Question

A function $g$ has domain $[0,2]$ and range $[-1,3]$. The domain and range of the function $f$ defined by $f(x)=3-2g\left(9x^2-1\right)$ are equal to

• A. $[0,2]$ and $[-1,3]$
• B. $\left[\frac{-1}{3},\frac{1}{3}\right]$ and $[-3,5]$
• C. $\left[\frac{-1}{\sqrt{3}},\frac{-1}{3}\right]Y\left[\frac{1}{3},\frac{1}{\sqrt{3}}\right]$ and $[-3,5]$
• D. $\left[\frac{-1}{\sqrt{3}},\frac{-1}{3}\right]$ and $[3,5]$

Answer 1

Answer: (C)

We have $f(x)=3-2g\left(9x^2-1\right)$
As domain of $g(x)=[0,2]$ so, for domain of $f(x)$,
$0\le 9x^2-1\le 2\Rightarrow \frac{1}{9}\le x^2\le \frac{1}{3}\Rightarrow x\in \left[\frac{-1}{\sqrt{3}},\frac{-1}{3}\right]\cup \left[\frac{1}{3},\frac{1}{\sqrt{3}}\right]=\text{ Domain of }f(x)\text{. }$
As range of $g(x)=[-1,3]$
$\Rightarrow -1\le g\le 3\Rightarrow -6\le -2g\le 2\Rightarrow -3\le 3-2g\le 5\Rightarrow -3\le f(x)\le 5$
So, range of $f(x)=[-3,5]$.