Question

Let $f\left(x\right)=2x-\sin x$ and $g\left(x\right)=\sqrt[3]{3x}$, then

• A. Range of $gof$ is $R$
• B. $gof$ is one-one
• C. both $f$ and $g$ are one-one
• D. both $f$ and $g$ are onto

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $gof$ is one-one
B both $f$ and $g$ are one-one
C Range of $gof$ is $R$
D both $f$ and $g$ are onto

Given : $f\left(x\right)=2x-\sin x$

$g\left(x\right)=3\sqrt{x}$

For all values of $x$ , in $f\left(x\right)andg\left(x\right)$ each element of the domain is mapped to exactly one element of the domain, Then it is known as Bijective Function.

$\Rightarrow f\left(x\right)=2x-\sin \left(x\right)$

$\Rightarrow f^1\left(x\right)=2-\cos x$ which is between 1 and 3 (inclusive for all x.Therefor $f\left(x\right)$ is strictly increasing which implies it is bijective (one-one and onto)

$\Rightarrow g\left(x\right)=\sqrt[3]{3x}={\left(x\right)}^{\frac{1}{3}}$

$\Rightarrow g^1\left(x\right)=\frac{1}{3}{\left(x\right)}^{\frac{1}{3}-1}=\frac{1}{3}{\left(x\right)}^{\frac{-2}{3}}$

Since for every $x$ there is distinct values of $g\left(x\right)$ which implies $g\left(x\right)$ is bijective (one - one and onto)

$gof=g\left(x\right)of\left(x\right)$

For $f=\sqrt[3]{3x}$ substitute x with $g\left(x\right)=2x-\sin x$

$gof=\sqrt[3]{32x-\sin \left(x\right)}$

Since $sinx$ function range is [-1,+1]

$\therefore 2x>\sin x$

$\Rightarrow (2x-\sin x)>0$

$gof=\sqrt[3]{32x-\sin \left(x\right)}$ will be in the range of real numbers (R) and one-one function ,since both $g\left(x\right)andf\left(x\right)$ are one-one function.

Therefore option A,B,C,D are correct.