JEE Mathematics Principles Of Mathematical Induction MCQs Set 07

Answer: (a) Both P(3) and P(5) are true.

Explanation:
1. \( P(3)=9-3+41=47 \), which is prime.
2. \( P(5)=25-5+41=61 \), which is also prime.
3. Both statements are true. Answer: (a)

Teacher's Note:
1. Substitute the values and test for primality.
2. Check divisibility by primes up to the square root.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Expanding, \( (n+1)^7-n^7-1={}^7C_1n^6+{}^7C_2n^5+\cdots+{}^7C_6n \). Every coefficient \( {}^7C_k \) for \( 1\le k\le6 \) is a multiple of 7, so the Assertion is true.
2. The Reason (Fermat's little theorem for 7, provable by induction using the same binomial expansion) is true.
3. Moreover \( (n+1)^7-n^7-1=[(n+1)^7-(n+1)]-[n^7-n] \), which is a difference of two multiples of 7 by the Reason. So R explains A. Answer: (a)

Teacher's Note:
1. Expand using the binomial theorem.
2. Write the expression as a difference of two terms of the form \( m^7-m \).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Check \( n=2 \): \( 1+0.707=1.707\gt\sqrt2=1.414 \). Assume it for \( n=k \); then the sum up to \( k+1 \) exceeds \( \sqrt k+\frac1{\sqrt{k+1}}=\frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}\gt\frac{k+1}{\sqrt{k+1}}=\sqrt{k+1} \), using \( \sqrt{k(k+1)}\gt k \). So the Assertion is true.
2. \( \sqrt{n(n+1)}\lt\frac{n+(n+1)}2\lt n+1 \) (AM-GM), so the Reason is true.
3. The proof of the Assertion needs \( \sqrt{k(k+1)}\gt k \), not the Reason's inequality \( \sqrt{n(n+1)}\lt n+1 \). So both are true but R is not the correct explanation. Answer: (b)

Teacher's Note:
1. Use induction on \( n \).
2. Compare which inequality is used in the inductive step.

Answer: (d) \( A^n=nA-(n-1)I \)

Explanation:
1. \( A^2=\begin{bmatrix}1&0\\2&1\end{bmatrix} \), \( A^3=\begin{bmatrix}1&0\\3&1\end{bmatrix} \), so \( A^n=\begin{bmatrix}1&0\\n&1\end{bmatrix} \) (provable by induction).
2. \( nA-(n-1)I=\begin{bmatrix}n&0\\n&n\end{bmatrix}-\begin{bmatrix}n-1&0\\0&n-1\end{bmatrix}=\begin{bmatrix}1&0\\n&1\end{bmatrix}=A^n \). Answer: (d) The other options fail at \( n=2 \).

Teacher's Note:
1. Compute the first few powers and spot the pattern.
2. Test each option for \( n=2 \).

Answer: (b) \( S(k)\Rightarrow S(k+1) \)

Explanation:
1. For \( k=1 \): LHS \( =1 \) and RHS \( =4 \), so \( S(1) \) is not correct; hence (a) and (d) are false.
2. If \( S(k) \) is true, adding \( (2k+1) \) to both sides gives \( 1+3+\cdots+(2k+1)=3+k^2+2k+1=3+(k+1)^2 \), so \( S(k+1) \) is true.
3. So \( S(k)\Rightarrow S(k+1) \) holds. Answer: (b)

Teacher's Note:
1. Test the base case and the inductive step separately.
2. The formula fails at \( k=1 \), so induction cannot prove it.