1

### JEE Main 2019 (Online) 10th January Evening Slot

Let f : ($-$1, 1) $\to$ R be a function defined by f(x) = max $\left\{ { - \left| x \right|, - \sqrt {1 - {x^2}} } \right\}.$ If K be the set of all points at which f is not differentiable, then K has exactly -
A
one element
B
three elements
C
five elements
D
two elements

## Explanation

f : ($-$ 1, 1) $\to$ R

f(x) = max {$-$ $\left| x \right|, - \sqrt {1 - {x^2}}$} Non-derivable at 3 points in ($-$1, 1)
2

### JEE Main 2019 (Online) 11th January Morning Slot

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 :
A
non continuous
B
differentiable at all points
C
not differentiable at two points
D
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
3

### JEE Main 2019 (Online) 11th January Morning Slot

Let [x] denote the greatest integer less than or equal to x. Then $\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$
A
equals $\pi$ + 1
B
equals 0
C
does not exist
D
equals $\pi$

## Explanation

R.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$

(as x $\to$ 0+ $\Rightarrow$  [x] $=$ 0)

$=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$

$=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$

L.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$

(as x $\to$ 0$-$ $\Rightarrow$  [x] $=$ $-$1)

$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi$

R.H.L.  $\ne$  L.H.L.
4

### JEE Main 2019 (Online) 11th January Evening Slot

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 :
A
{0, $\pi$}
B
$\phi$ (an empty set)
C
{ r }
D
{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 = $\phi$