JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 2)
If the function f : R $$ \to $$ R is defined by f(x) = |x| (x $$-$$ sin x), then which of the following statements is TRUE?
f is one-one, but NOT onto
f is onto, but NOT one-one
f is BOTH one-one and onto
f is NEITHER one-one NOR onto
Explanation
The given function f : R $$ \to $$ R is
$$f(x) = |x|(x - \sin x)$$ .....(i)
$$ \because $$ The function 'f' is a odd and continuous function and as $$\mathop {\lim }\limits_{x \to \infty } f(x) = \infty $$ and $$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty $$, so range is R, therefore, 'f' is a onto function.
$$ \because $$ $$f(x) = \left[ \matrix{ x(x - \sin x),\,x \ge \,0 \hfill \cr - x(x - \sin x),\,x\, < \,0 \hfill \cr} \right.$$
$$ \therefore $$ $$f'(x) = \left[ \matrix{ 2x - \sin x - x\cos x,\,x > \,0 \hfill \cr - 2x + \sin x + x\cos x,\,x\, < \,0 \hfill \cr} \right.$$
$$\left[ \matrix{ (x - \sin x) + x(1 - \cos x),\,x > \,0 \hfill \cr ( - x + \sin x) - x(1 - \cos x),\,x\, < \,0 \hfill \cr} \right.$$
$$ \because $$ for x > 0, x $$-$$ sin x > 0 and x(1 $$-$$ cos x) > 0
$$ \therefore $$ $$f'(x) > 0\forall x \in (0,\infty )$$
$$ \Rightarrow $$ f is strictly increasing function. $$\forall x \in (0,\infty )$$.
Similarly, for x < 0, $$-$$x + sin x > 0 and ($$-$$ x) (1 $$-$$ cos x) > 0, therefore, $$f'(x) > 0\forall x \in ( - \infty ,0)$$
$$ \Rightarrow $$ f is strictly increasing function, $$\forall x \in $$(0, $$\infty $$)
Therefore 'f' is a strictly increasing function for x$$ \in $$R and it implies that f is one-one function.
$$f(x) = |x|(x - \sin x)$$ .....(i)
$$ \because $$ The function 'f' is a odd and continuous function and as $$\mathop {\lim }\limits_{x \to \infty } f(x) = \infty $$ and $$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty $$, so range is R, therefore, 'f' is a onto function.
$$ \because $$ $$f(x) = \left[ \matrix{ x(x - \sin x),\,x \ge \,0 \hfill \cr - x(x - \sin x),\,x\, < \,0 \hfill \cr} \right.$$
$$ \therefore $$ $$f'(x) = \left[ \matrix{ 2x - \sin x - x\cos x,\,x > \,0 \hfill \cr - 2x + \sin x + x\cos x,\,x\, < \,0 \hfill \cr} \right.$$
$$\left[ \matrix{ (x - \sin x) + x(1 - \cos x),\,x > \,0 \hfill \cr ( - x + \sin x) - x(1 - \cos x),\,x\, < \,0 \hfill \cr} \right.$$
$$ \because $$ for x > 0, x $$-$$ sin x > 0 and x(1 $$-$$ cos x) > 0
$$ \therefore $$ $$f'(x) > 0\forall x \in (0,\infty )$$
$$ \Rightarrow $$ f is strictly increasing function. $$\forall x \in (0,\infty )$$.
Similarly, for x < 0, $$-$$x + sin x > 0 and ($$-$$ x) (1 $$-$$ cos x) > 0, therefore, $$f'(x) > 0\forall x \in ( - \infty ,0)$$
$$ \Rightarrow $$ f is strictly increasing function, $$\forall x \in $$(0, $$\infty $$)
Therefore 'f' is a strictly increasing function for x$$ \in $$R and it implies that f is one-one function.
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