JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 14)
Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\pi $$) cos |x| is not differentiable. Then the set K is equal to :
{0, $$\pi $$}
$$\phi $$ (an empty set)
{ r }
{0}
Explanation
f(x) = sin$$\left| x \right| - \left| x \right|$$ + 2(x $$-$$ $$\pi $$) cosx
$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0
$$ \therefore $$ k = $$\phi $$
$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0
$$ \therefore $$ k = $$\phi $$
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