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1

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

Match the statements given in Column I with the intervals/union of intervals given in Column II :

A
(A) $$\to$$ (S), (B) $$\to$$ (T), (C) $$\to$$ (P), (D) $$\to$$ (Q)
B
(A) $$\to$$ (S), (B) $$\to$$ (T), (C) $$\to$$ (R), (D) $$\to$$ (P)
C
(A) $$\to$$ (S), (B) $$\to$$ (T), (C) $$\to$$ (R), (D) $$\to$$ (R)
D
(A) $$\to$$ (P), (B) $$\to$$ (Q), (C) $$\to$$ (R), (D) $$\to$$ (R)

Explanation

(A) $$z = {{2i(x + iy)} \over {1 - {{(x + iy)}^2}}} = {{2i(x + iy)} \over {1 - ({x^2} - {y^2} + 2ixy)}}$$

Using $$1 - {x^2} = {y^2}$$, we get

$$Z = {{2ix - 2y} \over {2{y^2} - 2ixy}} = - {1 \over y}$$

Since $$ - 1 \le y \le 1 \Rightarrow - {1 \over y} \le - 1$$ or $$ - {1 \over y} \ge 1$$.

(B) For domain :

$$ - 1 \le {{8({3^{x - 2}})} \over {1 - {3^{2(x - 1)}}}} \le 1 \Rightarrow - 1 \le {{{3^x} - {3^{x - 2}}} \over {1 - {3^{2x - 2}}}} \le 1$$

Case 1 : $${{{3^x} - {3^{x - 2}}} \over {1 - {3^{2x - 2}}}} - 1 \le 0$$.

$$ \Rightarrow {{({3^x} - 1)({3^{x - 2}} - 1)} \over {({3^{2x - 2}} - 1)}} \ge 0$$

$$ \Rightarrow x \in ( - \infty ,0] \cup (1,\infty )$$

Case 2 : $${{{3^x} - {3^{x - 2}}} \over {1 - {3^{2x}} - 2}} + 1 \ge 0$$

$$ \Rightarrow {{({x^{x - 2}} - 1)({3^x} + 1)} \over {({3^x}\,.\,{3^{x - 2}} - 1)}} \ge 0$$

$$ \Rightarrow x \in ( - \infty ,1) \cup [2,\infty )$$

So, $$x \in ( - \infty ,0] \cup [2,\infty )$$.

(C) R1 $$\to$$ R1 + R3 :

$$f(\theta ) = \left| {\matrix{ 0 & 0 & 2 \cr { - \tan \theta } & 1 & {\tan \theta } \cr { - 1} & { - \tan \theta } & 1 \cr } } \right| = 2({\tan ^2}\theta + 1) = 2{\sec ^2}\theta $$

(D) $$f'(x) = {3 \over 2}{(x)^{1/2}}(3x - 10) + {(x)^{3/2}} \times 3 = {{15} \over 2}{(x)^{1/2}}(x - 2)$$

Increasing, when $$x \ge 2$$.

2

IIT-JEE 2011 Paper 2 Offline

MCQ (Single Correct Answer)

Let f(x) = x2 and g(x) = sin x for all x $$\in$$ R. Then the set of all x satisfying $$(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)$$, where $$(f \circ g)(x) = f(g(x))$$, is

A
$$ \pm \sqrt {n\pi } ,\,n \in \{ 0,1,2,....\} $$
B
$$ \pm \sqrt {n\pi } ,\,n \in \{ 1,2,....\} $$
C
$${\pi \over 2} + 2n\pi ,\,n \in \{ ....., - 2, - 1,0,1,2,....\} $$
D
$$2n\pi ,n \in \{ ....., - 2, - 1,0,1,2,....\} $$

Explanation

$$f(x) = {x^2}$$, $$g(x) = \sin x$$

$$(g \circ f)(x) = \sin {x^2}$$

$$g \circ (g \circ f)(x) = \sin (\sin {x^2})$$

$$(f \circ g \circ g \circ f)(x) = {(\sin (\sin {x^2}))^2}$$ ..... (i)

Again, $$(g \circ f)(x) = \sin {x^2}$$

$$(g \circ g \circ f)(x) = \sin (\sin {x^2})$$ ..... (ii)

Given, $$(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)$$

$$ \Rightarrow {(\sin (\sin {x^2}))^2} = \sin (\sin {x^2})$$

$$ \Rightarrow \sin (\sin {x^2})\{ \sin (\sin {x^2}) - 1\} = 0$$

$$ \Rightarrow \sin (\sin {x^2}) = 0$$ or $$\sin (\sin {x^2}) = 1$$

$$ \Rightarrow \sin {x^2} = 0$$ or $$\sin {x^2} = {\pi \over 2}$$

$$\therefore$$ $${x^2} = n\pi $$ (i.e. not possible as $$ - 1 \le \sin \theta \le 1$$)

$$x = \pm \sqrt {n\pi } $$

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