Functions

(fog) (x)   =   x

$ \Rightarrow $$\,\,\,$ f (g(x))   =   x

$ \Rightarrow $$\,\,\,$ f (3¹⁰. x $-$ 1)   =   x    [ as    g(x) = 3¹⁰. x $-$ 1]

$ \Rightarrow $$\,\,\,$ 2¹⁰ . (3¹⁰ . x $-$ 1) + 1   =   x

$ \Rightarrow $$\,\,\,$ 3¹⁰ . x $-$ 1 + 2^$-$10   =   x . 2^$-$10   [dividing by 2¹⁰]

$ \Rightarrow $$\,\,\,$3¹⁰ . x $-$ 2^$-$10 . x   =   1 $-$ 2^$-$10

$ \Rightarrow $$\,\,\,$ x (3¹⁰ $-$ 2^{$-$ 10})   =   1$-$ 2^$-$10

$ \Rightarrow $ $\,\,\,$ x = ${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$

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

$ \therefore $ $f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)$

$ \therefore $ f(x) is many-one function.

Now let y = f(x) = ${x \over {1 + {x^2}}}$

$ \Rightarrow $ y + x²y = x

$ \Rightarrow $ yx - x + y = 0

As x $ \in $ R

$ \therefore $ (-1)² - 4(y)(y) $ \ge $ 0

$ \Rightarrow $ 1 - 4y² $ \ge $ 0

$ \Rightarrow $ y $ \in $ $\left[ { - {1 \over 2},{1 \over 2}} \right]$

$ \therefore $ Range = Codomain = $\left[ { - {1 \over 2},{1 \over 2}} \right]$

So, f(x) is surjective.

$ \therefore $ f(x) is surjective but not injective

f(x) = ax² + bx + c

f(1) = a + b + c = 3 $ \Rightarrow $ f (1) = 3

Now f(x + y) = f(x) + f(y) + xy ...(1)

Put x = y = 1 in eqn (1)

f(2) = f(1) + f(1) + 1

= 2f(1) + 1

$ \Rightarrow $ f(2) = 7

Similarly f(3) = 12

f(4) = 18

$\sum\limits_{n = 1}^{10} {f(n)} $ = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 + 63 + 75 = 330

As   f_n+1(x) = f₀(f_n(x))

$ \therefore $   f₁(x) = f₀₊₁(x) = f₀(f₀(x)) = ${1 \over {1 - {1 \over {1 - x}}}}$ = ${{x - 1} \over x}$

f₂(x) = f₁₊₁(x) = f₀(f₁(x)) = ${1 \over {1 - {{x - 1} \over x}}}$ = x

f₃(x) = f₂₊₁(x) = f₀(f₂(x)) = ${1 \over {1 - x}}$

f₄(x) = f₃₊₁(x) = f₀(f₃(x)) = ${1 \over {1 - {1 \over {1 - x}}}}$ = ${{x - 1} \over x}$

$ \therefore $ f₀(x) = f₃(x) = f₆(x) = . . . . . = ${1 \over {1 - x}}$

f₁(x) = f₄(x) = f₇(x) = . . . . . .= ${{x - 1} \over x}$

f₂(x) = f₅(x) = f₈(x) = . . . . . . = x

So, f₁₀₀(3) = ${{3 - 1} \over 3}$ = ${2 \over 3}$

f₁$\left( {{2 \over 3}} \right)$ = ${{{2 \over 3} - 1} \over {{2 \over 3}}}$ = $-$ ${1 \over 2}$

f₂$\left( {{3 \over 2}} \right)$ = ${{3 \over 2}}$

$ \therefore $    f₁₀₀(3) + f₁$\left( {{2 \over 3}} \right)$ + f₂$\left( {{3 \over 2}} \right)$

= ${2 \over 3}$ $-$ ${1 \over 2}$ + ${3 \over 2}$

= ${5 \over 3}$

We have, $f(x)+2 f\left(\frac{1}{x}\right)=3 x, \quad x \neq 0$   $\ldots$ (i)

On replacing $x$ by $\frac{1}{x}$ in the above equation, we get

$ \begin{aligned} & f\left(\frac{1}{x}\right)+2 f(x) =\frac{3}{x} \\\\ \Rightarrow & \,\, 2 f(x)+f\left(\frac{1}{x}\right) =\frac{3}{x} \,\,\,\,\,...(ii) \end{aligned} $

On multiplying Eq. (ii) by 2, we get

$ 4 f(x)+2 f\left(\frac{1}{x}\right)=\frac{6}{x}\quad...(iii) $

and subtracting Eq. (i) from Eq. (iii), we get

$[4 f(x)+2 f\left(\frac{1}{x}\right)] - [f(x)+2 f\left(\frac{1}{x}\right)]=\frac{6}{x} - 3x$

$\Rightarrow {3 f(x)=\frac{6}{x}-3 x}$

$\Rightarrow f(x)=\frac{2}{x}-x$

Now, consider $\quad f(x)=f(-x)$

$ \begin{aligned} &\Rightarrow \frac{2}{x}-x =-\frac{2}{x}+x \\\\ &\Rightarrow \frac{4}{x} =2 x \\\\ &\Rightarrow 2 x^2 =4 \\\\ &\Rightarrow x^2 =2 \\\\ &\Rightarrow x =\pm \sqrt{2} \end{aligned} $

Hence, $S$ contains exactly two elements.

$f\left( x \right) = {1 \over {\sqrt {\left| x \right| - x} }},\,\,$ define if $\,\,\,\left| x \right| - x > 0$

$ \Rightarrow \left| x \right| > x,\,\,\, \Rightarrow x < 0$

Hence domain of $f\left( x \right)$ is $\left( { - \infty ,0} \right)$

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 $\,\, - {\pi \over 2} < x < {\pi \over 2}$

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

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)$

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