Functions

(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.

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

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.

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

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}} $

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.

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.

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

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

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

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

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