JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 17)
Let $$f\left( x \right) = \left\{ {\matrix{
{ - 1} & { - 2 \le x < 0} \cr
{{x^2} - 1,} & {0 \le x \le 2} \cr
} } \right.$$ and
$$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$$
Then, in the interval (–2, 2), g is :
$$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$$
Then, in the interval (–2, 2), g is :
non continuous
differentiable at all points
not differentiable at two points
not differentiable at one point
Explanation
$$\left| {f\left( x \right)} \right| = \left\{ {\matrix{
1 & , & { - 2 \le x < 0} \cr
{1 - {x^2}} & , & {0 \le x < 1} \cr
{{x^2} - 1} & , & {1 \le x \le 2} \cr
} } \right.$$
and $$f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$$
Hence $$g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$$
It is not differentiable at x = 1
and $$f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$$
Hence $$g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$$
It is not differentiable at x = 1
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