\(n(A \cap B)=1\)

\(f(x) =\frac{\left(x^{2}+\cos x\right)\left(1+x^{4}\right)}{(x-\sin x)\left(2 x-x^{3}\right)}+e^{-|x|} \)
\(f(-x) =\frac{\left[(-x)^{2}+\cos (-x)\right]\left[1+(-x)^{4}\right]}{[-x-\sin (-x)]\left[-2 x-\left(-x^{3}\right)\right]}+e^{+x \mid} \)
\(=\frac{\left(x^{2}+\cos x\right)\left(1+x^{4}\right)}{(x-\sin x)\left(2 x-x^{3}\right)}+e^{-|x|} \)
\(=f(x) \)
Hence f(x) is an even function

\(f(x) =\sin x+\cos x \)
\(f(-x) =\sin (-x)+\cos (-x) \)
\(=-\sin x+\cos x \)
\(-f(-x) =\sin x-\cos x \)
\(f(x) \neq-f(-x) \)
f(x) is neither odd function nor even function.

\(\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} \text { is defined by }\)
\(\mathrm{f}(x)=1-|x|\)
\(\text { The range is }(-\infty, 1] \text { as } f(-\infty)=-\infty\)
\(f(0)=1\)
\(f(\infty) =-\infty\)

\(\text { i) Let } y=x \text { then } x=y \Rightarrow f^{-1}(x)=x\)
\(\text { ii) Let } y=x^{2} \text { then } y=\sqrt{x} \Rightarrow f^{-1}(x)=\sqrt{x}\)
\(\text { iii) Let } y=8 \sqrt{x} \text { then } \frac{y^{2}}{64}=x \Rightarrow \mathrm{f}^{-1}(x)=\frac{x^{2}}{64}\)
\(\therefore \mathrm{f}^{-1}(x)= \begin{cases}x & x<1 \\ \sqrt{x} & 1 \leq x \leq 16 \\ \frac{x^{2}}{64} & x>16\end{cases}\)

It is not a function since it has two images

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)