ⁻¹.

⇒k =1.

16. Answer: (c) 11

Explanation: Let P(n) =2.4²ⁿ⁺¹ + 3³ⁿ⁺¹

Then P(1)=2.4³+3⁴ =209, which is divisible by 11 but not divisible by 2, 7 or 27.

Further, let P(k)=2.4^2k+1 + 3^3k+1 is divisible by 11, i.e., 2.4^2k+1+ 3^3k+1=11q for some integer q.

Now P(k+1)=2.4^2k+3 + 3^3k+4

=2.4^2k+1.4² + 3^3k+1.3³

=16.2.4^2k+1+27.3^3k+1

=16.2.2^2k+1+(16+11).3^3k+1

=16[2.4^2k+1+3^3k+1]+11.3^3k+1

=16.11q+11.3^3k+1

=11(16q+3^3k+1)

=11m

where m=16q+33k+1 is another integer.

∴p(k+1) is divisible by 11.

∴P(n)=2.4²ⁿ⁺¹+3³ⁿ⁺¹ is divisible by 11

for all n∈N.

17. Answer: (c) for all n ≥ 5

Explanation: Since P(5) is true and P(k + 1) is true, whenever P (k) is true.

18. Answer: (c) 24

Explanation: The product of r consecutive integers is divisible by r ! . Thus n (n+ 1 ) (n + 2) (n + 3) is divisible by 4 ! = 24.

19. Answer: (a) 3

Explanation: Let P(n) = n(n + 1)(n + 5)

Substituting n = 1, 2, 3,….

P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6

P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7

P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12

20. Answer: (b) 4

Explanation: Let P(n) = 7ⁿ – 3ⁿ

Substituting n = 1, 2, 3,…

P(1) = 71 – 31 = 7 – 3 = 4

P(2) = 72 – 32 = 49 – 9 = 40

P(3) = 73 – 33 = 343 – 27 = 316

Thus, for every positive integer n, 7n – 3n is divisible by 4.

Click here to practice MCQ Questions for Principle of Mathematical Induction class 11

Correct option is B) 25%

A) Women at home

Correct option is C) Primary

correct option is B) Agriculture

C) Gross Domestic Product (G.D.P.)

correct option is D) forest

(a) It is called the water on the earth’s surface.

(a) River Kaveri – Krishna raj sagar in 2nd century AD

(b) Yash: It is necessary to use water sparingly.

(d) Ukaikakrapara: River Tapi

(d) Bhakra Nangal, Chambal Valley, Narmada Valley, Nagaijunasagar

Correct option is (a) 4, 3, 2, 1

C) Organised sector

Different methods reveal different results. Hence, it becomes difficult for economists to compare the growth in national income among different countries.

A meter scale is larger then a match stick. So the center of gravity for meter scale is higher than a matchstick when we keep it vertically. It is easier to balance the object whose center of gravity is higher than the object whose centro of gravity is lower. So, it is hard to balance a match stick than a meter scale.