1

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

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
A
equals $\left\{ { - 2, - 1,1,2} \right\}$
B
equals $\left\{ { - 2, - 1,0,1,2} \right\}$
C
equals $\left\{ { - 2,2} \right\}$
D
is an empty set

## Explanation

$f\left( x \right) = \left\{ {\matrix{ {8 + 2x,} & { - 4 \le x \le - 2} \cr {{x^2},} & { - 2 \le x \le - 1} \cr {\left| x \right|,} & { - 1 < x < 1} \cr {{x^2},} & {1 \le x \le 2} \cr {8 - 2x,} & {2 < x \le 4} \cr } } \right.$

f(x) is not differentiable at

x = $\left\{ { - 2, - 1,0,1,2} \right\}$

$\Rightarrow$  S = {$-$2, $-$ 1, 0, 1, 2}

2

### 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)
3

### 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
4

### 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.