SOLUTION : (i) $f(x)=1-|x-2| \quad \Rightarrow \quad|x-2| \in[0, \infty) \quad \therefore \quad f(x) \in(-\infty, 1]$
(ii) $\quad f(x)=\frac{1}{\sqrt{16-4^{x^2-x}}} \quad$ for function to be defined
$\begin{array}{lll}16-4^{x^2-x}>0 & \Rightarrow & 16>4^{x^2-x} \\ x^2-x-2<0 & \Rightarrow \quad(x-2)(x+1)<0\end{array} \quad \Rightarrow \quad 4^{x^1-x<4^2}$
so $n \in(-1,2) \quad \Rightarrow \quad n \in(-1,2) \quad \Rightarrow \quad x^2-x \in\left[-\frac{1}{4}, 2\right]$
$4^{x^2-x} \in\left[\frac{1}{\sqrt{2}}, 16\right) \quad \Rightarrow \quad \sqrt{16-4^{x^2-x}} \in\left(0, \sqrt{16-\frac{1}{\sqrt{2}}}\right]$
So range of $\frac{1}{\sqrt{16-4^{x^2-x}}} \in\left[\frac{1}{\sqrt{16-\frac{1}{\sqrt{2}}}}, \infty\right]$
(iii) $f(x)=\frac{1}{2-\cos 3 x} \quad$ range of $\cos 3 x$ is $[-1,1] \quad \cos 3 x[-1,1]$
$\therefore \quad f(x) \in\left[\frac{1}{3}, 1\right]$
(iv) $f(x)=\frac{x+2}{x^2-8 x-4}=y \quad \Rightarrow \quad x+2=y x^2-8 y x-4 y \\$
$\text { or } \quad y x^2-x(8 y+1)-(4 y+2)=0 \\$
$\text { for } x \text { to be real } D \geq 0 \\$
$(8 y+1)^2+4 y(4 y+2) \geq 0 \quad \Rightarrow \quad 64 y^2+16 y+1+16 y^2+8 y \geq 0 \\$
$80 y^2+24 y+1 \geq 0$
$y \in\left(-\infty,-\frac{1}{4}\right] \cup\left[-\frac{1}{20}, \infty\right)$
(v) $f(x)=\frac{x^2-2 x+4}{x^2+2 x+4}=y \Rightarrow x^2-2 x+4=y x^2+2 x y+4 y \\$
$x^2(1-y)-2 x(1+y)+4(1-y)=0 \\$
$D \geq 0$
$4(1+y)^2-16(1-y)^2 \geq 0 \quad \text { or } \quad y \in\left[\frac{1}{3}, 3\right]$
(vi) $f(x)=3 \sin \sqrt{\frac{\pi^2}{16}-x^2} \Rightarrow D: x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \quad \Rightarrow \quad \sqrt{\frac{\pi^2}{16}-x^2} \in\left[0, \frac{\pi}{4}\right]$
$\therefore \quad f(x) \in\left[0, \frac{3}{\sqrt{2}}\right]$
(vii) $f(x)=x^3-2 x^2+5=\left(x^2-1\right)^2+4 \quad \Rightarrow \quad R:[4, \infty)$
(viii) $\quad f(x)=x^3-12 x, x \in[-3,1]=x\left(x^2-12\right) \Rightarrow f^{\prime}(x)=3 x^2-12=0$ or $\quad x= \pm 2$
$\mathrm{R}:[-11,16]$
(ix) $\quad f(x)=\sin ^2 x+\cos ^4 x=\sin ^2 x+1+\sin ^4 x-2 \sin ^2 x=\sin ^4 x-\sin ^2 x+1=\left(\sin ^2 x-\frac{1}{2}\right)+\frac{3}{4}$
$R:\left[\frac{3}{4}, 1\right]$.
$f(x)=[\sin x]+[\tan x]+[\cos x]+[\sec x] \\$
$\because x \in(0, \pi / 4) \Rightarrow \sin x \in\left(0, \frac{1}{\sqrt{2}}\right) \Rightarrow[\sin x]=0$
$\Rightarrow \cos x \in\left(\frac{1}{\sqrt{2}}, 1\right) \Rightarrow[\cos x]=0 \\$
$\tan x \in(0,1) \Rightarrow[\tan x]=0 \Rightarrow \sec x \in(1, \sqrt{2}) \Rightarrow[\sec x]=1 \\$
$\text { Range of } f(x)=\{1\}$
(xi) Domain is $R-(2 n+1) \frac{\pi}{2}$ and $-\sqrt{2} \leq \sin x+\cos x \leq \sqrt{2} \cdot \sin (\sin x+\cos$
(x) $\neq \pm \sin 1$ But these values will come at $x=0, \pi$ so cannot be excluded.
