Functions

(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} $

Given that $f\left( x \right) = {\left( {x + 1} \right)^2} - 1,\,x \ge - 1$

Clearly ${D_f} = \left[ { -1 ,\infty } \right)$ but co-domain is not given

$\therefore$ $f(x)$ need not be necessarily onto.

But if $f(x)$ is onto then as $f\left( x \right)$ is one one also,

$(x+1)$ being something $+ve,$ ${f^{ - 1}}\left( x \right)$ will exist where

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

$ \Rightarrow x + 1 = \sqrt {y + 1} $

$\,\,\,\,\,$ $\left( { + ve} \right.$ square root as $x + 1 \ge 0$$\left. {} \right)$

$ \Rightarrow x = - 1 + \sqrt {y + 1} $

$ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1$

Then $\,\,\,\,\,\,\,\,$ $f\left( x \right) = {f^{ - 1}}\left( x \right)$

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

$ \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} $

$ \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right)$

$ \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0$

$ \Rightarrow x = - 1,0$

$\therefore$ The statement -$1$ is correct but statement- $2$ is false.

Given that $f\left( x \right) = {x^3} + 5x + 1$

$\therefore$ $\,\,\,\,\,$ $f'\left( x \right) = 3{x^2} + 5 > 0,\,\,\,\forall x \in R$

$ \Rightarrow f\left( x \right)\,\,$ is strictly increasing on $R$

$ \Rightarrow f\left( x \right)$ is one one

$\therefore$ $\,\,\,\,\,\,\,$ Being a polynomial $f(x)$ is cont. and inc.

on $R$ with $\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = - \infty $

and $\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = \infty $

$\therefore$ $\,\,\,\,\,\,\,$ Range of $f = \left( { - \infty ,\infty } \right) = R$

Hence $f$ is onto also, So, $f$ is one and onto $R.$

Clearly $f$ is one one and onto, so invertible

Also $f\left( x \right) = 4x + 3 = y \Rightarrow x = {{y - 3} \over 4}$

$\therefore$ $\,\,\,\,g\left( y \right) = {{y - 3} \over 4}$

$f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)$

$f\left( x \right)$ is defined if $ - 1 \le \left( {{x \over 2} - 1} \right) \le 1$ and $\cos \,x > 0$

or $\,\,\,\,0 \le {x \over 2} \le 2\,\,$ and $\,\, - {\pi \over 2} < x < {\pi \over 2}$

or $\,\,\,$ $0 \le x \le 4$ $\,\,$ and $\,\, - {\pi \over 2} < x < {\pi \over 2}$

$\therefore$ $\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)$

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$

Given $\,\,f\left( x \right) = {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x$

for $x \in \left( { - 1,1} \right)$

If$\,\,x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 4},{\pi \over 4}} \right)$

$ \Rightarrow 2{\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)$

Clearly, range of $f\left( x \right) = \left( { - {\pi \over 2},{\pi \over 2}} \right)$

For $f$ to be onto, co-domain $=$ range

$\therefore$ Co-domain of function $ = B = \left( { - {\pi \over 2},{\pi \over 2}} \right).$

$f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right)$

$ = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right)$

$ = f\left( a \right)f\left( {x - a} \right) - f\left( x \right)$

$ = - f\left( x \right)$

$\left[ {} \right.$ as $x = 0,y = 0,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right)$

$\,\,\,\,\,\,\,\, \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( a \right) = \left. 0 \right]$

$ \Rightarrow f\left( {2a - x} \right) = - f\left( x \right)$

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

$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$ is defined

if $(i)$ $\,\,\, - 1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4$

and $(ii)$ $9 - {x^2} > 0 \Rightarrow - 3 < x < 3$

Taking common solution of $\left( i \right)$ and $\left( {ii} \right),$

we get $2 \le x < 3$

$\therefore$ Domain $ = \left[ {2,\left. 3 \right)} \right.$

$f\left( x \right)$ is onto

$\therefore$ $S=$ range of $f(x)$

Now $f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1$

$ = 2\sin \left( {x - {\pi \over 3}} \right) + 1$

As $1 - \le \sin \left( {x - {\pi \over 3}} \right) \le 1$

$ - 1 \le 2\sin \left( {x - {\pi \over 3}} \right) + 1 \le 3$

$\therefore$ $f\left( x \right) \in \left[ { - 1,3} \right] = S$

The range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.

The domain of $f(x)$ is the set of all real numbers such that

(i) $x \geq 3$ (since the permutation function is only defined for non-negative integers)

(ii) 7 - x > 0 $ \Rightarrow $ x < 7

(iii) x - 3 $ \le $ 7 - x $ \Rightarrow $ 2x $ \le $ 10 $ \Rightarrow $ x $ \le $ 5.

To find the range of $f(x)$, we need to consider what values the expression ${}^{7-x}P_{x-3}$ can take as $x$ varies over its domain.

For $x=3$, we have ${}^{7-x}P_{x-3} = {}^{4}P_{0} = 1$.

For $x=4$, we have ${}^{7-x}P_{x-3} = {}^{3}P_{1} = 3$.

For $x=5$, we have ${}^{7-x}P_{x-3} = {}^{2}P_{2} = 2$.

Therefore, the range of $f(x)$ is the set of all non-negative integers less than or equal to 3, i.e., ${1, 2, 3}$.

Let us consider a graph symm. with respect to line $x=2$ as shown in the figure.



From the figure

$f\left( {{x_1}} \right) = f\left( {{x_2}} \right),$

where ${x_1} = 2 - x$ and ${x_2} = 2 + x$

$\therefore$ $\,\,\,\,f\left( {2 - x} \right) = f\left( {2 + x} \right)$

We have $f:N \to I$

If $x$ and $y$ are two even natural numbers,

then $f\left( x \right) = f\left( y \right) \Rightarrow {{ - x} \over 2} = {{ - y} \over 2} \Rightarrow x = y$

Again if $x$ and $y$ are two odd natural numbers then

$f\left( x \right) = f\left( y \right) \Rightarrow {{x - 1} \over 2} = {{y - 1} \over 2} \Rightarrow x = y$

$\therefore$ $f$ is onto.

Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.

$\therefore$ $f$ is onto.

Hence $f$ is one one and onto both.

$f\left( x \right) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$

$f\left( { - x} \right) = \log \left\{ { - x + \sqrt {{x^2} + 1} } \right\}$

$ = \log \left\{ {{{ - {x^2} + {x^2} + 1} \over {x + \sqrt {{x^2} + 1} }}} \right\}$

$ = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f\left( x \right)$

$ \Rightarrow f\left( x \right)\,\,\,\,$ is an odd function.

$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right).$

Function should be $f(x)=mx$

$f\left( 1 \right) = 7;$

$\therefore$ $m=7,$ $f\left( x \right) = 7x$

$\sum\limits_{r = 1}^n {f\left( r \right)} = 7\sum\limits_1^n {r = {{7n\left( {n + 1} \right)} \over 2}} $

$f\left( x \right) = {3 \over {4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)$

$4 - {x^2} \ne 0;\,\,\,{x^3} - x > 0;$

$x \ne \pm \sqrt 4 $ and $ - 1 < x < 0$

or $\,\,\,1 < x < \infty $



$\therefore$ $D = \left( { - 1,0} \right) \cup \left( {1,\infty } \right) - \left\{ {\sqrt 4 } \right\}$

$D = \left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right).$

The period of ${\sin ^2}\theta $ is = $\pi $

Note :
(1) When $n$ is odd then the period of ${\sin ^n}\theta $, ${\cos ^n}\theta $, ${\csc ^n}\theta $, ${\sec ^n}\theta $ = $2\pi $

(2) When $n$ is even then the period of ${\sin ^n}\theta $, ${\cos ^n}\theta $, ${\csc ^n}\theta $, ${\sec ^n}\theta $ = $\pi $

(3) When $n$ is even/odd then the period of ${\tan ^n}\theta $, ${\cot ^n}\theta $ = $\pi $

(3) When $n$ is even/odd then the period of $\left| {{{\sin }^n}\theta } \right|$, $\left| {{{\cos }^n}\theta } \right|$, $\left| {{{\csc }^n}\theta } \right|$, $\left| {{{\sec }^n}\theta } \right|$, $\left| {{{\tan }^n}\theta } \right|$, $\left| {{{\cot }^n}\theta } \right|$ = $\pi $

$f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {{x \over 3}} \right)} \right)$ exists

if $\,\,\,\, - 1 \le {\log _3}\left( {{x \over 3}} \right) \le 1$

$ \Leftrightarrow {3^{ - 1}} \le {x \over 3} \le {3^1}$

$ \Leftrightarrow 1 \le x \le 9$

or $\,\,\,\,x \in \left[ {1,9} \right]$

$\sqrt x $ is non periodic function and $\cos \left( {something} \right)$ is a periodic function so here in $\cos \sqrt x $ $ \to $ inside periodic function there is non periodic function which always produce non periodic function.

${{{\cos }^2}x}$ is a periodic function with period $\pi $

Note :
(1) When $n$ is odd then the period of ${\sin ^n}\theta $, ${\cos ^n}\theta $, ${\csc ^n}\theta $, ${\sec ^n}\theta $ = $2\pi $

(2) When $n$ is even then the period of ${\sin ^n}\theta $, ${\cos ^n}\theta $, ${\csc ^n}\theta $, ${\sec ^n}\theta $ = $\pi $

(3) When $n$ is even/odd then the period of ${\tan ^n}\theta $, ${\cot ^n}\theta $ = $\pi $

(3) When $n$ is even/odd then the period of $\left| {{{\sin }^n}\theta } \right|$, $\left| {{{\cos }^n}\theta } \right|$, $\left| {{{\csc }^n}\theta } \right|$, $\left| {{{\sec }^n}\theta } \right|$, $\left| {{{\tan }^n}\theta } \right|$, $\left| {{{\cot }^n}\theta } \right|$ = $\pi $

$\cos \sqrt x + {\cos ^2}x$ = non periodic function + periodic function = non periodic function