JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 10)
Consider the function f : R $$ \to $$ R defined by
$$f(x) = \left\{ \matrix{ \left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \hfill \cr 0,\,\,x = 0 \hfill \cr} \right.$$. Then f is :
$$f(x) = \left\{ \matrix{ \left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \hfill \cr 0,\,\,x = 0 \hfill \cr} \right.$$. Then f is :
not monotonic on ($$-$$$$\infty $$, 0) and (0, $$\infty $$)
monotonic on (0, $$\infty $$) only
monotonic on ($$-$$$$\infty $$, 0) only
monotonic on ($$-$$$$\infty $$, 0) $$\cup$$ (0, $$\infty $$)
Explanation
$$f(x) = \left\{ {\matrix{
{ - \left( {2 - \sin {1 \over x}} \right)x} & , & {x < 0} \cr
0 & , & {x = 0} \cr
{\left( {2 - \sin {1 \over x}} \right)x} & , & {x > 0} \cr
} } \right.$$
$$f'(x) = \left\{ \matrix{ - x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) - \left( {2 - \sin {1 \over x}} \right),x < 0 \hfill \cr x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) + \left( {2 - \sin {1 \over x}} \right),x > 0 \hfill \cr} \right.$$
= $$\left\{ \matrix{ - {1 \over x}\cos {1 \over x} + \sin {1 \over x} - 2,x < 0 \hfill \cr {1 \over x}\cos {1 \over x} - \sin {1 \over x} + 2,x > 0 \hfill \cr} \right.$$
$$ \therefore $$ f'(x) is an oscillating function which is non-monotonic on ($$-$$$$\infty $$, 0) and (0, $$\infty $$).
$$f'(x) = \left\{ \matrix{ - x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) - \left( {2 - \sin {1 \over x}} \right),x < 0 \hfill \cr x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) + \left( {2 - \sin {1 \over x}} \right),x > 0 \hfill \cr} \right.$$
= $$\left\{ \matrix{ - {1 \over x}\cos {1 \over x} + \sin {1 \over x} - 2,x < 0 \hfill \cr {1 \over x}\cos {1 \over x} - \sin {1 \over x} + 2,x > 0 \hfill \cr} \right.$$
$$ \therefore $$ f'(x) is an oscillating function which is non-monotonic on ($$-$$$$\infty $$, 0) and (0, $$\infty $$).
Comments (0)
