1

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

Let f : R $\to$ R be a function such that f(x) = x3 + x2f'(1) + xf''(2) + f'''(3), x $\in$ R. Then f(2) equals -
A
30
B
$-$ 2
C
$-$ 4
D
8

## Explanation

f(x) = x3 + x2f '(1) + xf ''(2) + f '''(3)

$\Rightarrow$  f '(x) = 3x2 + 2xf '(1) + f ''(x)     . . . . . (1)

$\Rightarrow$  f ''(x) = 6x + 2f '(1)     . . . . . . (2)

$\Rightarrow$  f '''(x) = 6      . . . . . .(3)

put x = 1 in equation (1) :

f '(1) = 3 + 2f '(1) + f ''(2)     . . . . .(4)

put x = 2 in equation (2) :

f ''(2) = 12 + 2f '(1)     . . . . .(5)

from equation (4) & (5) :

$-$3 $-$ f '(1) = 12 + 2f'(1)

$\Rightarrow$  3f '(1) = $-$ 15

$\Rightarrow$  f '(1) = $-$ 5 $\Rightarrow$  f ''(2) = 2      . . . . .(2)

put x = 3 in equation (3) :

f ''' (3) = 6

$\therefore$  f(x) = x3 $-$ 5x2 + 2x + 6

f(2) = 8 $-$ 20 + 4 + 6 = $-$ 2
2

### 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

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

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

$=$ $\mathop {\lim }\limits_{x \to {1^ + }} \left( {1 - {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right)\left( { - 1} \right) = \left( {1 - 1} \right)\left( { - 1} \right) = 0$
3

### 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}

4

### 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)