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1

### JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)

Let $$f(x) = \sin \left( {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right)$$ for all $$x \in R$$ and g(x) = $${{\pi \over 2}\sin x}$$ for all x$$\in$$R. Let $$(f \circ g)(x)$$ denote f(g(x)) and $$(g \circ f)(x)$$ denote g(f(x)). Then which of the following is/are true?

A
Range of f is $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$.
B
Range of f $$\circ$$ g is $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$.
C
$$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}} = {\pi \over 6}$$.
D
There is an x$$\in$$R such that (g $$\circ$$ f)(x) = 1.

## Explanation

(a) $$f(x) = \sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right],\,x \in R$$

$$= \sin \left( {{\pi \over 6}\sin \theta } \right),\,\theta \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$,

where, $$\theta = {\pi \over 2}\sin x$$

$$= \sin \alpha ,\alpha \in \left[ { - {\pi \over 6},{\pi \over 6}} \right]$$,

where, $$\alpha = {\pi \over 6}\sin \theta$$

$$\therefore$$ $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$

Hence, range of $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$

So, option (a) is correct.

(b) $$f\{ g(x)\} = f(t),t \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$

$$\Rightarrow f(t) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$

$$\therefore$$ Option (b) is correct.

(c) $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}}$$

$$= \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 2}(\sin x)}}$$

$$= \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)}}\,.\,{{{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \over {\left( {{\pi \over 2}\sin x} \right)}}$$

$$= 1 \times {\pi \over 6} \times 1 = {\pi \over 6}$$

$$\therefore$$ Option (c) is correct.

(d) $$g\{ f(x)\} = 1$$

$$\Rightarrow {\pi \over 2}\sin \{ f(x)\} = 1$$

$$\Rightarrow \sin \{ f(x)\} = {2 \over \pi }$$ ..... (i)

But, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right] \subset \left[ { - {\pi \over 6},{\pi \over 6}} \right]$$

$$\therefore$$ $$\sin \{ f(x)\} \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$ ..... (ii)

$$\Rightarrow \sin \{ f(x)\} \ne {2 \over \pi }$$, [from Eqs. (i) and (ii)]

i.e. No solution.

$$\therefore$$ Option (d) is not correct.

2

### JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
Let $$f:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R$$ be given by $$f(x) = {[\log (\sec x + \tan x)]^3}$$. Then,
A
f(x) is an odd function
B
f(x) is a one-one function
C
f(x) is an onto function
D
f(x) is an even function

## Explanation

$$f:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R$$

$$f(x) = {[\log (\sec x + \tan x)]^3}$$

$$f( - x) = {[\log (\sec x - \tan x)]^3}$$

$$= {\left[ {\log \left( {{{(\sec x - \tan x)(\sec x + \tan x)} \over {\sec x + \tan x}}} \right)} \right]^3}$$

$$= {\left[ {\log \left( {{1 \over {\sec x + \tan x}}} \right)} \right]^3} = {[ - \log (\sec x + \tan x)]^3}$$

$$= - {[\log (\sec x + \tan x)]^3} = - f(x)$$

$$\therefore$$ f is an odd function. (a) is correct and (d) is not correct.

Also,

$$f'(x) = 3{[\log (\sec x + \tan x)]^2}\,.\,{{\sec x\tan x + {{\sec }^2}x} \over {\sec x + \tan x}}$$

$$= 3\sec x{[\log (\sec x + \tan x)]^2} > 0\,\forall x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$

$$\therefore$$ f is increasing on $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$

We know that strictly increasing function is one one.

$$\therefore$$ f is one one

$$\therefore$$ (b) is correct.

$$(\sec x + \tan x) = \tan \left( {{\pi \over 4} + {\pi \over 2}} \right)$$

as $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, then

$$0 < \tan \left( {{\pi \over 4} + {\pi \over 2}} \right) < \infty$$

$$0 < \sec x + \tan x < \infty$$

$$\Rightarrow - \infty < \ln (\sec x + \tan x) < \infty$$

$$- \infty < {[\ln (\sec x + \tan x)]^3} < \infty$$

$$\Rightarrow - \infty < f(x) < \infty$$

Range of f(x) is R and thus f(x) is an onto function.

$$\therefore$$ (c) is correct.
3

### JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
For every pair of continuous function f, g : [0, 1] $$\to$$ R such that max {f(x) : x $$\in$$ [0, 1]} = max {g(x) : x $$\in$$ [0, 1]}. The correct statement(s) is (are)
A
[f(c)]2 + 3f(c) = [g(c)]2 + 3g(c) for some c $$\in$$ [0, 1]
B
[f(c)]2 + f(c) = [g(c)]2 + 3g(c) for some c $$\in$$ [0, 1]
C
[f(c)]2 + 3f(c) = [g(c)]2 + g(c) for some c $$\in$$ [0, 1]
D
[f(c)]2 = [g(c)]2 + 3g(c) for some c $$\in$$ [0, 1]

## Explanation

Suppose f(x) is maximum at c1 and g(x) is maximum at c2. When f(x) is maximum g(x) may or may not be maximum.

Therefore, in the function h(x) = f(x) $$-$$ g(x), we get

$$h({c_1}) = f({c_1}) - g({c_1}) \ge 0$$ and $$h({c_2}) = f({c_2}) - g({c_2}) \ge 0$$.

Therefore, h(x) = 0 for some c $$\in$$ [0, 1].

Therefore, $$h(c) = 0 \Rightarrow f(c) - g(c) = 0$$.

Therefore, $$f(c) = g(c)$$.

Option (a) $$\Rightarrow {f^2}(c) - {g^2}(c) + 3[f(c) - g(c)] = 0$$ which is true from Eq. (i).

Option (d) $$\Rightarrow {f^2}(c) - {g^2}(c) = 0$$ which is true from Eq. (i)

Now, if we take

f(x) = 1 and g(x) = 1, $$\forall$$x $$\in$$[0, 1]

Option (b) and (c) does not hold. Hence, option (a) and (d) are correct.
4

### IIT-JEE 2012 Paper 2 Offline

MCQ (More than One Correct Answer)

Let $$f:( - 1,1) \to R$$ be such that $$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$ for $$\theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 4},{\pi \over 2}} \right)$$. Then the value(s) of $$f\left( {{1 \over 3}} \right)$$ is(are)

A
$$1 - \sqrt {{3 \over 2}}$$
B
$$1 + \sqrt {{3 \over 2}}$$
C
$$1 - \sqrt {{2 \over 3}}$$
D
$$1 + \sqrt {{2 \over 3}}$$

## Explanation

$$f(\cos 4\theta ) = {2 \over {2 - {{\sec }^2}\theta }}$$

Let $$\cos 4\theta = t$$

$$\Rightarrow 2{\cos ^2}2\theta - 1 = t \Rightarrow {\cos ^2}2\theta = {2 \over 3}$$

For $$t = {1 \over 3}$$ we have $${\cos ^2}2\theta = {2 \over 3}$$

$$\cos 2\theta = \sqrt {{2 \over 3}}$$ or $$\cos 2\theta = - \sqrt {{2 \over 3}}$$

$$f(\cos 4\theta ) = {2 \over {2 - {1 \over {{{\cos }^2}\theta }}}} = {{2{{\cos }^2}\theta } \over {2{{\cos }^2}\theta - 1}} = {{1 + \cos 2\theta } \over {\cos 2\theta }} = 1 + {1 \over {\cos 2\theta }}$$

Hence, $$f\left( {{1 \over 3}} \right) = 1 + \sqrt {{3 \over 2}}$$ or $$1 - \sqrt {{3 \over 2}}$$

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