JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 8)
Let $$f\left( x \right) = \left\{ {\matrix{
{\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr
{8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr
} } \right.$$
Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S
Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S
equals $$\left\{ { - 2, - 1,1,2} \right\}$$
equals $$\left\{ { - 2, - 1,0,1,2} \right\}$$
equals $$\left\{ { - 2,2} \right\}$$
is an empty set
Explanation
$$f\left( x \right) = \left\{ {\matrix{
{8 + 2x,} & { - 4 \le x \le - 2} \cr
{{x^2},} & { - 2 \le x \le - 1} \cr
{\left| x \right|,} & { - 1 < x < 1} \cr
{{x^2},} & {1 \le x \le 2} \cr
{8 - 2x,} & {2 < x \le 4} \cr
} } \right.$$
_10th_January_Morning_Slot_en_8_1.png)
f(x) is not differentiable at
x = $$\left\{ { - 2, - 1,0,1,2} \right\}$$
$$ \Rightarrow $$ S = {$$-$$2, $$-$$ 1, 0, 1, 2}
f(x) is not differentiable at
x = $$\left\{ { - 2, - 1,0,1,2} \right\}$$
$$ \Rightarrow $$ S = {$$-$$2, $$-$$ 1, 0, 1, 2}
Comments (0)
