$$\left| {\sin \frac{x}{2}} \right|$$ is a periodic function of period $$2\pi $$ and $$\left| {\cos \,x} \right|$$ is a periodic function of period $$\pi .$$
2.
The domain of $$f\left( x \right) = \frac{1}{{\sqrt {\left| {\cos \,x} \right| + \cos \,x} }}$$ is :
A.
$$\left[ { - 2n\pi ,\,2n\pi } \right]$$
B.
$$\left( {2n\pi ,\,\overline {2n + 1} \,\pi } \right)$$
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 $$f\left( x \right)$$ 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 one and onto $$R$$.
6.
The domain of the function $$\sqrt {{x^2} - 5x + 6} + \sqrt {2x + 8 - {x^2}} $$ is :
A.
$$\left[ {2,\,3} \right]$$
B.
$$\left[ { - 2,\,4} \right]$$
C.
$$\left[ { - 2,\,2} \right] \cup \left[ {3,\,4} \right]$$
D.
$$\left[ { - 2,\,1} \right] \cup \left[ {2,\,4} \right]$$
$$f\left( x \right) = \sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} + \sqrt { - \left( {x - 4} \right)\left( {x + 2} \right)} $$
The first part is real outside $$\left( {2,\,3} \right)$$ and the second is real in $$\left[ { - 2,\,4} \right]$$ so that the domain is $$\left[ { - 2,\,2} \right] \cup \left[ {3,\,4} \right].$$
7.
The domain of the function $$f\left( x \right) = \frac{1}{{\sqrt {\left| x \right| - x} }}$$ is
$$\eqalign{
& f\left( x \right) = \frac{1}{{\sqrt {\left| x \right| - x} }},{\text{define}}\,{\text{if}}\,\left| x \right| - x > 0 \cr
& \Rightarrow \left| x \right| > x, \Rightarrow x < 0 \cr} $$
Hence domain of $$f\left( x \right)$$ is $$\left( { - \infty ,0} \right)$$
8.
If $$f\left( x \right) = {x^n},\,n\, \in \,N$$ and $$\left( {g\,o\,f} \right)\left( x \right) = ng\left( x \right)$$ then $$g\left( x \right)$$ can be :
$$g\left\{ {f\left( x \right)} \right\} = g\left( {{x^n}} \right) = ng\left( x \right).$$ Also $$\log \,{x^n} = n\,\log \,\left| x \right|.$$ So, $$g\left( x \right) = \log \,\left| x \right|$$ is possible.
9.
If $$f\left( x \right) = 4x - {x^2},\,x\, \in \,R,$$ then $$f\left( {a + 1} \right) - f\left( {a - 1} \right)$$ is equal to :
10.
Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \geqslant - 1.$$ Then the set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\}$$ is