Match the Statements/Expressions in Column I with the Statements/Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the matrix given in the ORS. [2008]

Question

Match the Statements/Expressions in Column I with the Statements/Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [2008]

	Column I		Column 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 the possible values of $k$ are	(q)	1
(C)	Let $a = \log_{3} \log_{5} 2 .$ An integer $k$ satisfying $1 < 2^{(- k + 3 - a)} < 2,$ must be less than	(r)	2
(D)	If $\sin θ = \cos ϕ,$ then the possible values of $\frac{1}{π} (θ \pm ϕ - \frac{π}{2})$ are	(s)	3

Smart Achievers · Accepted Answer

(2)

A → r, B → q, s; C → r, s; D → p, r

$4 \times 4$ Let $y = \frac{x^{2} + 2 x + 4}{x + 2}$ $\Rightarrow \frac{d y}{d x} = \frac{x^{2} + 4 x}{{(x + 2)}^{2}}$

Now put $\frac{d y}{d x} = 0$ $\Rightarrow \frac{x^{2} + 4 x}{{(x + 2)}^{2}} = 0$ $\Rightarrow x = 0, - 4$

Now, $\frac{d^{2} y}{d x^{2}} = \frac{8}{{(x + 2)}^{3}}$

At $x = 0$ , $\frac{d^{2} y}{d x^{2}} = 1 > 0$

$∴$ $y is minimum when x = 0$

$k$ $1 < 2^{(- k + 3 - a)} < 2,$

$∴$ $(A) \to (r)$

$(B)$ Since $A$ is a symmetric and $B$ is a skew-symmetric matrix,

$∴$ $A^{t} = A and B^{t} = - B$ $\dots (i)$

Now, $(A + B) (A - B) = (A - B) (A + B)$

$\Rightarrow A^{2} - A B + B A - B^{2} = A^{2} + A B - B A - B^{2}$

$\Rightarrow 2 B A = 2 A B$ $\Rightarrow A B = B A$ $\dots (i i)$

Also, ${(A B)}^{t} = {(- 1)}^{k} A B$

$\Rightarrow {(B A)}^{t} = {(- 1)}^{k} A B$ $(using (ii))$

$\Rightarrow A^{t} B^{t} = {(- 1)}^{k} A B$

$\Rightarrow - A B = {(- 1)}^{k} A B$ $[using (i)]$

$\Rightarrow k should be an odd number$

$∴$ $(B) \to (q), (s)$

$(C)$ $a = \log_{3} \log_{3} 2$ $\Rightarrow \log_{3} 2 = 3^{a}$

$\Rightarrow \frac{1}{\log_{2} 3} = 3^{a}$ $\Rightarrow \log_{2} 3 = 3^{- a}$

$∴$ $3 = 2^{3^{- a}}$ $\dots (i)$

Now, $1 < 2^{(- k + 3^{- a})} < 2$ $\Rightarrow 1 < 2^{- k} 2^{3^{- a}} < 2$

$\Rightarrow 1 < 2^{- k} \cdot 3 < 2$ $(using (i))$

$\Rightarrow \frac{1}{3} < 2^{- k} < \frac{2}{3}$ $\Rightarrow \frac{3}{2} < 2^{k} < 3$ $\Rightarrow k = 1$

$∴$ $k is less than 2 and 3$

$∴$ $(C) \to (r), (s)$

$(D)$ $\sin θ = \cos ϕ$ $\Rightarrow \cos (\frac{π}{2} - θ) = \cos ϕ$

$\Rightarrow \frac{π}{2} - θ = 2 n π \pm ϕ, n \in ℤ$ $\Rightarrow θ \pm ϕ - \frac{π}{2} = - 2 n π$

$\Rightarrow \frac{1}{π} (θ \pm ϕ - \frac{π}{2}) = - 2 n$

For $n = 0, - 1$ , possible values of $\frac{1}{π} (θ \pm ϕ - \frac{π}{2})$ are 0 and 2.

$∴$ $D \to (p), (r)$

Solution

1 A → p, r, B → q, s; C → r, s; D → r

2 A → r, B → q, s; C → r, s; D → p, r

3 A → p, r, B → r, s; C → q, s; D → r

4 A → q, s, B → r, s; C → p, r; D → r

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