Functions

$ S=(0,1) \cup(1,2) \cup(3,4) \text { and } T=\{0,1,2,3\} \text {. } $

Let domain and co-domain of a function $y=f(x)$ are $S$ and $T$ respectively.

(A) There are infinitely many elements in domain and four elements in co-domain.

$\Rightarrow$ There are infinitely many functions from $S$ to $T$.

$\Rightarrow$ Option $(A)$ is correct

(B) If number of elements in domain is greater than number of elements in co-domain, then number of strictly increasing function is zero.

$\Rightarrow$ Option (B) is incorrect

(C) Maximum number of continuous functions = $4 \times 4 \times 4=64$

(Every subset $(0,1),(1,2),(3,4)$ has four choices)

$\because 64 < 120 \Rightarrow$ option (C) is correct.

(D) For every point at which $f(x)$ is continuous, $f(x)=0$

$\Rightarrow$ Every continuous function from $S$ to $T$ is differentiable.

Option (D) is correct.

$\begin{aligned} \text { Given } f:[0,1] & \rightarrow[0,1] \\\\ f(x) & =\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36} \\\\ f^{\prime}(x) & =\frac{3 x^2}{3}-2 x+\frac{5}{9}\end{aligned}$

$\begin{aligned} & \quad f^{\prime}(x)=0 \\\\ & 9 x^2-18 x+5=0 \\\\ & \Rightarrow 9 x^2-15 x-3 x+5=0 \\\\ & \Rightarrow 3 x(3 x-5)-1(3 x-5)=0 \\\\ & \Rightarrow(3 x-5)(3 x-1)=0 \\\\ & \Rightarrow x=\frac{1}{3} \text { or } \frac{5}{3} \\\\ & f^{\prime \prime}(x)=2 x-2 \\\\ & f^{\prime \prime}\left(\frac{1}{3}\right)=\frac{2}{3}-2<0 \text { point of maxima }\end{aligned}$


$\begin{aligned} \text { Area }_{\text {red }} & =\int_0^1 f(x) d x \\\\ & =\left[\frac{x^4}{12}-\frac{x^3}{3}+\frac{5 x^2}{18}+\frac{17 x}{36}\right]_0^1 \\\\ & =\frac{1}{12}-\frac{1}{3}+\frac{5}{18}+\frac{17}{36} \\\\ & =\frac{3-12+10+17}{36} \\\\ & =\frac{18}{36}=\frac{1}{2}=0.5 \\\\ \therefore (\text { Area })_{\text {green }} & =1-\frac{1}{2}=0.5\end{aligned}$

(A) $1-h=h-\frac{1}{2} \Rightarrow h=\frac{3}{4}, \frac{3}{4}>\frac{2}{3}$ option (A) is incorrect

(B) $h=\frac{1}{2}-h \Rightarrow h=\frac{1}{4} \Rightarrow$ option (B) is correct.

(C) $\int\limits_0^1 f(x) d x=\frac{1}{2}, \int\limits_0^1 \frac{1}{2} d x=\frac{1}{2} \Rightarrow \int\limits_0^1\left(f(x)-\frac{1}{2}\right) d x=0$

$\Rightarrow h=\frac{1}{2} \Rightarrow$ option (C) is correct.

(D) $\because$ Option (C) is correct $\Rightarrow$ option (D) is also correct.

Given,

$ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} \sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _{e}\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _{e}\left(\frac{\pi}{4}\right) & \tan \pi \end{array}\right| $

Here $\cos \left(\theta+\frac{\pi}{4}\right)=-\sin \left(\theta-\frac{\pi}{4}\right)$

and $\tan \left(\theta-\frac{\pi}{4}\right)=-\cot \left(\theta+\frac{\pi}{4}\right)$

and $\log _e\left(\frac{4}{\pi}\right)=-\log _e\left(\frac{\pi}{4}\right)$

Also $\sin \pi=-\cos \frac{\pi}{2}=\tan \pi=0$

$ \Rightarrow $ $ f(\theta)=\frac{1}{2}\left|\begin{array}{ccc} 2 & \sin \theta & 1 \\ 0 & 1 & \sin \theta \\ 0 & -\sin \theta & 1 \end{array}\right|+\left|\begin{array}{ccc} 0 & -\sin \left(\theta-\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & 0 & \log _{\mathrm{e}}\left(\frac{4}{\pi}\right) \\ -\tan \left(\theta-\frac{\pi}{4}\right) & -\log _{\mathrm{e}}\left(\frac{4}{\pi}\right) & 0 \end{array}\right| $

$ \begin{aligned} & f(\theta)=\left(1+\sin ^2 \theta\right)+0 \text { (skew symmetric) } \\\\ & g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1} \\\\ & =|\sin \theta|+|\cos \theta| \quad \\\\ & g(\theta) \in[1, \sqrt{2}] \text { for } \theta \in\left[0, \frac{\pi}{2}\right] \end{aligned} $

Again let $\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}-\sqrt{2})(\mathrm{x}-1)$

$ \begin{aligned} & 2-\sqrt{2}=\mathrm{k}(2-\sqrt{2})(2-1) \\\\ & \Rightarrow \mathrm{k}=1 \quad(\mathrm{P}(2)=2-\sqrt{2} \text { given }) \\\\ & \therefore \mathrm{P}(\mathrm{x})=(\mathrm{x}-\sqrt{2})(\mathrm{x}-1) \end{aligned} $

For option (A) $P\left(\frac{3+\sqrt{2}}{4}\right)<0$ Correct.

For option (B) $P\left(\frac{1+3 \sqrt{2}}{4}\right)<0$ Incorrect.

For option (C) $P\left(\frac{5 \sqrt{2}-1}{4}\right)>0$ Correct.

For option (D) $P\left(\frac{5-\sqrt{2}}{4}\right)>0$ Incorrect.

(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.

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.

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

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

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

$ = {\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.

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

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}} $