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JEE Advanced · Mathematics · 1. Basic Math

Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.
Column IColumn II
(A) The minimum value of \(\frac{x^2+2 x+4}{x+2}\) is(p) 0
(B) Let \(A\) and \(B\) be \(3 \times 3\) matrices of real numbers, where \(A\) is symmetric, \(B\) is skew-symmetric and \((A+B)(A-B) =\) \((A-B)(A+B)\). If \((A B)^t=(-1)^k\) \(A B\), where \((A B)^t\) is the transpose of the matrix \(A B\), then possible values of \(k\) are(q) 1
(C) Let \(a=\log _3 \log _3 2\). An integer \(k\) satisfying \(1<2^{\left(-k+3^{-a}\right)}<2\), must be less than(r) 2
(D) If \(\sin \theta=\cos \phi\), then the possible values of \(\frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right)\) are(s) 3

  1. A (A) r, (B) q,r, (C) s, (D) p,r
  2. B (A) s, (B) q,s, (C) r, (D) p,q
  3. C (A) s, (B) q,r, (C) s, (D) p,q
  4. D (A) r, (B) q,s, (C) r, (D) p,r
Verified Solution

Answer & Solution

Correct Answer

(D) (A) r, (B) q,s, (C) r, (D) p,r

Step-by-step Solution

Detailed explanation

(A) Let \(y=\frac{x^2+2 x+4}{x+2}\)
\(\Rightarrow x^2+(2-y) x+(4-2 y)=0 \)
\( \Rightarrow (2-y)^2-4(4-2 y) \geq 0 \)
\( \Rightarrow y^2+4 y-12 \geq 0 \)
\( \Rightarrow y \leq-6, y \geq 2\)
\(\therefore\) Minimum value of \(y\) is 2 .
(B) Since, \((A+B)(A-B)=(A-B)(A+B)\)
\(\Rightarrow A^2-A B+B A-B^2=A^2+A B\) \(-~B A-B^2 \)
\( \Rightarrow A B=B A \)
\( \text { and } (A B)^t=(-1)^k A B \)
\( \Rightarrow B^t A^t=(-1)^k A B \)
\( \Rightarrow -B A=(-1)^k A B [\because B^t=-B, A^t\) \(=A] \)
\( \Rightarrow B A=(-1)^{k+1} A B \)
\( \Rightarrow (-1)^{k+1}=1\)
\(\therefore k+1\) is even or \(k\) is odd.
(C) \(1 < 2^{\left(-k+3^{-a}\right)} < 2 \Rightarrow 0 < -k+3^{-a} < 1\)
Given, \(a=\log _3 \log _3 2 \Rightarrow 3^a=\log _3 2\)
\(\Rightarrow 3^{-a}=\log _2 3 \)
\( \therefore k < \log _2 3 < 2 \)
\( \text { and } 1+k>\log _2 3>1 \Rightarrow k>0\)
From Eqs. (ii) and (iii), \(0 < k < 2 \Rightarrow k=1\)
\([\because k\) is an integer]
\(\text { (D) } \sin \theta =\cos \phi \)
\( \Rightarrow \cos \left(\frac{\pi}{2}-\theta\right) =\cos \phi \)
\( \Rightarrow \frac{\pi}{2}-\theta =2 n \pi \pm \phi, n \in Z \)
\( \Rightarrow \theta \pm \phi-\frac{\pi}{2} =-2 n \pi, n \in Z \)
\( \Rightarrow \frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right) =-2 n, n \in Z\)
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