Functions

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

$f\left( x \right)$ 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