1

### JEE Main 2019 (Online) 9th January Morning Slot

$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$
A
exists and equals ${1 \over {2\sqrt 2 }}$
B
exists and equals ${1 \over {4\sqrt 2 }}$
C
exists and equals ${1 \over {2\sqrt 2 (1 + \sqrt {2)} }}$
D
does not exists

## Explanation

$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$

If you put y = 0 at ${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$ it is in ${0 \over 0}$ form. So we can use L' Hospital's Rule.

= $\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$

= $\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$

= ${1 \over {4\sqrt 2 }}$
2

### JEE Main 2019 (Online) 9th January Evening Slot

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

Then $\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$ is equal to :
A
$-$ sin 1
B
1
C
sin 1
D
0

## Explanation

$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$

$= \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$

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

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

### JEE Main 2019 (Online) 9th January Evening Slot

Let f be a differentiable function from

R to R such that $\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$

for all  $x,y \in$ R.

If   $f\left( 0 \right) = 1$

then   $\int\limits_0^1 {{f^2}} \left( x \right)dx$  is equal to :
A
1
B
2
C
${1 \over 2}$
D
0

## Explanation

$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$

$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$

$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$

$\Rightarrow \left| {f'\left( x \right)} \right| \le 0$  $\Rightarrow f'\left( x \right) = 0$

$\Rightarrow f\left( x \right) =$ constant

as  $f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$

$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$
4

### 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