JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 22)

Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

	Column I		Column II
(A)	The minimum value of $${{{x^2} + 2x + 4} \over {x + 2}}$$ 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 (AB)$$^t$$ = ($$-1$$)$$^k$$ AB, where (AB)$$^t$$ is the transpose of the matrix AB, then the possible values of k are	(Q)	1
(C)	Let $$a=\log_3\log_3 2$$. An integer k satisfying $$1 < {2^{( - k + 3 - a)}} < 2$$, must be less than	(R)	2
(D)	If $$\sin \theta = \cos \varphi $$, then the possible values of $${1 \over \pi }\left( {\theta + \varphi - {\pi \over 2}} \right)$$ are	(S)	3

A - iii; B - ii, iv; C - iii, iv; D - i, iii

A - iii; B - ii; C - iii, iv; D - i, iii

A - ii; B - ii, iv; C - iii, iv; D - i

A - ii; B - ii, iv; C - iii, iv; D - i, iii

(A) Let $$y = {{{x^2} + 2x + 4} \over {x + 2}}$$

$${{dy} \over {dx}} = {{{x^2} + 4x} \over {{{(x + 2)}^2}}} = 0$$

$$x = 0, - 4$$

$${{{d^2}y} \over {d{x^2}}} = {8 \over {{{(x + 2)}^3}}}$$

At $$x = 0,{{{d^2}y} \over {d{x^2}}}$$ is true

$$\therefore$$ y is min. when $$x = 0$$

$$\therefore$$ $${y_{\min }} = 2$$

(A) - (iii)

(B) As A is symmetric and B is skew symmetric matrix

We should have

A$$^+$$ = A and B$$^+$$ = $$-$$B ...... (i)

Also given that

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

A$$^2$$ $$-$$ AB + BA $$-$$ B$$^2$$ = A$$^2$$ + AB $$-$$ AB $$-$$ B$$^2$$

2BA = 2AB or AB = BA ...... (ii)

Now, given that

(AB)$$^t$$ = ($$-$$1)$$^k$$AB

(BA)$$^t$$ = ($$-$$1)$$^k$$AB (Using equation (i))

$$\Rightarrow$$ K should be an odd no.

$$\therefore$$ B - (ii, iv)

$$a = {\log _3}{\log _3}2$$

$$ \Rightarrow {\log _3}2 = {3^a} \Rightarrow {{{1_x}} \over {{{\log }_2}^3}} = {3^a}$$

Or $${\log _2}^3 = {3^{ - a}}$$

$$3 = {2^{(3 - a)}}$$

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

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

$$ = {1 \over 3}\,.\, < {2^{ - k}} < {2 \over 3} \Rightarrow {3 \over 2} < {2^k} < 3$$

$$ \Rightarrow k = 1$$

$$\therefore$$ k is less than 2 and 3.

$$\therefore$$ (C) - (iii, iv)

(D) Given that,

$$\sin \theta = \cos \phi $$

$$\cos \left( {{\pi \over 2} - \theta } \right) = \cos \phi $$

$$ = {\pi \over 2} - \theta = 2n\pi \pm \phi ,n \in Z$$

$$ \Rightarrow \theta \pm \phi - {\pi \over 2} = - 2n\pi $$

$$ = {1 \over \pi }\left( {\theta \pm \phi - {\pi \over 2}} \right) = - 2n$$

$$\therefore$$ Here, possible value of $${1 \over \pi }\left( {\theta \pm \phi - {\pi \over 2}} \right)$$ are 0 and 2 for $$n = 0, - 1$$

$$\therefore$$ (D) - p, r