Vidaara.orgClass 11 · Mathematics
CodeVID-M11-15-INEQ-01
Inequalities & Applications by Induction — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
$2^n > n$ holds for:
- A.$n \ge 1$
- B.$n \ge 5$
- C.$n \ge 3$
- D.no $n$
2.
Bernoulli's inequality is:
- A.$(1+x)^n \le 1+nx$
- B.$(1+x)^n \ge 1+nx$
- C.$(1+x)^n = 1+nx$
- D.$(1+x)^n \ge nx$
3.
$2^n > n^2$ first holds at:
- A.$n=2$
- B.$n=3$
- C.$n=4$
- D.$n=5$
4.
$n! > 2^{\,n-1}$ holds for:
- A.$n \ge 1$
- B.$n \ge 2$
- C.$n \ge 3$
- D.$n \ge 4$
5.
The number of subsets of an $n$-element set is:
- A.$n^2$
- B.$2^n$
- C.$n!$
- D.$2n$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Verify the base case of $2^n > n$.
7.
State Bernoulli's inequality with its condition on $x$.
8.
Give the base case for $2^n > n^2$.
9.
In the Bernoulli step, name the term that is dropped.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Show that if $2^k > k$ then $2^{k+1} > k+1$.
11.
Prove the step for $2^n \ge n+1$.
12.
Explain why $n! > 2^{\,n-1}$ cannot start at $n=1$.
13.
Expand $(1+x)(1+kx)$ and state the inequality used.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Prove by induction that $(1+x)^n \ge 1+nx$ for all $n\in\mathbb{N}$ and $x>-1$.
15.
Prove by induction that $n! > 2^{\,n-1}$ for all $n \ge 3$.
Answer Key
Section A — Multiple Choice Questions
- (A) $n \ge 1$
- (B) $(1+x)^n \ge 1+nx$
- (D) $n=5$
- (C) $n \ge 3$
- (B) $2^n$
Section B — Short Answer (2 marks)
- $n=1$: $2 > 1$; true.
- $(1+x)^n \ge 1+nx$ for $x > -1$, $n\in\mathbb{N}$.
- $P(5)$: $32 > 25$.
- $kx^2$ (it is $\ge 0$).
Section C — Short Answer (3 marks)
- $2^{k+1}=2\cdot2^k>2k\ge k+1$ since $k\ge1$.
- $2^{k+1}=2\cdot2^k\ge2(k+1)\ge(k+1)+1$.
- At $n=1,2$ the two sides are equal, so the strict inequality fails; it first holds at $n=3$.
- $1+(k+1)x+kx^2\ge1+(k+1)x$ since $kx^2\ge0$.
Section D — Long Answer (5 marks)
- Base $n=1$: equality; step multiplies by $1+x>0$ to get $(1+x)^{k+1}\ge1+(k+1)x+kx^2\ge1+(k+1)x$; by PMI it holds for all $n$.
- Base $n=3$: $6>4$; step $(k+1)!=(k+1)k!>(k+1)2^{k-1}\ge2\cdot2^{k-1}=2^k$; by PMI it holds for all $n\ge3$.
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