1

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

For each t $\in$ R , let [t] be the greatest integer less than or equal to t

Then  $\mathop {\lim }\limits_{x \to 1 + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$
A
equals $-$ 1
B
equals 1
C
equals 0
D
does not exist

## Explanation

$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}$

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

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

### 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}

3

### 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)
4

### 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