If the function f : R → R is defined by f(x) = |x|(x - x), then which of the following statements is TRUE? [JEE Adv 2020 P1]

Correct Answer is : f is BOTH one-one and onto

Correct Answer is : f is BOTH one-one and onto

Test Index

Functions

Section: Mathematics

Question : 12 of 30

Marks: +1, -0

f(x) = |x|(x - \sin x),

Solution:

f(x) = |x|(x - \sin x)

f(x)=\begin{cases} -x(x-\sin x), & x < 0 \\ x(x-\sin x), & x \ge 0 \end{cases}

f(x)=\begin{cases} -x^2 + x\sin x, & x < 0 \\ x^2 - x\sin x, & x \ge 0 \end{cases}

f'(x)=\begin{cases} -2x + x\cos x + \sin x, & x<0 \\ 2x - x\cos x - \sin x, & x>0 \end{cases}

f'(x)=\begin{cases} -(x-\sin x)-x(1-\cos x), & x<0 \\ (x-\sin x)+x(1-\cos x), & x>0 \end{cases}

\because x-\sin x<0

x<0

and

1-\cos x>0, \forall x \in R

\therefore f'(x)>0

x<0

and

f'(x)>0

x>0

\therefore f(x)

is increasing function in

x \in R

and

\lim\limits_{x \rightarrow \infty} f(x)=\lim\limits_{x \rightarrow \infty} |x|(x-\sin x)

=\lim\limits_{x \rightarrow \infty} (x^{2}-x \sin x )=\infty

and

\lim\limits_{x \rightarrow \infty} f(x)

=\lim\limits_{x \rightarrow \infty} |x|(x-\sin x)=-\infty

∴ Range of f(x) = R.

∴ f is both one-one and onto.

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