JEE Advanced Maths PYP Composite Functions & Relations

Q 1 :

Let $f (x) = x^{2}$ 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 [2011]

$\pm \sqrt{n π}, n \in {0, 1, 2, \dots}$
$\pm \sqrt{n π}, n \in {1, 2, \dots}$
$\frac{π}{2} + 2 n π, n \in {\dots, - 2, - 1, 0, 1, 2, \dots}$
$2 n π, n \in {\dots, - 2, - 1, 0, 1, 2, \dots}$

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(1)

Given : $f (x) = x^{2}$ and $g (x) = \sin x,$ $\forall x \in R$

$∴$ $(g \circ f) (x) = \sin x^{2}$

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

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

Since given that $(f \circ g \circ g \circ f) (x) = (g \circ g \circ f) (x),$

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

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

$\Rightarrow \sin x^{2} = n π$ or $\frac{(4 n + 1) π}{2},$ where $n \in Z$

$\Rightarrow \sin x^{2} = 0$ $(∵ \sin x^{2} \in [- 1, 1]) \Rightarrow x^{2} = n π$

$∴$ $x = \pm \sqrt{n π},$ where $n \in W$

Q 2 :

$X$ and $Y$ are two sets and $f : X \to Y$ . If ${f (c) = y; c \subset X, y \subset Y}$ and ${f^{- 1} (d) = x; d \subset Y, x \subset X},$ then the true statement is [2005]

$f (f^{- 1} (b)) = b$
$f^{- 1} (f (a)) = a$
$f (f^{- 1} (b)) = b, b \subset Y$
$f^{- 1} (f (a)) = a, a \subset X$

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(4)

Given that $X$ and $Y$ are two sets and $f : X \to Y$ .

${f (c) = y; c \subset X, y \subset Y}$ and ${f^{- 1} (d) = x; d \subset Y, x \subset X}$

The pictorial representation of the given information is as shown:

[IMAGE 1209]

$Since f^{- 1} (d) = x \Rightarrow f (x) = d$

Now if $a \subset x \Rightarrow f (a) \subset f (x) = d \Rightarrow f^{- 1} (f (a)) = a$

Hence, $f^{- 1} (f (a)) = a,$ $a \subset x$ is the correct option.

Q 3 :

If $f (x) = \sin x + \cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain [2004]

$[0, \frac{π}{2}]$
$[- \frac{π}{4}, \frac{π}{4}]$
$[- \frac{π}{2}, \frac{π}{2}]$
$[0, π]$

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(2)

Given: $f (x) = \sin x + \cos x$ and $g (x) = x^{2} - 1$

$\Rightarrow g (f (x)) = {(\sin x + \cos x)}^{2} - 1 = \sin 2 x$

Clearly, $g (f (x))$ is invertible in $- \frac{π}{2} \leq 2 x \leq \frac{π}{2}$

$(∵ \sin θ is invertible in - \frac{π}{2} \leq θ \leq \frac{π}{2})$

$\Rightarrow - \frac{π}{4} \leq x \leq \frac{π}{4}$

Q 4 :

If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $\min f (x) > \max g (x),$ then the relation between $b$ and $c$ is [2003]

no real value of $b$ & $c$
$0 < c < b \sqrt{2}$
$| c | < | b | \sqrt{2}$
$| c | > | b | \sqrt{2}$

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(4)

$f (x) = x^{2} + 2 b x + 2 c^{2}$ $\Rightarrow f (x) = {(x + b)}^{2} + 2 c^{2} - b^{2}$

$\Rightarrow f_{\min} = 2 c^{2} - b^{2}$

and $g (x) = - x^{2} - 2 c x + b^{2}$

$\Rightarrow g (x) = - {(x + c)}^{2} + b^{2} + c^{2}$ $\Rightarrow g_{\max} = b^{2} + c^{2}$

For $f_{\min} > g_{\max}$ $\Rightarrow 2 c^{2} - b^{2} > b^{2} + c^{2}$

$\Rightarrow c^{2} > 2 b^{2}$

$\Rightarrow | c | > | b | \sqrt{2}$

Q 5 :

Domain of definition of the function $f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ for real valued $x$ , is [2003]

$[- \frac{1}{4}, \frac{1}{2}]$
$[- \frac{1}{2}, \frac{1}{2}]$
$(- \frac{1}{2}, \frac{1}{9})$
$[- \frac{1}{4}, \frac{1}{4}]$

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(1)

For $f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ to be defined and real, $\sin^{- 1} (2 x) + \frac{π}{6} \geq 0$

$\Rightarrow \sin^{- 1} (2 x) \geq - \frac{π}{6}$ $\dots (i)$

But, $- \frac{π}{2} \leq \sin^{- 1} (2 x) \leq \frac{π}{2}$ $\dots (i i)$

On combining (i) and (ii), we get

$- \frac{π}{6} \leq \sin^{- 1} (2 x) \leq \frac{π}{2}$

$\Rightarrow \sin (- \frac{π}{6}) \leq 2 x \leq \sin (\frac{π}{2})$

$\Rightarrow - \frac{1}{2} \leq 2 x \leq 1$

$\Rightarrow - \frac{1}{4} \leq x \leq \frac{1}{2}$

$∴$ $Domain = [- \frac{1}{4}, \frac{1}{2}]$

Q 6 :

Let $f (x) = \frac{α x}{x + 1},$ $x \neq - 1 .$ Then, for what value of $α$ is $f (f (x)) = x ?$ [2001]

$\sqrt{2}$
$- \sqrt{2}$
$1$
$- 1$

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(4)

$f (x) = \frac{α x}{x + 1},$ $x \neq - 1$

Now, $f (f (x)) = x$ $\Rightarrow \frac{α (\frac{α x}{x + 1})}{\frac{α x}{x + 1} + 1} = x$

$\Rightarrow \frac{α^{2} x}{(α + 1) x + 1} = x$ $\Rightarrow (α + 1) x^{2} + (1 - α^{2}) x = 0$

$\Rightarrow α + 1 = 0$ and $1 - α^{2} = 0$

$∴$ $α = - 1$

Q 7 :

The domain of definition of $f (x) = \frac{\log_{2} (x + 3)}{x^{2} + 3 x + 2}$ is [2001]

$R \ {- 1, - 2}$
$(- 2, \infty)$
$R \ {- 1, - 2, - 3}$
$(- 3, \infty) - {- 1, - 2}$

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(4)

For domain of $f (x) = \frac{\log_{2} (x + 3)}{x^{2} + 3 x + 2},$

we require

$x^{2} + 3 x + 2 \neq 0$ and $x + 3 > 0$

$\Rightarrow x \neq - 1, - 2$ and $x > - 3$

$∴$ $Domain of f (x) = (- 3, \infty) - {- 1, - 2}$

Q 8 :

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + \frac{1}{x},$ then $f^{- 1} (x)$ equals [2001]

$\frac{x + \sqrt{x^{2} - 4}}{2}$
$\frac{x}{1 + x^{2}}$
$\frac{x - \sqrt{x^{2} - 4}}{2}$
$1 + \sqrt{x^{2} - 4}$

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(1)

$Given: f (x) = x + \frac{1}{x} = y$

$(let) \Rightarrow x^{2} - y x + 1 = 0$

$\Rightarrow x = \frac{y \pm \sqrt{y^{2} - 4}}{2}$

$∴$ $x = \frac{y + \sqrt{y^{2} - 4}}{2}$ $[∵ x \geq 1 and y \geq 2]$

$∴$ $f^{- 1} (x) = \frac{x + \sqrt{x^{2} - 4}}{2}$

Q 9 :

Let $g (x) = 1 + x - [x]$ and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$ $.$ Then for all $x, f (g (x))$ is equal to [2001]

$x$
$1$
$f (x)$
$g (x)$

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(2)

$g (x) = 1 + x - [x]$

and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$

For integral values of $x$ , $g (x) = 1$

For $x < 0$ (but not integral value), $x - [x] > 0 \Rightarrow g (x) > 1$

For $x > 0$ (but not integral value), $x - [x] > 0 \Rightarrow g (x) > 1$

$∴$ $g (x) \geq 1, \forall x \Rightarrow f (g (x)) = 1, \forall x$

Q 10 :

The value of $({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$ is ________. [2018]

(8)

$({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$

$= {(\log_{2} 9)}^{2 \times \log_{(\log_{2} 9)} 2} \times 7^{\frac{1}{2} \times \log_{7} 4}$

$= {(\log_{2} 9)}^{\log_{(\log_{2} 9)} 4} \times 7^{\log_{7} 2} = 4 \times 2 = 8$

Q 11 :

Let $f : [0, 4 π] \to [0, π]$ be defined by $f (x) = \cos^{- 1} (\cos x) .$ The number of points $x \in [0, 4 π]$ satisfying the equation $g (x) = \frac{10 - x}{10}$ is [2014]

(3)

$Given: f : [0, 4 π] \to [0, π] defined by$

$f (x) = \cos^{- 1} (\cos x)$

and $g (x) = \frac{10 - x}{10} = 1 - \frac{x}{10}$

The graph of $y = f (x)$ and $y = g (x)$ are as follows.

[IMAGE 1210]

Q 12 :

Let $ℝ$ denote the set of all real numbers. Let $f : ℝ \to ℝ$ and $g : ℝ \to (0, 4)$ be functions defined by

$f (x) = \log_{e} (x^{2} + 2 x + 4)$ and $g (x) = \frac{4}{1 + e^{- 2 x}} .$

Define the composite function $f \circ g^{- 1}$ by $(f \circ g^{- 1}) (x) = f (g^{- 1} (x)),$ where $g^{- 1}$ is the inverse of the function $g$ .

Then the value of the derivative of the composite function $f \circ g^{- 1}$ at $x = 2$ is ________. [2025]

(0.25)

$Let h (x) = f (g^{- 1} (x)) and g (0) = 2$

$\Rightarrow h' (x) = f' (g^{- 1} (x)) \cdot (g^{- 1} (x))'$

$\Rightarrow h' (2) = f' (g^{- 1} (2)) \cdot (g^{- 1})' (2)$ $(put x = 2) = f' (0) \cdot (g^{- 1})' (2)$

Now, $f (x) = \log_{e} (x^{2} + 2 x + 4)$

$f' (x) = \frac{2 x + 2}{x^{2} + 2 x + 4} \Rightarrow f' (0) = \frac{1}{2}$

$g (x) = \frac{4}{1 + e^{- 2 x}},$ $g (0) = 2$

$∴$ $g' (x) = \frac{8 e^{- 2 x}}{{(1 + e^{- 2 x})}^{2}}$

$g' (0) = \frac{8}{4} = 2$ $∵$ $g^{- 1} (g (x)) = x$

$(g^{- 1})' (g (x)) g' (x) = 1$

$(g^{- 1})' (2) = \frac{1}{g' (0)} = \frac{1}{2}$

$∴$ $h' (2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$

Clearly, $f (x) = g (x)$ has 3 solutions.

Q 13 :

Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)),$ for all $x \in R$ and $g (x) = \frac{π}{2} \sin x,$ 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? [2015]

Range of $f$ is $[- \frac{1}{2}, \frac{1}{2}]$
Range of $f \circ g$ is $[- \frac{1}{2}, \frac{1}{2}]$
$\lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{π}{6}$
There is an $x \in R$ such that $(g \circ f) (x) = 1$

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Select one or more options

(1, 2, 3)

$f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$

$- 1 \leq \sin x \leq 1 \Rightarrow - \frac{π}{2} \leq \frac{π}{2} \sin x \leq \frac{π}{2}$

$\Rightarrow - 1 \leq \sin (\frac{π}{2} \sin x) \leq 1$ $\Rightarrow - \frac{π}{6} \leq \frac{π}{6} \sin (\frac{π}{2} \sin x) \leq \frac{π}{6}$

$\Rightarrow - \frac{1}{2} \leq \sin [\frac{π}{6} \sin (\frac{π}{2} \sin x)] \leq \frac{1}{2}$

$∴$ $Range of f = [\frac{- 1}{2}, \frac{1}{2}]$

Now, $(f \circ g) (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$

$Range of (f \circ g) = [- \frac{1}{2}, \frac{1}{2}]$

Now, $\lim_{x \to 0} \frac{f (x)}{g (x)} = \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))}{\frac{π}{2} \sin x}$

$= \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))}{\frac{π}{6} \sin (\frac{π}{2} \sin x)} \times \frac{\frac{π}{6} \sin (\frac{π}{2} \sin x)}{\frac{π}{2} \sin x}$

$= 1 \times \frac{π}{6} = \frac{π}{6}$

Now, $(g \circ f) (x) = \frac{π}{2} \sin (\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)))$ $- \frac{π}{2} \sin (\frac{1}{2}) \leq g (f (x)) \leq \frac{π}{2} \sin (\frac{1}{2})$

Let $\frac{π}{2} \sin (\frac{1}{2}) = p$

Clearly, $0 < p < 1$

$∴$ $- \frac{π}{2} \sin (\frac{1}{2}) \leq g (f (x)) \leq \frac{π}{2} \sin (\frac{1}{2})$

$- p \leq g (f (x)) \leq p \Rightarrow 0 < p < 1$

$∴$ $(g \circ f) (x) \neq 1 for any x \in R .$

Q 14 :

Let $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ be given by $f (x) = (\log {(\sec x + \tan x))}^{3} .$ Then [2014]

$f (x)$ is an odd function.
$f (x)$ is a one-one function.
$f (x)$ is an onto function.
$f (x)$ is an even function.

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Select one or more options

(1, 2, 3)

Given: $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ is given by

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

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

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

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

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

$∴$ $f (x) is an odd function.$

$∴$ option (1) is correct and (4) is not correct.

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

$= 3 \sec x {[\log (\sec x + \tan x)]}^{2} > 0, ∀ x \in (- \frac{π}{2}, \frac{π}{2})$

$∴$ $f (x) is increasing on (- \frac{π}{2}, \frac{π}{2})$

We know that a strictly increasing function is one-one.

$∴$ $f$ is one-one, hence (2) is the correct option.

Also, $\lim_{x \to {\frac{π}{2}}^{-}} {[\log (\sec x + \tan x)]}^{3} = \infty$

and $\lim_{x \to {\frac{π}{2}}^{+}} {[\log (\sec x + \tan x)]}^{3} = - \infty$

$∴$ $Range of f = (- \infty, \infty) = R = domain$

$∴$ $f is an onto function.$

$∴$ option (3) is correct.

Q 15 :

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x},$ where $b$ is a constant such that $0 < b < 1$ . Then [2011]

$f$ is not invertible on (0, 1)
$f \neq f^{- 1}$ on (0, 1) and $f' (b) = \frac{1}{f' (0)}$
$f = f^{- 1}$ on (0, 1) and $f' (b) = \frac{1}{f' (0)}$
$f^{- 1}$ is differentiable on (0, 1)

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Select one or more options

(1, 2)

$Given: f (x) = \frac{b - x}{1 - b x}, 0 < b < 1$

Let $f (x_{1}) = f (x_{2})$ $\Rightarrow \frac{b - x_{1}}{1 - b x_{1}} = \frac{b - x_{2}}{1 - b x_{2}}$

$\Rightarrow b - b^{2} x_{2} - x_{1} + b x_{1} x_{2} = b - x_{2} - b^{2} x_{1} + b x_{1} x_{2}$

$\Rightarrow x_{2} (1 - b^{2}) = x_{1} (1 - b^{2})$ $\Rightarrow x_{1} = x_{2}$ as $1 - b^{2} \neq 0$

$∴$ $f is one-one.$

Also, $\frac{b - x}{1 - b x} = y$ $\Rightarrow b - x = y - b x y$

$\Rightarrow (b y - 1) x = y - b$ $\Rightarrow x = \frac{y - b}{b y - 1}$

For $y = \frac{1}{b},$ $x$ is not defined.

$∴$ $f is not onto and hence not invertible.$

Also, $f' (x) = \frac{- 1 (1 - b x) - (- b) (b - x)}{{(1 - b x)}^{2}} = \frac{b^{2} - 1}{{(1 - b x)}^{2}}$

$∴$ $f' (b) = \frac{1}{b^{2} - 1}$ and $f' (0) = b^{2} - 1$ $\Rightarrow f' (b) = \frac{1}{f' (0)}$

$∴$ $(1) and (2) are the correct options.$

Q 16 :

Let $E_{1} = {x \in ℝ : x \neq 1 and \frac{x}{x - 1} > 0}$ and

$E_{2} = {x \in E_{1} : \sin^{- 1} (\log_{e} (\frac{x}{x - 1})) is a real number} .$

(Here, the inverse trigonometric function $\sin^{- 1} x$ assumes values in $[- \frac{π}{2}, \frac{π}{2}]$ ).

Let $f : E_{1} \to ℝ$ be the function defined by $f (x) = \log_{e} (\frac{x}{x - 1})$ and $g : E_{2} \to ℝ$ be the function defined by $g (x) = \sin^{- 1} (\log_{e} (\frac{x}{x - 1})) .$ [2018]

LIST-I LIST-II

P. The range of $f$ is 1. $(- \infty, \frac{1}{1 - e}] \cup [\frac{e}{e - 1}, \infty)$

Q. The range of $g$ contains 2. $(0, 1)$

R. The domain of $f$ contains 3. $[- \frac{1}{2}, \frac{1}{2}]$

S. The domain of $g$ is 4. $(- \infty, 0) \cup (0, \infty)$

5. $(- \infty, \frac{e}{e - 1}]$

6. $(- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}]$

The correct option is:

P → 4; Q → 2; R → 1; S → 1
P → 3; Q → 3; R → 6; S → 5
P → 4; Q → 2; R → 1; S → 6
P → 4; Q → 3; R → 6; S → 5

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(1)

For $E_{1}$ , $\frac{x}{x - 1} > 0$ and $x \neq 1$ $\Rightarrow x \in (- \infty, 0) \cup (1, \infty)$

For $E_{2}$ , $- 1 \leq \log_{e} (\frac{x}{x - 1}) \leq 1$ $\Rightarrow \frac{1}{e} \leq \frac{x}{x - 1} \leq e$

$\Rightarrow \frac{1}{e} \leq 1 + \frac{1}{x - 1} \leq e$ $\Rightarrow \frac{1}{e} - 1 \leq \frac{1}{x - 1} \leq e - 1$

$\Rightarrow (x - 1) \in (- \infty, \frac{e}{1 - e}] \cup [\frac{1}{e - 1}, \infty)$

$\Rightarrow x \in (- \infty, \frac{1}{e - 1}] \cup [\frac{e}{e - 1}, \infty)$

For $E_{1}$ , $\frac{x}{x - 1} \in (0, \infty) - {1}$

$\Rightarrow \log_{e} (\frac{x}{x - 1}) \in (- \infty, \infty) - {0}$

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

$g (x) = \sin^{- 1} (\log_{e} (\frac{x}{x - 1})) \in [- \frac{π}{2}, \frac{π}{2}] - {0}$

Topic Question Set

Q 1 :

Let $f (x) = x^{2}$ 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 [2011]

Q 2 :

$X$ and $Y$ are two sets and $f : X \to Y$ . If ${f (c) = y; c \subset X, y \subset Y}$ and ${f^{- 1} (d) = x; d \subset Y, x \subset X},$ then the true statement is [2005]

Q 3 :

If $f (x) = \sin x + \cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain [2004]

Q 4 :

If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $\min f (x) > \max g (x),$ then the relation between $b$ and $c$ is [2003]

Q 5 :

Domain of definition of the function $f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ for real valued $x$ , is [2003]

Q 6 :

Let $f (x) = \frac{α x}{x + 1},$ $x \neq - 1 .$ Then, for what value of $α$ is $f (f (x)) = x ?$ [2001]

Q 7 :

The domain of definition of $f (x) = \frac{\log_{2} (x + 3)}{x^{2} + 3 x + 2}$ is [2001]

Q 8 :

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + \frac{1}{x},$ then $f^{- 1} (x)$ equals [2001]

Q 9 :

Let $g (x) = 1 + x - [x]$ and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$ $.$ Then for all $x, f (g (x))$ is equal to [2001]

Q 10 :

The value of $({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$ is ________. [2018]

Q 11 :

Let $f : [0, 4 π] \to [0, π]$ be defined by $f (x) = \cos^{- 1} (\cos x) .$ The number of points $x \in [0, 4 π]$ satisfying the equation $g (x) = \frac{10 - x}{10}$ is [2014]

Q 13 :

Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)),$ for all $x \in R$ and $g (x) = \frac{π}{2} \sin x,$ 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? [2015]

Q 14 :

Let $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ be given by $f (x) = (\log {(\sec x + \tan x))}^{3} .$ Then [2014]

Q 15 :

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x},$ where $b$ is a constant such that $0 < b < 1$ . Then [2011]

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	LIST-I		LIST-II
P.	The range of $f$ is	1.	$(- \infty, \frac{1}{1 - e}] \cup [\frac{e}{e - 1}, \infty)$
Q.	The range of $g$ contains	2.	$(0, 1)$
R.	The domain of $f$ contains	3.	$[- \frac{1}{2}, \frac{1}{2}]$
S.	The domain of $g$ is	4.	$(- \infty, 0) \cup (0, \infty)$
		5.	$(- \infty, \frac{e}{e - 1}]$
		6.	$(- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}]$

Topic Question Set

Q 1 : Let f(x)=x2 and g(x)=sinx for all x∈R. Then the set of all x satisfying (f∘g∘g∘f)(x)=(g∘g∘f)(x), where (f∘g)(x)=f(g(x)), is [2011]

Q 2 : X and Y are two sets and f:X→Y. If {f(c)=y; c⊂X, y⊂Y} and {f-1(d)=x; d⊂Y, x⊂X}, then the true statement is [2005]

Q 3 : If f(x)=sinx+cosx, g(x)=x2-1, then g(f(x)) is invertible in the domain [2004]

Q 4 : If f(x)=x2+2bx+2c2 and g(x)=-x2-2cx+b2 such that min f(x)>max g(x), then the relation between b and c is [2003]

Q 5 : Domain of definition of the function f(x)=sin-1(2x)+π6 for real valued x, is [2003]

Q 6 : Let f(x)=αxx+1, x≠-1. Then, for what value of α is f(f(x))=x? [2001]

Q 7 : The domain of definition of f(x)=log2(x+3)x2+3x+2 is [2001]

Q 8 : If f:[1,∞)→[2,∞) is given by f(x)=x+1x, then f-1(x) equals [2001]

Q 9 : Let g(x)=1+x-[x] and f(x)={-1,x<00,x=01,x>0. Then for all x, f(g(x)) is equal to [2001]

Q 10 : The value of ((log29)2)1log2(log29)×(7)1log47 is ________. [2018]

Q 11 : Let f:[0,4π]→[0,π] be defined by f(x)=cos-1(cosx). The number of points x∈[0,4π] satisfying the equation g(x)=10-x10 is [2014]

Q 13 : Let f(x)=sin(π6sin(π2sinx)), for all x∈R and g(x)=π2sinx, for all x∈R. Let (f∘g)(x) denote f(g(x)) and (g∘f)(x) denote g(f(x)). Then which of the following is (are) true? [2015]

Q 14 : Let f:(-π2,π2)→R be given by f(x)=(log(secx+tanx))3. Then [2014]

Q 15 : Let f:(0,1)→R be defined by f(x)=b-x1-bx, where b is a constant such that 0<b<1. Then [2011]

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Q 1 :

Let $f (x) = x^{2}$ 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 [2011]

Q 2 :

$X$ and $Y$ are two sets and $f : X \to Y$ . If ${f (c) = y; c \subset X, y \subset Y}$ and ${f^{- 1} (d) = x; d \subset Y, x \subset X},$ then the true statement is [2005]

Q 3 :

If $f (x) = \sin x + \cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain [2004]

Q 4 :

If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $\min f (x) > \max g (x),$ then the relation between $b$ and $c$ is [2003]

Q 5 :

Domain of definition of the function $f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ for real valued $x$ , is [2003]

Q 6 :

Let $f (x) = \frac{α x}{x + 1},$ $x \neq - 1 .$ Then, for what value of $α$ is $f (f (x)) = x ?$ [2001]

Q 7 :

The domain of definition of $f (x) = \frac{\log_{2} (x + 3)}{x^{2} + 3 x + 2}$ is [2001]

Q 8 :

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + \frac{1}{x},$ then $f^{- 1} (x)$ equals [2001]

Q 9 :

Let $g (x) = 1 + x - [x]$ and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$ $.$ Then for all $x, f (g (x))$ is equal to [2001]

Q 10 :

The value of $({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$ is ________. [2018]

Q 11 :

Let $f : [0, 4 π] \to [0, π]$ be defined by $f (x) = \cos^{- 1} (\cos x) .$ The number of points $x \in [0, 4 π]$ satisfying the equation $g (x) = \frac{10 - x}{10}$ is [2014]

Q 13 :

Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)),$ for all $x \in R$ and $g (x) = \frac{π}{2} \sin x,$ 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? [2015]

Q 14 :

Let $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ be given by $f (x) = (\log {(\sec x + \tan x))}^{3} .$ Then [2014]

Q 15 :

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x},$ where $b$ is a constant such that $0 < b < 1$ . Then [2011]