Relations and Functions Test - 74

Let $$y=f(x)=3x-2$$ and now solve for $$x$$ to find the inverse of the given function.

Given $$f(x) = \begin{cases} 1+x, & 0\leq x\leq 2 \\ 3-x, & 2<x\leq 3 \end{cases}$$

Case I: When $$0\leq x\leq 2$$
$$(fof)(x)=f[f(x)]=f(1+x)$$
Since,$$0\leq x\leq 2$$
$$\Rightarrow 1\leq 1+x\leq 3$$
$$(fof)(x)=1+1+x=2+x$$ when $$1\leq 1+x\leq 2$$
$$\Rightarrow (fof)(x)=2+x$$ when $$0\leq x\leq 1$$
And $$(fof)(x)=3-(1+x)=2-x $$when $$2<1+x\leq 3$$
$$\Rightarrow (fof)(x)=2-x$$ when $$1<x \le 2$$

Case I: When $$2< x\leq 3$$
$$(fof)(x)=f[f(x)]=f(3-x)$$
Since,$$0\leq x\leq 2$$
$$\Rightarrow -2< -x \leq 0$$
$$\Rightarrow 1< 3-x \le 3$$
$$(fof)(x)=1+3-x=4-x$$ when $$1\le  3-x\leq 2$$
$$\Rightarrow (fof)(x)=4-x$$ when $$1\le x\leq 2$$
And $$(fof)(x)=3-(3-x)=x $$when $$2<3-x\leq 3$$
$$\Rightarrow (fof)(x)=x$$ when $$0<x \le 1$$

$$f(x)\, =\, -1\, + |x\, -\, 2|, \, 0\, \leq\, x\, \leq\, 4$$
$$g(x)\, =\, 2\, -\, |x|,\, -1\, \leq\, x\, \leq\, 3$$

Let $$\displaystyle F\left ( x \right )=h\left ( x \right )-h\left ( 1 \right )=f\left ( g\left ( x \right ) \right )-h\left ( 1 \right )$$ $$\displaystyle F'\left ( x \right )=f'\left ( g\left ( x \right ) \right ).g'\left ( x \right )=\left ( + \right )\left ( - \right )=-ive.$$ (As f is increasing function $$\displaystyle f'\left ( g\left ( x \right ) \right )$$ is +ive and as g is decreasing function $$\displaystyle g'\left ( x \right )$$ is-ive.) 
Since F'(x) is-ive therefore F(x)i.e.h(x)-h(1) is decreasing function.
Now split the interval $$\displaystyle I=\left [ 0,\infty  \right ]$$ into two intervals $$\displaystyle I_{1},0\leq x< 1and I_{2}, 1\leq x< \infty ,$$ 
Apply the definition of decreasing function on $$\displaystyle h\left ( x \right )-h\left ( 1 \right ):$$ on $$\displaystyle I_{1},0\leq x< 1,\underset{\left ( Big \right )}{h\left ( x \right )}-\underset{\left ( Less \right )}{h\left ( 1 \right )}=+ive$$ On $$\displaystyle I_{2},1\leq x< \infty ,\underset{\left ( Less \right )}{h\left ( x \right )}-\underset{\left ( Big \right )}{h\left ( 1 \right )}=-ive$$ Hence for $$\displaystyle I,h\left ( x \right )-h\left ( 1 \right )$$ is neither always zero nor always +ive nor always-ive,nor strictly increasing throughout.
Hence (v) is the correct answer.

$$ gof(x) = g(f(x)) = \sqrt {{(x+5)}^{2} -9} = \sqrt {x^2 + 25 + 10x -9} $$
           $$ = \sqrt {x^2 + 16 + 10x} = \sqrt {(x+8)(x+2)} $$

Now, for the square root to have a real value, both $$ (x+8)\ and\ (x+2) $$ should be positive or equal to zero or both should be negative or equal to zero.

Case 1: When both
$$ x+8 \ge 0 $$   and $$ x + 2 \ge 0 $$
$$ => x \ge -8 $$  and $$ x\ge -2 $$

$$ => x \ge -2 $$ is the solution for this case

Case 2: When both
$$ x+8 \le 0 $$   and $$ x + 2 \le 0 $$
$$ => x \le -8 $$  and $$ x\le -2 $$

$$ => x \le -8 $$ is the solution for this case.

So, combining these both, we get the domain, $$ x \le -8\ and\   x \ge -2$$

$$f(x)=\ln x$$ and $$g(x)\, =\, \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3}$$

Now, $$f(g(x))=f \left( \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} \right)$$

$$\Rightarrow f(g(x))=\ln\left( \displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} \right)$$

For log to be defined 
$$\displaystyle \frac{x^{4}\, -\, x^{3}\, +\, 3x^{2}\, -\, 2x\, +\, 2}{2x^{2}\, -\, 2x\, +\, 3} >0$$
The above inequality holds for all $$x\in R$$

$$f(x)=\begin{cases} x+1,\quad \quad if\quad x\, \leq \, 1 \\ 5-x^{ 2 }\quad \quad if\quad x>1 \end{cases},$$

Let $$y=f(x)={e}^{x-1}+3$$ and now solve for $$x$$ to find the inverse of the given function.

$$\Rightarrow g(3) = \sqrt3$$ ..... given $$g(x) = \sqrt x $$
$$\Rightarrow f(g(3)) = f(\sqrt3) = 3 + \sqrt3 $$  ..... given $$f(x) = { x }^{ 2 }+x$$

Let $$y=f(x)=\ln { (x+2) }-3$$ and now solve for $$x$$ to find the inverse of the given function.