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}} $

Here, $f(x) = {{b - x} \over {1 - bx}}$

where, 0 < b < 1, 0 < x < 1

For function to be invertible it should be one-one onto.

$\therefore$ Check range :

Let $f(x) = y \Rightarrow y = {{b - x} \over {1 - bx}}$

$ \Rightarrow y - bxy = b - x \Rightarrow x(1 - by) = b - y$

$ \Rightarrow x = {{b - y} \over {1 - by}}$

where, 0 < x < 1

$\therefore$ $0 < {{b - y} \over {1 - by}} < 1$

${{b - y} \over {1 - by}} > 0$ and ${{b - y} \over {1 - by}} < 1$

$ \Rightarrow y < b$ or $y > {1 \over b}$ .... (i)

${{(b - 1)(y + 1)} \over {1 - by}} < - 1 < y < {1 \over b}$ ..... (ii)

From Eqs. (i) and (ii), we get

$y \in \left( { - 1,{1 \over b}} \right) \subset $ Codomain

Thus, f(x) is not invertible.

The given function f : R $ \to $ R is

$f(x) = |x|(x - \sin x)$ .....(i)

$ \because $ The function 'f' is a odd and continuous function and as $\mathop {\lim }\limits_{x \to \infty } f(x) = \infty $ and $\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty $, so range is R, therefore, 'f' is a onto function.

$ \because $ $f(x) = \left[ \matrix{ x(x - \sin x),\,x \ge \,0 \hfill \cr - x(x - \sin x),\,x\, < \,0 \hfill \cr} \right.$

$ \therefore $ $f'(x) = \left[ \matrix{ 2x - \sin x - x\cos x,\,x > \,0 \hfill \cr - 2x + \sin x + x\cos x,\,x\, < \,0 \hfill \cr} \right.$

$\left[ \matrix{ (x - \sin x) + x(1 - \cos x),\,x > \,0 \hfill \cr ( - x + \sin x) - x(1 - \cos x),\,x\, < \,0 \hfill \cr} \right.$

$ \because $ for x > 0, x $-$ sin x > 0 and x(1 $-$ cos x) > 0

$ \therefore $ $f'(x) > 0\forall x \in (0,\infty )$

$ \Rightarrow $ f is strictly increasing function. $\forall x \in (0,\infty )$.

Similarly, for x < 0, $-$x + sin x > 0 and ($-$ x) (1 $-$ cos x) > 0, therefore, $f'(x) > 0\forall x \in ( - \infty ,0)$

$ \Rightarrow $ f is strictly increasing function, $\forall x \in $(0, $\infty $)

Therefore 'f' is a strictly increasing function for x$ \in $R and it implies that f is one-one function.

We have,

${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$

$ \therefore $ ${E_1} = {x \over {x - 1}} > 0$



${E_1} = x \in ( - \infty ,0) \cup (1,\infty )$

and ${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$

${E_2} = - 1 \le {\log _e}{x \over {x - 1}} \le 1$

$ \Rightarrow {e^{ - 1}} \le {x \over {x - 1}} \le e$

Now, ${x \over {x - 1}} \ge {e^{ - 1}} \Rightarrow {x \over {x - 1}} - {1 \over e} \ge 0$

$ \Rightarrow {{ex - x + 1} \over {e(x - 1)}} \ge 0 \Rightarrow {{x(e - 1) + 1} \over {(x - 1)e}} \ge 0$



$ \Rightarrow x \in \left( { - \infty ,\,{1 \over {1 - e}}} \right) \cup (1,\infty )$

Also, ${x \over {x - 1}} \le e$

$ \Rightarrow {{(e - 1)x - e} \over {x - 1}} \ge 0$



$ \Rightarrow x \in ( - \infty ,\,1) \cup \left[ {{e \over {e - 1}},\infty } \right)$

So, ${E_2} = \left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$

$ \therefore $ The domain of f and g are

$\left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$

and Range of ${x \over {x - 1}}$ is R⁺ $-$ {1}

$ \Rightarrow $ Range of f is R $-$ {0} or ($-$$\infty $, 0) $ \cup $ (0, $\infty $)

Range of g is $\left[ { - {\pi \over 2},{\pi \over 2}} \right] - \{ 0\} $ or $\left[ { - {\pi \over 2},0} \right) \cup \left( {0,{\pi \over 2}} \right]$

Now, P $ \to $ 4, Q $ \to $ 2, R $ \to $ 1, S $ \to $ 1

Hence, option (a) is correct answer.

${N_i} = {}^5{C_k} \times {}^4{C_{5 - k}}$

$\eqalign{ & {N_1} = 5 \times 1 \cr & {N_2} = 10 \times 4 \cr & {N_3} = 10 \times 6 \cr & {N_4} = 5 \times 4 \cr & {N_5} = 1 \cr & {N_1} + {N_2} + {N_3} + {N_4} + {N_5} = 126 \cr} $

$\text { Given } f_2(x)=x^2 \text { for } x \in[0, \infty) \text { and co-domain } \mathrm{R} \text {. }$

It is clear that $f_2(x)$ is continuous, differential and one-one but not onto because range of $f_2(x)$ is $[0, \infty)$ and given co-domain is R.

Hence $S$ match with 4.

Given, $f_3(x)=\left\{\begin{array}{cc}\sin x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{array}\right.$. Domain of $f_2(x)$ is $R$ and co-domain $R$.

Draw the graph of $f_3(x)$

Range of $f_3(x) \in(-\infty, 1] \neq$ co-domain of $f_3(x)$

So, $f_3(x)$ is not onto.

$\because \sin x$ is not one-one function for $x \geq 0$

$\therefore f_3(x)$ is not one-one function.

We know $y=\sin x$ and $y=x$ both are continuous in the given domain.

At $\quad x=0$

$\begin{aligned} \lim _{x \rightarrow 0^{+}} f_3(x) & =\lim _{x \rightarrow 0^{+}} \sin x=0 \\ \lim _{x \rightarrow 0^{-}} f_3(x) & =\lim _{x \rightarrow 0^{-}} x=0 \\ f_3(0) & =\sin 0=0 \\ \Rightarrow \lim _{x \rightarrow 0^{+}} f_3(x) & =\lim _{x \rightarrow 0^{-}} f_3(x)=f_3(0) \end{aligned}$

$\therefore f_3(x)$ is continuous for all $x \in \mathrm{R}$.

We know that $y=\sin x$ and $y=x$ both are differentiable in the given domain R.H.D at $x=0$ is

$\lim _\limits{h \rightarrow 0} \frac{f_3(0+h)-f_3(0)}{h}=\lim _\limits{h \rightarrow 0} \frac{\sin h-0}{h}=1$

L.H.D at $x=0$ is

$\lim _\limits{h \rightarrow 0} \frac{f_3(0-h)-f_3(0)}{-h}=\lim _\limits{h \rightarrow 0} \frac{-h-0}{-h}=1$

$\Rightarrow$ R.H.D $=$ L.H.D at $x=0$

$\therefore f_3(x)$ is differentiable for all $x \in R$.

Hence, Q match with 3.

Now, $f_2{ }^{\circ} f_1(x)= \begin{cases}x^2, & x<0 \\ e^{2 x}, & x \geq 0\end{cases}$

Draw the graph of $y=f_2{ }^{\circ} f_1(x)$

It is clear that $f_2{ }^{\circ} f_1(x)$ is neither continuous nor one-one.

Hence, R match with 2.

Given, $f_4: \mathrm{R} \rightarrow[0, \infty)$ and

$f_4(x)=\left\{\begin{array}{l} f_2\left(f_1(x)\right), \text { if } x<0 \\ f_2\left(f_1(x)\right)-1, \text { if } x \geq 0 \end{array}\right.$

Draw the graph of $y=f_4(x)$

Range of $f_4(x)=$ co-domain of $f_4(x)=[0, \infty)$

$\therefore \quad f_4(x)$ is an onto function

$f_4(x)$ is continuous, on to but not one-one.

Hence, P match with 1.

Hint:

(i) If $f(x)$ is continuous at $x=a$, if and only if $\lim _\limits{x \rightarrow a^{+}} f(x)=\lim _\limits{x \rightarrow a^{-}} f(x)=f(a)$.

(ii) If $f(x)$ is differentiable at $x=a$ if and only if $\lim _\limits{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\lim _\limits{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$.

(iii) If Range of $f(x)=\operatorname{co-domain}$ of $f(x)$, then $f(x)$ is called an onto function.

(iv) If $f\left(x_1\right)=f\left(x_2\right)$ is possible only for $x_1=x_2$, then $f(x)$ is called an one-one function.

The function $f:[0,3] \to [1,29]$, defined by $f(x) = 2x^3 - 15x^2 + 36x + 1$, is :

First, calculate the derivative of $ f $ :

$ f'(x) = 6x^2 - 30x + 36 $

Which simplifies to :

$ f'(x) = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3) $

For the given domain $[0, 3]$, the function $ f(x) $ is both increasing and decreasing. Therefore, $ f(x) $ is many-to-one.

Setting $ f'(x) = 0 $ gives the critical points :

$ x = 2, 3 $

We now evaluate the function at these points within the domain to determine the range :

$ f(0) = 1 $

$ f(2) = 29 $

$ f(3) = 28 $

Thus, the range of $ f $ is $[1, 29]$, indicating that the function is onto.

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

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

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

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

(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 $x \ge 2$.

To solve this problem, we need to determine the total number of unordered pairs of disjoint subsets of the set $ S = \{1, 2, 3, 4\} $.

Understanding the Problem:

Disjoint Subsets: Two subsets $ A $ and $ B $ are disjoint if they have no elements in common, i.e., $ A \cap B = \emptyset $.

Unordered Pairs: Pairs where $ \{A, B\} $ is considered the same as $ \{B, A\} $.

Approach:

Counting Ordered Pairs of Disjoint Subsets:

For each element in $ S $, it can be in:

Subset $ A $ only,

Subset $ B $ only,

Neither $ A $ nor $ B $.

Note: An element cannot be in both $ A $ and $ B $ because $ A $ and $ B $ are disjoint.

Therefore, each element has 3 choices.

Total number of ordered pairs $ (A, B) $ is $ 3^4 = 81 $.

Identifying Ordered Pairs where $ A = B $:

Since $ A $ and $ B $ are disjoint and $ A = B $, the only possibility is when both are the empty set.

So, there's 1 ordered pair where $ A = B = \emptyset $.

Calculating Unordered Pairs:

Ordered Pairs with $ A \ne B $: $ 81 - 1 = 80 $.

Unordered Pairs from Ordered Pairs with $ A \ne B $: Each unordered pair corresponds to 2 ordered pairs (since $ (A, B) $ and $ (B, A) $ are different but represent the same unordered pair).

Number of unordered pairs from $ A \ne B $ is $ \frac{80}{2} = 40 $.

Include the pair where $ A = B = \emptyset $: Add $ 1 $.

Total Unordered Pairs: $ 40 + 1 = 41 $.

Conclusion:

The total number of unordered pairs of disjoint subsets of $ S $ is 41.

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Answer: Option D

Given, $f(x) = 4{x^3} + 3{x^2} + 2x + 1$

$f'(x) = 2(6{x^2} + 3x + 1)$

$D = 9 - 24 < 0$

Hence, f(x) = 0 has only one real root.

$f\left( { - {1 \over 2}} \right) = 1 - 1 + {3 \over 4} - {4 \over 8} > 0$

$f\left( { - {3 \over 4}} \right) = 1 - {6 \over 4} + {{27} \over {16}} - {{108} \over {64}}$

$ = {{64 - 96 + 108 - 108} \over {64}} < 0$

f(x) changes its sign in $\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$,

hence f(x) = 0 has a root in $\left( { - {3 \over 4},{{ - 1} \over 2}} \right)$.

$ \begin{aligned} & f(x)=e^{x^2}+e^{-x^2} \\\\ & \begin{aligned} f^{\prime}(x) & =e^{x^2} \frac{d}{d x}\left(x^2\right)+e^{-x^2} \frac{d}{d x}\left(-x^2\right) \\\\ & =e^{x^2}(2 x)+e^{-x^2}(-2 x) \\\\ & =2 x\left(e^{x^2}-e^{-x^2}\right) \geq 0 \forall x \in[0,1] \\\\ g(x) & =x e^{x^2}+e^{-x^2} \\\\ h(x) & =x^2 e^{x^2}+e^{-x^2} \end{aligned} \end{aligned} $

Clearly for $0 \leq x \leq 1 \quad f(x) \geq g(x) \geq h(x)$

$\because f(1)=g(1)=h(1)=e+\frac{1}{e}$ and $f(1)$ is greatest

$ \because a=b=c=e+\frac{1}{e} $

If $f''(x)=-f(x)$

$ f'(x)=g(x)$

$ f''(x) f'(x)=-f(x) \cdot g(x)$

$ f''(x) f'(x)+f(x) \cdot g(x)=0$

$f''(x) f'(x)+f(x) \cdot f'(x)=0 \quad \because f'(x)=g(x)$

$\frac{d}{d x}\left[\left(f^{\prime}(x)\right)^{2}+(f(x))^{2}\right]=c$

$\Rightarrow \quad\left[f^{\prime}(x)\right]^{2}+[f(x)]^{2}=c$

$\Rightarrow \quad[f(x)]^{2}+[g(x)]^{2}=$ Constant

$\Rightarrow \quad f(x)=$ Constant

$\because \quad f(5)=5$

Therefore $ f(10)=5$

Range of $t:\left[-\frac{\pi}{2}, \frac{-\pi}{10}\right] \cup\left[\frac{3 \pi}{10}, \frac{\pi}{2}\right]$.

$2 \sin t=\frac{1-2 x+5 x^{2}}{3 x^{2}-2 x-1}, t \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Let, $y=2 \sin t$

then, $y=\frac{1-2 x+5 x^{2}}{3 x^{2}-2 x-1}$

$\Rightarrow y\left(3 x^{2}-2 x-1\right)=1-2 x+5 x^{2}$

$\Rightarrow x^{2}(5-3 y)-2 x(1-y)+(1+y)=0$

Here, the roots of above quadratic must be real then Discriminant $\geq 0$

$[-2(1-y)]^{2}-4(5-3 y)(1+y) \geq 0$

$\Rightarrow 1+y^{2}-2 y-\left(5-3 y^{2}+2 y\right) \geq 0$

$\Rightarrow 4 y^{2}-4 y-4 \geq 0$

$\Rightarrow y^{2}-y-1 \geq 0$

$\Rightarrow y \geq \frac{1 \pm \sqrt{5}}{2}$

$\Rightarrow\left[y-\left(\frac{1+\sqrt{5}}{2}\right)\right]\left[y-\left(\frac{1-\sqrt{5}}{2}\right)\right] \geq 0$

$\Rightarrow y \leq\left(\frac{1-\sqrt{5}}{2}\right) \text { and } y \geq\left(\frac{\sqrt{5}+1}{2}\right)$

$\Rightarrow 2 \sin t \leq\left(\frac{1-\sqrt{5}}{2}\right) \text { and } 2 \sin t \geq\left(\frac{\sqrt{5}+1}{2}\right)$

$\Rightarrow \sin t \leq\left(\frac{1-\sqrt{5}}{4}\right) \text { and } \sin t \geq\left(\frac{\sqrt{5}+1}{4}\right)$

$\Rightarrow t \leq-\sin 18^{\circ}$ and $t \geq \sin 54^{\circ}$

or $t \leq-\sin \frac{\pi}{10}$ and $t \geq \sin \left(\frac{3 \pi}{10}\right)$

So, after plotting the point on graph

Range of $t:\left[-\frac{\pi}{2}, \frac{-\pi}{10}\right] \cup\left[\frac{3 \pi}{10}, \frac{\pi}{2}\right]$