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

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

Edited By Ramraj Saini | Updated on Sep 21, 2023 11:06 PM IST

Principle of Mathematical Induction Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction are provided here. These NCERT solutions are created according to latest syllabus and pattern of CBSE 2023-24. The faculty at Careers360 provides clear and comprehensive principle of mathematical induction class 11 solutions, enabling students to understand the concept and its application in detail. Through this chapter, students can learn about the Principle of Mathematical Induction and its practical uses. By practicing all the problems included in the NCERT solutions, students can achieve maximum marks in their exams with ease.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

This Story also Contains
  1. Principle of Mathematical Induction Class 11 Questions And Answers
  2. Principle of Mathematical Induction Class 11 Questions And Answers PDF Free Download
  3. Principle of Mathematical Induction Class 11 Solutions - Important Formulae And Points
  4. Principle of Mathematical Induction Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Key Features Of Principle Of Mathematical Induction Class 11 NCERT Solutions
  6. NCERT Solutions for Class 11 Mathematics - Chapter Wise
  7. Summary Of Class 11 Principle Of Mathematical Induction NCERT Solutions
  8. NCERT Solutions for Class 11- Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction

The principle of mathematical induction class 11 is a technique that is used to prove mathematically accepted statements in algebra and other mathematical applications, employing both inductive and deductive reasoning. The NCERT Solutions provided by Careers360 cover all the concepts related to this principle, which can help students score full marks in this chapter. These solutions are also beneficial for those preparing for competitive exams and for further studies. With their accuracy and comprehensiveness, NCERT solutions for class 11 make it easier for students to achieve good marks and succeed in their exams.

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Principle of Mathematical Induction Class 11 Questions And Answers PDF Free Download

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Principle of Mathematical Induction Class 11 Solutions - Important Formulae And Points

Important Terms:

Statement: A sentence is called a statement if it is either true or false.

Motivation: Motivation is tending to initiate an action. Here basis step motivates us for mathematical induction.

Principle of Mathematical Induction: The principle of mathematical induction is a tool used to prove various mathematical statements. Each statement, denoted as P(n) for a positive integer n, is examined for correctness when n = 1. Then, assuming the truth of P(k) for some positive integer k, the truth of P(k + 1) is established.

principle of mathematical induction class 11 questions and answers

Step 1: Show that the given statement is true for n = 1.

Step 2: Assume that the statement is true for n = k.

Step 3: Using the assumption made in Step 2, show that the statement is true for n = k + 1. We have proved the statement is true for n = k. According to Step 3, it is also true for k + 1 (i.e., 1 + 1 = 2). By repeating the above logic, it is true for every natural number.

Free download NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction for CBSE Exam.

Principle of Mathematical Induction Class 11 NCERT Solutions (Intext Questions and Exercise)

NCERT class 11 maths chapter 4 question answer - Exercise: 4.1

Question:1 Prove the following by using the principle of mathematical induction for all n\in N : 1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}

Answer:

Let the given statement be p(n) i.e.
p(n):1+3+3^2+...+3^{n-1}=\frac{(3^n-1)}{2}
For n = 1 we have
p(1): 1=\frac{(3^1-1)}{2}=\frac{3-1}{2}= \frac{2}{2}=1 , which is true

For n = k we have
p(k):1+3+3^2+...+3^{k-1}=\frac{(3^k-1)}{2} \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1+3+3^2+...+3^{k+1-1}= 1+3+3^2+...+3^{k-1}+3^{k}
= (1+3+3^2+...+3^{k-1})+3^{k}
= \frac{(3^k-1)}{2}+3^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Using \ (i))
= \frac{3^k-1+2.3^k}{2}
= \frac{3^k(1+2)-1}{2}
= \frac{3.3^k-1}{2}
= \frac{3^{k+1}-1}{2}
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:2 Prove the following by using the principle of mathematical induction for all n\in N : 1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2

Answer:

Let the given statement be p(n) i.e.
p(n):1^3+2^3+3^3+...+n^3=\left (\frac{n(n+1)}{2} \right )^2
For n = 1 we have
p(1):1=\left (\frac{1(1+1)}{2} \right )^2= \left ( \frac{1(2)}{2} \right )^2=(1)^2=1 , which is true

For n = k we have
p(k):1^3+2^3+3^3+...+k^3=\left (\frac{k(k+1)}{2} \right )^2 \ \ \ \ \ \ \ \ \ \ - (i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1^3+2^3+3^3+...+(k+1)^3=1^3+2^3+3^3+...+k^3+(k+1)^3
=(1^3+2^3+3^3+...+k^3)+(k+1)^3
=\left ( \frac{k(k+1)}{2} \right )^2+(k+1)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k^2(k+1)^2+4(k+1)^3}{4}
=\frac{(k+1)^2(k^2+4(k+1))}{4}
=\frac{(k+1)^2(k^2+4k+4)}{4}
=\frac{(k+1)^2(k+2)^2}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because a^2+2ab+b^2=(a+b)^2)
=\left ( \frac{(k+1)(k+2)}{2} \right )^2
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:3 Prove the following by using the principle of mathematical induction for all n\in N :

1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}

Answer:

Let the given statement be p(n) i.e.
p(n):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+n)}=\frac{2n}{(n+1)}
For n = 1 we have
p(1):1=\left (\frac{2(1)}{1+1} \right )= \left ( \frac{2}{2} \right )=1 , which is true

For n = k we have
p(k):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)}=\frac{2k}{(k+1)} \ \ \ \ -(i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k+1)} =\left ( 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...+k)} \right )+\frac{1}{(1+2+3+...+k+k+1)}

=\frac{2k}{k+1}+\frac{1}{(1+2+3+...+k+(k+1))} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+1+1)}{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because1+2+....+n = \frac{n(n+1)}{2} )
=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}
=\frac{2}{k+1}\left (k+\frac{1}{k+2} \right )
=\frac{2}{k+1}\left ( \frac{k^2+2k+1}{k+2} \right )
=\frac{2}{k+1}.\frac{(k+1)^2}{k+2}
=\frac{2(k+1)}{k+2}
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:4 Prove the following by using the principle of mathematical induction for all n\in N : 1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}

Answer:

Let the given statement be p(n) i.e.
p(n):1.2.3+2.3.4+...+n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}
For n = 1 we have
p(1):6=\left (\frac{1(1+1)(1+2)(1+3)}{4} \right )= \left ( \frac{1.2.3.4}{4} \right )=6 , which is true

For n = k we have
p(k):1.2.3+2.3.4+...+k(k+1)(k+2)=\frac{k(k+1)(k+2)(k+3)}{4} \ \ \ \ \ \ -(i) , Let's assume that this statement is true

Now,
For n = k + 1 we have
p(k+1):1.2.3+2.3.4+...+k(k+1)(k+2) + (k+1)(k+2)(k+3) =(1.2.3+2.3.4+...+k(k+1)(k+2)) + (k+1)(k+2)(k+3)

=\frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3) \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3) }{4}
=\frac{(k+1)(k+2)(k+3)(k+4) }{4}
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:5 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.3+2.3^2+3.3^3+...+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}

Answer:

Let the given statement be p(n) i.e.
p(n):1.3+2.3^2+3.3^3+...+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}
For n = 1 we have
p(1):3=\frac{(2(1)-1)3^{1+1}+3}{4}= \frac{(2-1)9+3}{4}=\frac{12}{4}=3 , which is true

For n = k we have

p(k):1.3+2.3^2+3.3^3+...+k.3^k=\frac{(2k-1)3^{k+1}+3}{4} \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):1.3+2.3^2+3.3^3+...+(k+1).3^{(k+1)}=1.3+2.3^2+3.3^3+...+k.3^k+(k+1).3^{(k+1)}

=\frac{(2k-1)3^{k+1}+3}{4}+(k+1).3^{(k+1)}
=\frac{(2k-1)3^{k+1}+3+4(k+1).3^{(k+1)}}{4}
=\frac{3^{k+1}((2k-1)+4(k+1))+3}{4}
=\frac{3^{k+1}(6k+3)+3}{4}
=\frac{3^{k+1}.3(2k+1)+3}{4}
=\frac{(2k+1)3^{k+2}+3}{4}
=\frac{(2(k+1)-1)3^{(k+1)+1}+3}{4}
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:6 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.2+2.3+3.4+...+n.(n+1)=\left [\frac{n(n+1)(n+2)}{3} \right ]

Answer:

Let the given statement be p(n) i.e.
p(n):1.2+2.3+3.4+...+n.(n+1)=\left [\frac{n(n+1)(n+2)}{3} \right ]
For n = 1 we have
p(1):2=\left [\frac{1(1+1)(1+2)}{3} \right ]= \frac{1.2.3}{3}=2 , which is true

For n = k we have

p(k):1.2+2.3+3.4+...+k.(k+1)=\left [\frac{k(k+1)(k+2)}{3} \right ] \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):1.2+2.3+3.4+...+(k+1).(k+2)=1.2+2.3+3.4+...+k(k+1)+(k+1).(k+2)

=\frac{k(k+1)(k+2)}{3}+(k+1).(k+2) \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(k+1)(k+2)+3(k+1).(k+2)}{3}
=\frac{(k+1)(k+2)(k+3)}{3}
=\frac{(k+1)(k+1+1)(k+1+2)}{3}

Thus, p(k+1) is true whenever p(k) is true
Hence, by principle of mathematical induction , statement p(n) is true for all natural numbers n

Question:7 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}

Answer:

Let the given statement be p(n) i.e.
p(n):1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}
For n = 1 we have
p(1):1.3=3=\frac{1(4(1)^2+6(1)-1)}{3}= \frac{4+6-1}{3}=\frac{9}{3}=3 , which is true

For n = k we have

p(k):1.3+3.5+5.7+...+(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3} \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true
Now,
For n = k + 1 we have
p(k+1):1.3+3.5+5.7+...+(2(k+1)-1)(2(k+1)+1)=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)

=\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3) \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(4k^2+6k-1)+3(2k+1)(2k+3)}{3}
=\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}
=\frac{(4k^3+6k^2-k+12k^2+28k+9)}{3}
=\frac{(4k^3+18k^2+23k+9)}{3}
=\frac{(4k^3+14k^2+9k+4k^2+14k+9)}{3}
=\frac{(k(4k^2+14k+9)+4k^2+14k+9)}{3}
=\frac{(4k^2+14k+9)(k+1)}{3}
=\frac{(k+1)(4k^2+8k+4+6k+6-1)}{3}
=\frac{(k+1)(4(k^2+2k+1)+6(k+1)-1)}{3}
=\frac{(k+1)(4(k+1)^2+6(k+1)-1)}{3}

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:8 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1.2+2.2^2+3.2^3+....+n.2^n=(n-1)n^{n+1}+2

Answer:

Let the given statement be p(n) i.e.
p(n):1.2+2.2^2+3.2^3+...+n.2^n=(n-1)2^{n+1}+2
For n = 1 we havep(k):1.2+2.2^2+3.2^3+...+k.2^k=(k-1)2^{k^+1}+2 \ \ \ \ \ \ -(i)p(1):1.2=2=(1-1)2^{1^+1}+2 = 2 , which is true

For n = k we have

p(k):1.2+2.2^2+3.2^3+...+k.2^k=(k-1)2^k^+^1+2 \ \ \ \ \ \ -(i)

, Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):1.2+2.2^2+3.2^3+...+(k+1).2^{k+1} =1.2+2.2^2+3.2^3+...+k.2^k+(k+1).2^{k+1}

=(k-1)2^{k+1}+2+(k+1).2^{k+1} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=2^{k+1}(k-1+k+1)+2
=2^{k+1}(2k)+2
=k.2^{k+2}+2
=(k+1-1).2^{k+1+1}+2

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:9 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}=1-\frac{1}{2^n}
For n = 1 we have
p(1):\frac{1}{2}=1-\frac{1}{2^1}= 1-\frac{1}{2} = \frac{1}{2} , which is true

For n = k we have

p(k):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}=1-\frac{1}{2^k} \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{k+1}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^k}+\frac{1}{2^{k+1}}

=1-\frac{1}{2^k}+\frac{1}{2^{k+1}} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=1-\frac{1}{2^k}\left (1-\frac{1}{2} \right )
=1-\frac{1}{2^k}\left (\frac{1}{2} \right )
=1-\frac{1}{2^{k+1}}

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:10 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}
For n = 1 we have
p(1):\frac{1}{2.5}= \frac{1}{10}=\frac{1}{(6(1)+4)}= \frac{1}{10} , which is true

For n = k we have

p(k):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)} \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3(k+1)-1)(3(k+1)+2)}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)(3(k+1)+2)}

=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{1}{3k+2}\left ( \frac{k}{2}+\frac{1}{3k+5} \right )
=\frac{1}{3k+2}\left ( \frac{k(3k+5)+2}{2(3k+5)} \right )
=\frac{1}{3k+2}\left ( \frac{3k^2+5k+2}{2(3k+5)} \right )
=\frac{1}{3k+2}\left ( \frac{3k^2+3k+2k+2}{2(3k+5)} \right )
=\frac{1}{3k+2}\left ( \frac{(3k+2)(k+1)}{2(3k+5)} \right )
=\frac{(k+1)}{6k+10}
=\frac{(k+1)}{6(k+1)+4}

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:11 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}
For n = 1 we have
p(1):\frac{1}{1.2.3}=\frac{1}{6}=\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4.2.3}=\frac{1}{6} , which is true

For n = k we have

p(k):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)} \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(k+1)(k+2)(k+3)}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}

=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \ \ \ \ \ \ (using \ (i))
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)}{4}+ \frac{1}{k+3} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k+3)^2+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+9+6k)+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+9k+6k^2+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k^3+2k^2+k+8k+4k^2+4}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)} \right )
=\frac{1}{(k+1)(k+2)}\left ( \frac{(k+1)^2(k+4)}{4(k+3)} \right )
= \frac{(k+1)((k+1)+3)}{4(k+1+1)(k+1+2)}

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:12 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}

Answer:

Let the given statement be p(n) i.e.
p(n):a+ar+ar^2+...+ar^{n-1}=\frac{a(r^n-1)}{r-1}
For n = 1 we have
p(1):a=\frac{a(r^1-1)}{r-1}=\frac{r-1}{r-1}=1 , which is true

For n = k we have

p(k):a+ar+ar^2+...+ar^{k-1}=\frac{a(r^k-1)}{r-1} \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):a+ar+ar^2+...+ar^{k}=a+ar+ar^2+...+ar^{k-1}+ar^{k}

=a.\frac{r^k-1}{r-1}+ar^{k} \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{a(r^k-1)+(r-1)ar^{k}}{r-1}
=\frac{ar^k(1+r-1)-a}{r-1}
=\frac{ar^k.r-a}{r-1}
=\frac{a(r^{k+1}-1)}{r-1}


Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:13 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2

Answer:

Let the given statement be p(n) i.e.
p(n):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2n+1)}{n^2} \right )=(n+1)^2
For n = 1 we have
p(1):\left ( 1+\frac{3}{1} \right )= 4=(1+1)^2=2^2=4 , which is true

For n = k we have

p(k):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2k+1)}{k^2} \right )=(k+1)^2 \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right ) =\left ( 1+\frac{3}{1} \right )\left ( 1+\frac{5}{4} \right )\left ( 1+\frac{7}{9} \right )..\left ( 1+\frac{2k+1}{k^2} \right )\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right )

=(k+1)^2\left ( 1+\frac{(2(k+1)+1)}{(k+1)^2} \right ) \ \ \ \ \ \ \ \ \ \ (using \ (i))
=(k+1)^2\left ( \frac{{}(k+1)^2+(2(k+1)+1)}{(k+1)^2} \right )
=(k^2+1+2k+2k+2+1)
=(k^2+4k+4)
=(k+2)^2
=(k+1+1)^2


Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:14 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : \left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{n} \right )=(n+1)

Answer:

Let the given statement be p(n) i.e.
p(n):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{n} \right )=(n+1)
For n = 1 we have
p(1):\left ( 1+\frac{1}{1} \right )=2=(1+1)=2 , which is true

For n = k we have

p(k):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k} \right )=(k+1) \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k+1} \right ) &nbsnbsp; =\left ( 1+\frac{1}{1} \right )\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )...\left ( 1+\frac{1}{k} \right )\left ( 1+\frac{1}{k+1} \right )

=(k+1)\left ( 1+\frac{1}{k+1} \right ) \ \ \ \ \ \ \ (using \ (i))
=(k+1)\left ( \frac{k+1+1}{k+1} \right )
=(k+2)
=(k+1+1)
Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:15 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}

Answer:

Let the given statement be p(n) i.e.
p(n):1^2+3^2+5^2+...+(2n-1)^2=\frac{n(2n-1)(2n+1)}{3}
For n = 1 we have
p(1):1^2=1=\frac{1(2(1)-1)(2(1)+1)}{3}= \frac{1.1.3}{3}=1 , which is true

For n = k we have

p(k):1^2+3^2+5^2+...+(2k-1)^2=\frac{k(2k-1)(2k+1)}{3} \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):1^2+3^2+5^2+...+(2(k+1)-1)^2=1^2+3^2+5^2+...+(2k-1)^2+(2(k+1)-1)^2

=\frac{k(2k-1)(2k+1)}{3}+(2(k+1)-1)^2 \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{k(2k-1)(2k+1)+3(2(k+1)-1)^2}{3}
=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}
=\frac{(2k+1)(k(2k-1)+3(2k+1))}{3}
=\frac{(2k+1)(2k^2-k+6k+3)}{3}
=\frac{(2k+1)(2k^2+5k+3)}{3}
=\frac{(2k+1)(2k^2+2k+3k+3)}{3}
=\frac{(2k+1)(2k+3)(k+1)}{3}
=\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}


Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:16 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}
For n = 1 we have
p(1):\frac{1}{1.4}=\frac{1}{4}=\frac{1}{(3(1)+1)}=\frac{1}{3+1}=\frac{1}{4} , which is true

For n = k we have

p(k):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{k}{(3k+1)} \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3(k+1)-2)(3(k+1)+1)}=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{(3(k+1)-2)(3(k+1)+1)}

=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)} \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{1}{3k+1}\left ( k+\frac{1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{k(3k+4)+1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{3k^2+4k+1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{3k^2+3k+k+1}{3k+4} \right )
=\frac{1}{3k+1}\left ( \frac{(3k+1)(k+1)}{3k+4} \right )
= \frac{(k+1)}{3k+4}
= \frac{(k+1)}{3(k+1)+1}


Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:17 Prove the following by using the principle of mathematical induction for all n\in N : \frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}

Answer:

Let the given statement be p(n) i.e.
p(n):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}
For n = 1 we have
p(1):\frac{1}{3.5}= \frac{1}{15}=\frac{1}{3(2(1)+3)}=\frac{1}{3.5}=\frac{1}{15} , which is true

For n = k we have

p(k):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)} \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2(k+1)+1)(2(k+1)+3)}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{(2(k+1)+1)(2(k+1)+3)}

=\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)} \ \ \ \ \ \ \ \ \ \ (using \ (i))
=\frac{1}{2k+3}\left ( \frac{k}{3}+\frac{1}{2k+5} \right )
=\frac{1}{2k+3}\left ( \frac{k(2k+5)+3}{3(2k+5)} \right )
=\frac{1}{2k+3}\left ( \frac{2k^2+5k+3}{3(2k+5)} \right )
=\frac{1}{2k+3}\left ( \frac{2k^2+2k+3k+3}{3(2k+5)} \right )
=\frac{1}{2k+3}\left ( \frac{(2k+3)(k+1)}{3(2k+5)} \right )
= \frac{(k+1)}{3(2k+5)}
= \frac{(k+1)}{3(2(k+1)+3)}


Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:18 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 1+2+3+...+n<\frac{1}{8}(2n+1)^2.

Answer:

Let the given statement be p(n) i.e.
p(n):1+2+3+...+n<\frac{1}{8}(2n+1)^2.
For n = 1 we have
p(1):1<\frac{1}{8}(2(1)+1)^2= \frac{1}{8}(3)^2=\frac{9}{8} , which is true

For n = k we have

p(k):1+2+3+...+k<\frac{1}{8}(2k+1)^2 \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this statement is true


Now,
For n = k + 1 we have
p(k+1):1+2+3+...+k+1=1+2+3+...+k+k+1

< \frac{1}{8}\left ( 2k+1 \right )^2+(k+1) \ \ \ \ \ \ \ \ \ (using \ (i))
< \frac{1}{8}\left ( (2k+1)^2+8(k+1) \right )
< \frac{1}{8}\left ( 4k^2+4k+1+8k+8 \right )
< \frac{1}{8}\left ( 4k^2+12k+9\right )
< \frac{1}{8}\left ( 2k+3\right )^2
< \frac{1}{8}\left ( 2(k+1)+1\right )^2


Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Question:19 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : n(n+1)(n+5) is a multiple of 3.

Answer:

Let the given statement be p(n) i.e.
p(n):n(n+1)(n+5)
For n = 1 we have
p(1):1(1+1)(1+5)=1.2.6=12 , which is multiple of 3, hence true

For n = k we have

p(k):k(k+1)(k+5) \ \ \ \ \ \ \ -(i) , Let's assume that this is multiple of 3 = 3m


Now,
For n = k + 1 we have
p(k+1):(k+1)((k+1)+1)((k+1)+5)=(k+1)(k+2)((k+5)+1)

=(k+1)(k+2)(k+5)+(k+1)(k+2)
=k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)
=3m+2(k+1)(k+5)+(k+1)(k+2) \ \ \ \ \ \ \ \ (using \ (i))
=3m+(k+1)(2(k+5)+(k+2)) \ \ \ \ \ \ \ \ (using \ (i))
=3m+(k+1)(2k+10+k+2)
=3m+(k+1)(3k+12)
=3m+3(k+1)(k+4)
=3(m+(k+1)(k+4) )
=3l Where \left ( l=(m+(k+1)(k+4) ) \right ) some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is multiple of 3 for all natural numbers n

Question:20 Prove the following by using the principle of mathematical induction for all n\in N : 10^{2n-1}+1 is a divisible by 11.

Answer:

Let the given statement be p(n) i.e.
p(n):10^{2n-1}+1
For n = 1 we have
p(1):10^{2(1)-1}+1= 10^{2-1}+1=10^1+1=11 , which is divisible by 11, hence true

For n = k we have

p(k):10^{2k-1}+1 \ \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by 11 = 11m


Now,
For n = k + 1 we have
p(k+1):10^{2(k+1)-1}+1=10^{2k+2-1}+1
nbsp;
=10^{2k+1}+1
=10^2(10^{2k-1}+1-1)+1
=10^2(10^{2k-1}+1)-10^2+1
=10^2(11m)-100+1 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=100(11m)-99
=11(100m-9)
=11l Where l= (100m-9) some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 11 for all natural numbers n

Question:21 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : x^2^n-y^2^n is divisible by x+y.

Answer:

Let the given statement be p(n) i.e.
p(n):x^2^n-y^2^n
For n = 1 we have
p(1):x^{2(1)}-y^{2(1)}= x^2-y^2=(x-y)(x+y) , which is divisible by (x+y) , hence true (using \ a^2-b^2=(a+b)(a-b))

For n = k we have

p(k):x^{2k}-y^{2k} \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by (x+y) =(x+y)m


Now,
For n = k + 1 we have
p(k+1):x^{2(k+1)}-y^{2(k+1)} =x^{2k}.x^2-y^{2k}.y^2

=x^2(x^{2k}+y^{2k}-y^{2k})-y^{2k}.y^2
=x^2(x^{2k}-y^{2k})+x^2.y^{2k}-y^{2k}.y^2
=x^2(x+y)m+(x^2-y^2)y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=x^2(x+y)m+((x-y)(x+y))y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ a^2-b^2=(a+b)(a-b))
=(x+y)\left ( x^2.m+(x-y).y^{2k} \right )
=(x+y)l where l = (x^2.m+(x-y).y^{2k}) some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by (x+y) for all natural numbers n

Question:22 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : 3^{2n+2}-8n-9 is divisible by 8.

Answer:

Let the given statement be p(n) i.e.
p(n):3^{2n+2}-8n-9
For n = 1 we have
p(1):3^{2(1)+2}-8(1)-9= 3^4-8-9=81-17=64=8\times 8 , which is divisible by 8, hence true

For n = k we have

p(k):3^{2k+2}-8k-9 \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by 8 = 8m


Now,
For n = k + 1 we have
p(k+1):3^{2(k+1)+2}-8(k+1)-9=3^{2k+2+2}-8(k+1)-9
=3^{2k+2}.3^2-8k-8-9
=3^2(3^{2k+2}-8k-9+8k+9)-8k-17
=3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17
=9\times 8m+9(8k+9)-8k-17 \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=9\times 8m+72k+81-8k-17
=9\times 8m+80k-64
=9\times 8m+8(10k-8)
=8(9m+10k-8)
=8l where l= 9m+10k-8 some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 8 for all natural numbers n

Question:23 Prove the following by using the principle of mathematical induction for all n\in\mathbb{N} : 41^n-14^n is a multiple of 27.

Answer:

Let the given statement be p(n) i.e.
p(n):41^n-14^n
For n = 1 we have
p(1):41^1-14^1= 41-14= 27 , which is divisible by 27, hence true

For n = k we have

p(k):41^k-14^k \ \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is divisible by 27 = 27m


Now,
For n = k + 1 we have
p(k+1):41^{k+1}-14^{k+1}=41^{k}.41-14^{k}.14
=41(41^{k}-14^k+14^k)-14^{k}.14
=41(41^{k}-14^k)+14^k.41-14^{k}.14
=41(27m)+14^k(41-14) \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
=41(27m)+14^k.27
=27(41m+14^k)
=27l where l = 41m+14^k some natural number

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is divisible by 27 for all natural numbers n

Question:24 Prove the following by using the principle of mathematical induction for all n\in \mathbb{N} : (2n+7)<(n+3)^2

Answer:

Let the given statement be p(n) i.e.
p(n):(2n+7)<(n+3)^2
For n = 1 we have
p(1):(2(1)+7)<(1+3)^2\Rightarrow 9< 16 , which is true

For n = k we have

p(k):(2k+7)<(k+3)^2 \ \ \ \ \ \ \ \ \ \ \ -(i) , Let's assume that this is true


Now,
For n = k + 1 we have
p(k+1):(2(k+1)+7)=(2k+2+7)
<(k+3)^2+2 \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
<k^2+9+6k+2
<k^2+6k+11
Now , \ <k^2+6k+11<k^2+8k+16
<(k+4)^2
<((k+1)+3)^2

Thus, p(k+1) is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n) is true for all natural numbers n

Key Features Of Principle Of Mathematical Induction Class 11 NCERT Solutions

Concept Clarity: The maths chapter 4 class 11 solution provide a clear understanding of the concepts related to the Principle of Mathematical Induction. Students can understand the topic better and can solve the problems easily.

Easy to Understand: The solutions of ch 4 maths class 11 are written in a simple language, making them easy to understand for the students. They can easily comprehend the concepts and solve the problems based on them.

Comprehensive Coverage: The class 11 maths ch 4 question answer cover all the important topics and subtopics of the chapter, which helps the students to get a comprehensive understanding of the subject matter.

Interested students can find maths chapter 4 class 11 all exercise below.

NCERT Solutions for Exercise 4.1

NCERT Solutions for Class 11 Mathematics - Chapter Wise

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

By studying the class 11 maths chapter 4 ncert solutions, students gain an understanding of the process of proof by induction and the motivation behind using this method by considering natural numbers as the least inductive subset of real numbers. The NCERT Solutions for this chapter provide a clear explanation of the principle of mathematical induction and its applications, enabling students to score well in exams. The following concepts are summarized in the solutions of this chapter:

  • Deductive reasoning is an essential basis for mathematical thinking, while inductive reasoning involves generalizing from particular cases or facts.
  • The principle of mathematical induction is a powerful tool for proving various mathematical statements. Each statement is represented as P(n) for a positive integer n, and the truth of the statement is established for n=1. Assuming the truth of P(k) for some positive integer k, the truth of P(k+1) is then established.

NCERT Solutions for Class 11- Subject Wise

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NCERT solutions for class 11 maths
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NCERT solutions for Class 11 physics

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Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Principle of Mathematical Induction ?

The important topics covered in the class 11 chapter 4 maths NCERT Solutions are:

  1. Introduction to Deductive Reasoning
  2. Motivation for Mathematical Induction
  3. The Principle of Mathematical Induction
  4. Example Problems using Principle of Mathematical Induction
  5. Application of Mathematical Induction to Prove Mathematical Statements
  6. Proofs using Weak Principle of Mathematical Induction
  7. Summary of Concepts and Formulas Used in Mathematical Induction
2. How does the NCERT solutions are helpful ?

If students are stuck while solving the NCERT problems, they can take help form these NCERT solutions which are created by expert team of creers360 and provided in a detailed manner. practicing these solutions of chapter 4 class 11  students can get good hold on the concepts which lead to obtain good marks in the exam. For ease students can study mathematical induction class 11 pdf both online and offline mode.

3. What are the most difficult chapters in the class 11 maths ?

Permutation and combination, trigonometry are considered to be difficult chapters by the most of students but with the rigorous practice students can get command on them also.

4. explain the first principle of Mathematical Induction in Chapter 4 of NCERT Solutions for Class 11 Maths.

By verifying both the base step and inductive step to be true, we can establish the first principle of Mathematical Induction, which in turn proves that the given statement holds true for all natural numbers. Class 11 students can practice a multitude of problems based on this concept from the NCERT textbook. To assist them in securing good grades in their annual examination, solutions for the exercise-wise problems are also provided in PDF format.

5. Where can I find the complete solutions of NCERT for class 11 maths ?

Here you will get the detailed NCERT solutions for class 11 maths  by clicking on the link.

6. How many chapters are there in CBSE class 11 maths ?

There are 16 chapters starting from set to probability in CBSE class 11 maths.

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2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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
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