NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

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NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:43 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 4 Principle of Mathematical Induction deals with various problems and finds its relevance through the method of deductive reasoning which has been considered as an important principle in mathematics for ages. NCERT Exemplar Class 11 Maths chapter 4 solutions covers the application of both deductive and inductive reasoning for proving the applicability of a generalized principle to any specific case. NCERT Exemplar solutions for Class 11 Maths chapter 4 also describes how mathematical induction can be used effectively to prove different mathematical statements. The principle of mathematical deduction was gradually developed over the years to help with the formulation and proving of various statements of science as well as mathematics.

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NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction
Important Notes from NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction
Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 4
NCERT Solutions for Class 11 Mathematics Chapters

NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

Question:1

Give an example of a statement $P(n)$ which it true for all $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are not true. Justify your answer.

Answer:

Given:
$P(n)$ which it true for all $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are not true.

Let us consider P (n) be $2^n< n!$

Examples of the given statements are-

$P (0) \rightarrow 2^0<0! i.e., 1<1 \rightarrow Not True$

$P (1) \rightarrow 2^1<1! i.e., 2<1 \rightarrow Not True$

$\\P (2) \rightarrow 2^2<2!\: \: i.e., 4<2 \rightarrow Not True\\ P (3) \rightarrow2^3 < 3!\: \: i.e., 8<6 \rightarrow Not True\\ P (4) \rightarrow 2^4< 4!\: \: i.e., 16<24 \rightarrow True\\ \&P (5) \rightarrow 2^5<5! \: \: i.e., 32<60 \rightarrow True...... \& so on.$

Question:2

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

Answer:

P (n) is true for all n …….. (Given)

Now, let P (n) be,

$1+ 2 + 3 + 4 +... + n = n (n+1)/2\\ P (0) \rightarrow0 = 0(0+1)/2 = 0 \rightarrow true\\ P (1) \rightarrow1 = 1(2+2)/2 = 1 \rightarrow true\\ P (2) \rightarrow 1+2 = 2(2+1)/2 \rightarrow true\\ P (k) \rightarrow 1+2+3+... +k = k (k+1)/2\\ P (k+1) \rightarrow 1+2+3+... +k+1 = k (k+1)/2 + k + 1 = (k+1) (k+2)/2 \rightarrow p (k+1)$

is true for all k

Thus, P (n) is true for all n.

Question:3

Prove each of the statements in Exercises 3 to 16 by the Principle of Mathematical Induction.
$4^n -1$ is divisible by 3, for each natural number n.

Answer:

P (n) = $4^n -1$ is divisible by 3 …….. (Given)

Now, we’ll substitute different values for n,

$P (0) = 4^0 - 1 = 0$ , viz. divisible by 3

$P (1) = 4^1 - 1 = 3$ , divisible by 3.

$P (2) = 4^2 - 1 = 15$ , divisible by 3

Now, let $P (k) = 4^k - 1$ , is divisible by 3

Thus, $4^k - 1 = 3x$

We also get that,

$P (k+1) = 4^{k+1} - 1 = 4(3x+1) - 1$

$= 12x + 3$ , is divisible by 3

Thus, P (k+1) is also true,

Hence, by mathematical induction,

For each natural no. n it is true that, $P (n) = 4n -1$ is divisible by 3

Question:4

$2^{3n} -1$ is divisible by7, for all natural numbers n.

Answer:

Now, we'll substitute different values for n,
$\\ \mathrm{P}(\mathrm{0})=2^{0}-1=0,\text{ divisible by 7} \\ P(1)=2^{3}-1=7,\text{ divisible by 7}\\ P(2)=2^{6}-1=63,$ divisible by 7\\ $P(3)=2^{9}-1=512, divisible by 7\\$ Now, let us consider,
$\\ P(k)=2^{3 k}-1\ be \ divisible \ by \ 7 \\\ Thus, \ 2^{2 k}-1=7 x\\$ We also get that,
$P(k+1)=2^{3(k+1)}-1=2^{3}(3 x+1)-1=56 x+7, is\ divisible \ by\ 7$

$Thus, \mathrm{P}(\mathrm{k}+1)\ is\: \: also\: \: true,\\$ Hence, by mathematical induction, For each natural no. n it is true that, $\mathrm{P}(\mathrm{n})=2^{3 \mathrm{n}}-1 \ is\ divisible \ by \ 7 .$

Question:5

$n^3 -7n + 3$ is divisible by 3, for all natural numbers n.

Answer:

$n^3 -7n + 3$ is divisible by 3... given

Now, we’ll substitute different values for n,

$P (0) = 0^3 -7 \times 0 + 3 = 3$ , is divisible by 3

$P (1) = 1^3 -7 \times 1 + 3 = -3$ , is divisible by 3

$P (2) = 2^3 -7 \times 2 + 3 = -3$ , is divisible by 3

Now, let us consider,

$P (k) = k^3 -7k + 3$ be divisible by 3

Thus, $k^3 -7k + 3 = 3x$

We also get that,

$P (k+1) = (k+1)^3 -7(k+1) + 3 = k^3 + 3k^2 + 3k +1 -7k -7 + 3$

= $3x + 3(k^2+k-2)$ is divisible by 3

Thus, P (k+1) is also true,

Hence, by mathematical induction,

For each natural no. n it is true that, $P (n) = n^3 -7n + 3$ is divisible by 3.

Question:6

$3^{2n -1}$ is divisible by 8, for all natural numbers n.

Answer:

p(n)= $3^{2n -1}$ is divisible by 8.......(given)

Now, we’ll substitute different values for n,

$p(0) = 3^0 -1 = 0$ , is divisible by 8

$P(1) = 3^2 -1 = 8$ , is divisible by 8

$P(2) = 3^4 -1 = 80$ , is divisible by 8

Now, let us consider,

$P(k) = 3^{2k} -1$ be divisible by 8

Thus, $P(k+1) = 3^{2(k+1)} -1 = 72x + 8$ , is divisible by 8

Hence, by mathematical induction,
For each natural no. n it is true that, $P(n) = 3^{2n} -1$ is divisible by 8.

Question:7

For any natural number n, $7^n -2^n$ is divisible by 5.

Answer:

P(n)= $7^n -2^n$ is divisible by 5. ...............(given)

Now, we will substitute different values for n,

$P(0) = 7^0 -2^0 = 0$ , is divisible by 5

$P(1) = 7^1 -2^1 = 5$ , is divisible by 5

$P(2) = 7^2 -2^2 = 45$ , is divisible by 5

$P(3) = 7^3 -2^3 = 335$ , is divisible by 5

Now, let us consider that,

$P(k) = 7^k -2^k$ be divisible by 5.

Thus, $7^k -2^k=5x$

We also get,

$P(k+1) = 7^{k+1} -2^{k+1} = (5+2)7^k -2(2)^k$

$= 5(7)^k + 2(5x)$ , is divisible by 5

Thus, P(k+1) is true

Hence, by mathematical induction,
For each natural no. n it is true that, $P(n) = 7^n -2^n$ is divisible by 5.

Question:8

For any natural number n, $x^n -y^n$ is divisible by x -y, where x integers with $x \neq y$ .

Answer:

P(n)= $x^n -y^n$ is divisible by x -y, where x integers with $x \neq y$ . ........(given)
Now, we’ll substitute different values for n,

$P(0) = x^0 -y^0 = 0$ , is divisible by x-y

$P(1) = x -y,$ is divisible by x-y

P(2) = x^2 -y^2 = (x+y)(x-y) , is divisible by x-y

Now, let us consider,

P(k) = x^k -y^k , is divisible by x-y

Thus, x^k -y^k= a(x-y)

We also get that

$P(k+1) = x^{k+1} -y^{k+1}$

= x^k(x-y) + y(x^k-y^k)

= x^k(x-y) + y a(x-y) , is divisible by x-y

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) = x^n - y^n is divisible by x-y, x integers with x≠y

Question:9

$n^3 -n$ is divisible by 6, for each natural number n. $n \geq2$

Answer:

P(n)= $n^3 -n$ is divisible by 6

Now, we’ll substitute different values for n,

$P(2) = 2^3 -2 = 6$ , is divisible by 6

Now, let us consider,

$P(k) = k^3 -k$ be divisible by 6

Thus, $k^3 -k = 6x$

We also get that,

$P(k+1) =(k+1)^3 -(k+1)$

$= (k+1)(k^2 + 2k + 1 -1)$

$= k^3 + 3k^2 + 2k$

$= 6x + 3k(k+1)$

$= 6x + 3 \times 2y$ , $\because k(k+1) \text{ is even}$

$= 6x + 3 \times 2y$ is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)= $n^3 -n$ is divisible by 6.

Question:10

$n(n^2 + 5)$ is divisible by 6, for each natural number n.

Answer:

P(n)= $n(n^2 + 5)$ is divisible by 6, for each natural number n.

Now, we’ll substitute different values for n,

$P(1) = 1(1^2 + 5) = 6$ , is divisible by 6

$P(2) = 2(2^{}2 + 5) = 18$ , is divisible by 6

Let us consider,

$P(k) = k(k^2 + 5)$ be divisible by 6

Thus, $k(k^2 + 5)=6x$

We also get that,

$P(k+1) = (k+1)[(k+1)^{2} + 5]$

$= (k+1)(k^2 + 2k + 6)$

$= k(k^2 + 5 ) + k(2k+1) + k^2+2k+ 6$

$= 6x + 3k^2 + 3k + 6$ $\because (k^2 + k ) \text{ is even }$

$= 6x + 3\times 2y + 6$ , is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)= $n(n^2 + 5)$ is divisible by 6.

Question:11

$n^2 < 2^n$ for all natural numbers $n \geq 5$ .

Answer:

P(n)= $n^2 < 2^n$ for $n \geq 5$ .....(given)

Let us consider

$P(k) = k^2< 2^k$ to be true,

Thus, $P(k+1) = (k+1)^2$

$= k^2 + 2k + 1$

$2^{k+1} = 2(2^k) > 2k^2$

Now, since, $n^2> 2n + 1$ for $n\geq 3$

$k^2 + 2k + 1 < 2k^2$

Thus, $(k+1)^2< 2^{k+1}$

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n²< 2ⁿ it is true that, P(n)= $n^2 < 2^n$ for $n \geq 5$ ..

Question:12

$2n < (n + 2)!$ for all natural number n.

Answer:

$2n < (n + 2)!$ .....(given)

Now, we will substitute different values for n,

$P(0) \rightarrow 0 < 2!$

$P(1) \rightarrow 2 < 3!$

$P(2) \rightarrow 4 < 4!$

$P(3) \rightarrow 6 < 5!$

Now, let us consider,

$P(k) = 2k < (k+2)!$ To be true,

Thus, $P(k+1) = 2(k+1)< [(k+1) + 2]!$

Now, we know that, $[(k+1) + 2]! = (k+3)! = [(k+3) (k+2)(k+1) ... 3 \times 2\times 1]$

But also, $= 2(k+1) \times (k +3)(k+2) ... 3 \times 1 > 2(k+1)$

Thus, $2(k+1) < [(k+1) + 2]!$

Thus, P(k+1) is true.

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) = $2n < (n + 2)!$

Question:13

$\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$ .

Answer:

P(n) is $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}$ ; $n \geq 2$ .
$\mathrm{P}(2) is \sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.414<1.7 .0 \ldots thus, it is true\\ \mathrm{P}(3)\ is \sqrt{3}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}=1.732<2.284,\ thus \ it is\ true\\ Now, let us consider P(k)=\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}},\ is \ true\\ Now, we will add \sqrt{k+1}\ on\ both \ sides,$

Thus,
$\begin{array}{l} \sqrt{k}+\sqrt{k+1}-\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}}+\sqrt{k+1}-\sqrt{k} \\ \text { Thus, } \sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \end{array}$

Hence, by mathematical induction,

For each natural no. $n \geq 2$ ,P(n) is $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}$

Question:14

$2 + 4 + 6 +...+ 2n = n^2 + n$ for all natural numbers n.

Answer:P(n)= $2 + 4 + 6 +...+ 2n = n^2 + n$

Now, we will substitute different values for n,

$P(0) = 0 = 0^2 + 0$ , is true

$P(1) = 2 = 1^2 + 1$ , is true

$P(2) = 2+4 = 2^2 + 2$ , is true

Now, let us consider,

$P(k) = 2 + 4 + 6 + .... + 2k = k^2 + k$ to be true,

Thus,

P(k+1) is $2 + 4 + 6 + .... + 2k + 2(k+1) = k^2 + k + 2k + 2$

$= (k^2 + 2k + 1) + (k+1)$

$= (k + 1)^2 + (k + 1)$

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)= $2 + 4 + 6 +...+ 2n = n^2 + n$

Question:15

$1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$ for all natural numbers n.

Answer:

P(n) is $1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$ ..................given

Now, we’ll substitute different values for n,

$P(0) = 1 = 2^{0+1} -1$ , is true

$P(1) = 1 + 2 = 3 = 2^{1+1} -1$ , is true

$P(2) = 1 + 2+ 2^2 = 7 = 2^{2+1} -1$ , is true

Now, let us consider,

$P(k) = 1 + 2 + 2^2 + .... 2^k = 2^{k+1} -1$ to be true

Thus,

P(k+1) is $1+2+2^2+....+2^k + 2^{k+1 }= 2^{k+1} -1 + 2^{k+1}$

$= 2 \times 2^{k+1 }-1$

$= 2^{k+1+1 }-1$

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) is $1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$ .

Question:16

$1 + 5 + 9 + ...+(4n -3) = n (2n -1)$ for all natural number n.

Answer:

P(n)= $1 + 5 + 9 + ...+(4n -3) = n (2n -1)$

Now, we’ll substitute different values for n,

P(1) = 1

= 1(2-1), is true

P(2) = 1+5 = 6

= 2(4-1), is true

Now, let us consider,

$P(k) = 1+5+9+.....+[4k-3] = k(2k-1)$
$P(k+1) = 1+5+9+.....+[4(k+1)-3] = k(2k-1) + 4(k+1)-3$

$= k(2k -1) + 4k + 1\\ (k+1)[2(k+1)-1]$

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)= $1 + 5 + 9 + ...+(4n -3) = n (2n -1)$

Question:17

A sequence $a_1, a_2, a_3 ...$ is defined by letting $a_1 = 3 \ and \ a_k = 7a_{k-1}$ for all natural numbers k ≥ 2. Show that $a_n = 3.7^{n-1 }$ for all natural numbers.

Answer:

Given
$\begin{aligned} &a_{1}=3 \& a_{k}=7 a_{k-1}\\ &\text { Now, we'll substitute different values for k }\\ &\begin{aligned} a_{2} &=7 \times a_{1} \\ &=7 \times 3=21 \\ &=3.7^{2-1} \\ a_{3} &=7 \times a_{2} \\ &=7^{2} \times a_{1}=49 \times 3 \\ &=147 \end{aligned} \end{aligned}$
$=3.7^{3-1}$
$\begin{aligned} a_{n} &=7 a_{n-1} \\ &=3.7^{n-1} \end{aligned} \\$
$\\ \text { Thus, } a_{n+1}=7 a_{n+1-1} \\ =7 a_{n} \\ =7 \times 3.7^{n-1} \\ =3.7^{(m+1)-1}$

Thus, a_n+1 is true if a_n is true

Hence, by mathematical induction,

For each natural no. n it is true that, $a_n = 3.7^{n-1 }$

Question:18

A sequence b0, b1, b2 ... is defined by letting $b_0=5$ and $b_k = 4 + b_{k -1}$ for all natural numbers k. Show that for all natural number n $b_n = 5 + 4n$ using mathematical induction.

Answer:

Given:

The sequence b₀, b₁, b₂,... if defined if we let, $b_0 = 5$ & for all natural numbers k. $b_k = 4 +b_{k-1}$ Thus,

$b_1 = 4 + b_0\\ = 4 + 5\\ = 9 \\ =5 + 4.1$

$b_2 = 4 + b_1\\ = 4 + 9\\ = 13\\ = 5 + 4.2\\$

Now, let us consider, $b_m = 4 + b_{m-1} = 5 + 4m$ to be true.

Thus,

$b_{m+1} = 4 + b_m = 4 + 5+ 4m \\= 5 + 4(m+1)\\$

Thus, b_m+1 is true if b_m is true

Hence, by mathematical induction,

For each natural no. n it is true that, $b_n = 5 + 4n$

Question:19

A sequence $\mathrm{d}_{1}, \mathrm{d}_{2}, \mathrm{d}_{3} \ldots . .$ Is defined by letting

$\mathrm{d}_{1}=2 and \mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}} \mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !} numbers, k \geq 2.$ Show that $\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !}$ for all n ϵ N.

Answer:

A sequence $\mathrm{d}_{1}, \mathrm{d}_{2}, \mathrm{d}_{3} \ldots . .$ Is defined by letting
$\mathrm{d}_{1}=2 \ and \ \mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}}\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !} \ numbers, k \geq 2$
$\mathrm{d}_{k}=\mathrm{d}_{k-1} / \mathrm{k}$
Now, we'll substitute different values for k
$\begin{aligned} \mathrm{d}_{2}=& \mathrm{d}_{1} / 2 \\ &=2 / 2=2 / 2 ! \\ \mathrm{d}_{3}=\mathrm{d}_{2} / 3 & \\ &=1 / 3=2 / 3 ! \end{aligned}$
Now, let us consider,

$\mathrm{d}_{m}=\mathrm{d}_{m-1} / \mathrm{m}=2 / \mathrm{m} !$ to be true.

$Thus, \\ d_{m+1}=d_{m+1-1} / m+1\\ \begin{array}{l} =\mathrm{d}_{\mathrm{m}} / \mathrm{m}+1 \\ =2 /(\mathrm{m}+1) \cdot \mathrm{m} ! \\ =2 /(\mathrm{m}+1) ! \end{array}$
Thus, $d_{m+1 }$ is true if $d_{m }$ is true
Hence, by mathematical induction,
For each natural no. n it is true that, $\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !}$

Question:20

Prove that for all n $\epsilon$ N
$\begin{array}{c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ =\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} \end{array}$

Answer:

Given data:
$\begin{array}{c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ =\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} \end{array}$
Now, we’ll substitute different values for n,
n = 1,
$\begin{aligned} \cos (\alpha+(1-1) \beta) &=\cos \alpha \\ &=\frac{\cos \left(\alpha+\left(\frac{1-1}{2}\right) \beta\right) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \end{aligned}$
Thus, it’s true
Now, n = 2,
$\begin{aligned} \cos \alpha+\cos (\alpha+(2-1) \beta)=& \cos \alpha+\cos (\alpha+\beta) \\ =& \frac{\cos \left(\alpha+\left(\frac{2-1}{2}\right) \beta\right) \sin \left(\frac{2 \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \\ 2 \cos (\alpha+\beta / 2) \cos (\beta / 2) \end{aligned}$

Thus, it’s true.
Now, let n = k
$\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$ be true
$Thus, n=k+1\\ \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)+\cos (\alpha+(k+1-1) \beta)= \frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}+\cos (\alpha+k \beta) \\=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)+\cos (\alpha+k \beta) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

$=\frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)+\sin (\alpha+(2 k-1) \beta / 2) \sin \left(\alpha+(2 k-1) \frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}=\frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}\\=\frac{\cos \left(\alpha+\left(\frac{k}{2}\right) \beta\right) \sin \left(\frac{(k-1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,

$\begin{array}{c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ =\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} \end{array}$

Question:21

Prove that, $\cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$ , for all n $\epsilon$ N.

Answer:

Given:
$\cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$
Now, we’ll substitute different values for n,
$\\\begin{aligned} n=1 \\ \cos \theta= 2 \sin \theta \cos \theta / 2 \sin \theta \\ &=\sin 2^{1} \theta / 2^{1} \sin \theta \end{aligned} \\ \: \: Thus, \: \: it's\: \: true. \\Now, n=2$

$\begin{aligned} \cos \theta \cos 2 \theta=& \sin 2^{2} \theta / 2^{2} \sin \theta \\ &=2 \sin 2 \theta \cos 2 \theta / 4 \sin \theta \\ &=4 \sin \theta \cos \theta \cos 2 \theta / 4 \sin \theta \end{aligned} $$\\ Thus,\: \: it\: \: is\: \: true.$
$n=3\\ \begin{array}{l} \cos \theta \cos 2 \theta \cos 4 \theta=\sin 2^{3} \theta / 2^{3} \sin \theta \\ =2 \sin 4 \theta \cos 4 \theta / 8 \sin \theta \\ =4 \sin 2 \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta \\ =8 \sin \theta \cos \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta \end{array}$

Now, let us assume that,
$n=k,\\ \cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta=\sin 2^{k} \theta / 2^{k} \sin \theta\ to \ be \ true.\\ Thus, at \: \: n=k+1,\\ \begin{aligned} \cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta \cos 2^{k+1-1} \theta &=\sin 2^{k} \theta / 2^{k} \sin \theta \cdot \cos 2^{k} \theta \\ &=2 \sin 2^{k} \theta \cos 2^{k} \theta / 2^{k+1} \sin \theta \\ &=\sin 2^{k+1} \theta / 2^{k+1} \sin \theta \end{aligned}$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$\cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$

Question:22

Prove that, $\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$ for all $n \in N$

Answer:

Given:

P(n) : $\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$
Now, we’ll substitute different values for n=1,
$\\\sin \theta=[\sin (1 \theta / 2) \sin (1+1) \theta / 2] / \sin \theta / 2$
Thus, it is true.Now,
let us assume that, for n=k is true
$\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta=[\sin (k \theta / 2) \sin (k+1) \theta / 2] / \sin \theta / 2$
At n = k+1
$\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta+\sin (k+1) \theta\\$
$\\= \sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$
$\\ =\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$
Thus, n=k+1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$

Question:23

Show that $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{15}$ is a natural number for all n $\epsilon$ N.

Answer:

Given data:
P(n) = $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{15}$
Now, we’ll substitute different values for n,
$\begin{aligned} \text { At } n=1 & \\ P(1)=\frac{1^{5}}{5}+\frac{1^{8}}{3}+\frac{7(1)}{15} \\ &=\frac{1}{5}+\frac{1}{3}+\frac{7}{15} \end{aligned}$
$\\=(3+5+7 )/ 15\\ =1 \: \: is\: \: true.\\\\\ At \: \: $n=2$\\\\ $$ P(2)=\frac{2^{5}}{5} +\frac{2^{3}}{3}+\frac{7(2)}{15} \\\\ =\frac{32}{5}+\frac{8}{3}+\frac{14}{15} \\\\ =(96+40+14 )/ 15 =10 \: \: is\: \: true$
At n=3
$\\P(3)=\frac{3^{5}}{5} +\frac{3^{3}}{3}+\frac{7(3)}{15} \\\\ =\frac{243}{5}+\frac{27}{3}+\frac{21}{15} \\\\ =(729+135+21 )/ 15 \\\\=59$ is true.
Let us consider,
$P(k)=\frac{k^{5}}{5}+\frac{k^{2}}{3}+\frac{7(k)}{15} $$\: \: to\: \: be\: \: true.$
Thus, at n=k+1

$\\ P(k+1)=\frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7(k+1)}{15} \\\\ =\frac{k^{5}+5 k^{4}+10 k^{2}+10 k^{2}+5 k+1}{5}+\frac{k^{3}+3 k^{2}+3 k+1}{3}+\frac{7(k+1)}{15} \\\\ =\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7(k)}{15}+k^{4}+2 k^{3}+3 k^{2}+2 k+1,\ is\ a\ natural\ number$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
P(n) = $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{15}$

Question:24

Prove that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$ ,for all natural numbers n > 1.

Answer:

Given:
P(n) = $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$
Now, we’ll substitute different values for n,
$\begin{aligned} \text { At } n=2 & \\ P(2)=1 / 2 &+1 / 4 \\ &=3 / 4>13 / 24 \end{aligned} \\ It\: \: is\: \: true.\\ At \: \: $n=3$ \\ \begin{aligned} P(3)=1 / 2 &+1 / 4+1 / 6 \\ &=11 / 12>13 / 24 \end{aligned}\\$ .
It is true.
Now, let us consider,
$\\P(k)=\frac{1}{k+1}+\frac{1}{k+2}+\ldots \ldots+\frac{1}{2 k}>\frac{13}{24}$ To be true.\\ Now, at $n=k+1$ ,\\$P(k+1)=\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+\ldots . .+\frac{1}{2(k+1)}$

$\begin{array}{l} =\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k+2} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2 k+1}+\frac{1}{2 k+2}-\frac{1}{k+1} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{2k+2+2k+1-4k-2}{2(2 k+1)(k+1)} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2(2 k+1)(k+1)}>\frac{13}{24} \end{array}$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural no. n > 1 it is true that,
P(n) = $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$

Question:25

Prove that number of subsets of a set containing n distinct elements is $2^n$ , for all n $\epsilon$ N.

Answer:

To Prove:
$2^n$ is the no. of subsets of a set containing n distinct elements.
Now, there is only one subset since for a null set there is only one element φ
Now, we’ll substitute different values for n,
At n = 0,
No. of subsets = 1 = $2^0$
If we consider a set that has 1 element then there will be 2 subsets that has 1 & φ
Thus, at n = 1, it is true
now at n=2
No. of subsets = 2 = $2^2$
Now, let us consider that there is a set $S_k$ has k elements.
At n = k,
No. of subsets = $2^k$ is true.
Now, for set $S_{k+1}$ , having k+1 elements,
Compared to the set $S_{k}$ , $S_{k+1}$ has an extra element and thus will form an extra collection of subsets when combined with the existing $2^k$ subsets & an extra $2^k$ subset will be formed.
Thus, at n = k+1,
No. of subsets = $2^k$ + $2^k$
= 2. $2^k$
= $2^{k+1}$
Thus, by mathematical Induction,
$2^n$ is the no. of subsets of a set containing n distinct elements.

Question:26

If $10^n + 3.4^{n+2} + k$ is divisible by 9 for all n ? N, then the least positive integral value of k is
A. 5
B. 3
C. 7
D. 1

Answer:

The answer is the option (A) 5

Given:

P(n) = $10^n + 3.4^{n+2} + k$ is divisible by 9

Now, we’ll substitute different values for n,

At n = 1,

$P(1) = 10^1 + 3.4^3+ k$

$= 202 + k$

Now, we know that sum of the digits of this no. should be 9 or multiples of 9, for it to be divisible by 9.

Thus $2 + 2 + k = 9$

Thus, $k = 5.$

Question:27

For all n ? N, $3.5^{2n + 1} + 2^{3n + 1}$ is divisible by
A. 19
B. 17
C. 23
D. 25

Answer:

The answer is the option (B) 17
Given:
P(n) = $3.5^{2n + 1} + 2^{3n + 1}$
Now, we’ll substitute different values for n,
At n = 1,
P(1) = $3.5^{2+1} + 2^{3+1}$
$\\ = 375 + 16\\ = 391 \\=17\times 23$
At n = 2,
$P(2) = 3.5^{4+1} + 2^{6+1}$
$\\= 9375 + 256 \\ = 9503 \\ = 17 \times 559$
Therefore, it is divisible by 17.

Question:28

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

Answer:

The answer is the option (A) 1
Given:
P(n) = $x^n � 1$ is divisible by $x - k$
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = x -1$
At n = 2,
$P(2) = x^2 -1$
$= (x -1) (x + 1)$
At n = 3,
$P(3) = x^3 -1$
$= (x -1) (x^2 + xy + 1)$
At n = 4,
$\\P(4) = x^4 -1\\ = (x^2 -1) (x^2 + 1) \\ = (x -1) (x + 1) (x^2 + 1)$

Therefore, ‘1’ is the least possible integral value of k.

Question:29

Fill in the blanks in the following:

If $P(n) : 2n < n!, n \in N$ , then P(n) is true for all $n\geq$ __________.

Answer:

$n\geq4$
Given:
$P(n) : 2n < n!, n \in N$
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = 2 \times 1 < 1!$ ,
$= 2 < 1$ , viz. not true.
At n = 2,
$P(2) = 2 \times 2 < 2!$
$= 4 < 2$ , viz. not true.
At n = 3,
$P(3) = 2\times 3 < 3!$
$= 6 < 6$ , viz. not true
At n = 4,
$P(4) = 2 \times 4 < 4!$
$= 8 < 24$ , viz. not true
At n = 5,
$P(5) = 2 \times 5 < 5!$
$= 10 < 120$ , viz. true
Therefore, $n\geq4$

Question:30

True or False
Let P(n) be a statement and let $P(k) \Rightarrow P(k + 1)$ , for some natural number k, then P(n) is true of all n ? N.

Answer:

True.

Explanation: It is the Principle of Mathematical Induction.

Important Notes from NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

Deductive reasoning has helped in the field of mathematics to a great extent in understanding and deriving various conclusions out of already proven statements. It helps in establishing a new relationship out of two different things relating to a single entity. For example- if A is the father of B and also A is the son of G then we can establish a relationship between B and G that B is the grandson of G or G is the grandfather of B.

As per NCERT Exemplar Class 11 Maths solutions chapter 4, the same principle is used in mathematics for checking the relevance of certain statements through the process of deductive reasoning. It means applying any established principle of a general case to a special case to check its applicability and relevance.

This whole process is known as induction and is used in mathematics for proving different statements by using any established universal principle such as proving that a statement is true for a natural number x, then will be true for the next number x+1 as well as for verifying if any statement is true for a natural number y+1 then it will also apply to y.

Class 11 Maths NCERT Exemplar solutions chapter 4 are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams. The step-wise explanation of the chapter makes it easy for the students to understand the concept and method easily.

*This chapter has been omitted according to the revised syllabus of CBSE for 2020-21.

Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 4

Introduction
Motivation
The Principle of Mathematical Induction

What will the students learn from NCERT Exemplar Class 11 Maths solutions chapter 4?

NCERT Exemplar Class 11 Maths chapter 4 solutions is straightforward yet a difficult one but will ultimately help students to develop skills regarding inductive and deductive reasoning used in almost every other field ranging from science to many other branches.
NCERT Exemplar solutions for Class 11 Maths chapter 4 will help students to develop their skills relating to mathematical inductions that have its footsteps in every branch of science and is also a part of various competitive exams.
It also develops skills about establishing relationships between different things relating to some common term and is also used in daily routine for understanding relationships and connections between different things.

NCERT Solutions for Class 11 Mathematics Chapters

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability

Important Topics to cover in NCERT Exemplar Class 11 Maths solutions chapter 4 Principle of Mathematical Induction

Although this chapter is omitted for the 2020-2021 examination, the process of inductive and deductive reasoning is important and so are the examples which you should pay attention to sharpen your mind and have more clarity on these types of questions.

Check Chapter-Wise NCERT Solutions of Book

Chapter-1	Sets
Chapter-2	Relations and Functions
Chapter-3	Trigonometric Functions
Chapter-4	Principle of Mathematical Induction
Chapter-5	Complex Numbers and Quadratic equations
Chapter-6	Linear Inequalities
Chapter-7	Permutation and Combinations
Chapter-8	Binomial Theorem
Chapter-9	Sequences and Series
Chapter-10	Straight Lines
Chapter-11	Conic Section
Chapter-12	Introduction to Three Dimensional Geometry
Chapter-13	Limits and Derivatives
Chapter-14	Mathematical Reasoning
Chapter-15	Statistics
Chapter-16	Probability

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Frequently Asked Questions (FAQs)

1. Who has prepared the solutions?

The NCERT Exemplar Class 11 Maths chapter 4 solutions are prepared by professional mathematicians on our team. The solutions are prepared after referring to various advanced-level books.

2. How many questions are there in NCERT Exemplar Class 11 Maths Solutions Chapter 4?

There are a total of 24 questions in one exercise. The questions are for problem-solving in this chapter.

3. How to prepare for maths exam?

You can practice maths by solving as many problems as you can. The more variety of questions you solve, the better prepared you are.

4. Are these solutions helpful for competitive examinations?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 4 are designed to develop your base and give you the best solutions to the competitive exam questions.

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NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

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NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

Important Notes from NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 4

NCERT Solutions for Class 11 Mathematics Chapters

NCERT Exemplar Class 11 Solutions

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