Question

(i) Find range of the function $f\left(x\right)=\frac{1}{5-cos3x}$

(ii) If $f\left(x\right)=\sqrt{x^2+1}$ and $g\left(x\right)=2x^2+3,$ find

(a) $(f+g)\left(x\right)$ (b) $f\left[g\right(x\left)\right]$

(c) $(f-g)x$ (d) $\left(\frac{f}{g}\right)\left(x\right)$

Answer 1

(i) Given, $f\left(x\right)=\frac{1}{5-cos3x}$

We know that, $-1\le cos\theta \le 1$

$\therefore -1\le cos3x\le 1$

$\Rightarrow -1\le -cos3x\le 1$

$\Rightarrow -1+5\le 5-cos3x\le 1+5$

$\Rightarrow 4\le f\left(x\right)\le 6$

$\therefore$ Ranged of $f\left(x\right)=[4,6]$

(ii) Given , $f\left(x\right)=\sqrt{x^2+1}andg\left(x\right)=2x^2+3$

(a) $(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)=\sqrt{x^2+1}+2x^2+3$

(b) $f\left[g\right(x\left)\right]=f(2x^2+3)=\sqrt{(2x^2+3{)}^2+1}$

=$\sqrt{4x^4+9+12x^2+1}$

=$\sqrt{4x^4+12x^2+10}$

(c) $f-g\left)\right(x)$ =$\sqrt{x^2+1}-(2x^2+3)=\sqrt{x^2+1}-2x^2-3$

(d) $\left(\frac{f}{g}\right)x=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\sqrt{x^2+1}}{2x^2+3}$