1

### 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.
2

### 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 = $\phi$
3

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

$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$ is equal to :
A
0
B
4
C
1
D
2

## Explanation

$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$
4

### JEE Main 2019 (Online) 12th January Morning Slot

Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then $\int\limits_0^a \,$f(x) g(x) dx is equal to :
A
4$\int\limits_0^a \,$f(x)dx
B
$-$ 3$\int\limits_0^a \,$f(x)dx
C
$\int\limits_0^a \,$f(x)dx
D
2$\int\limits_0^a \,$f(x)dx

## Explanation

${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx}$

${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx}$

${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx}$

${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}}$

$\Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx}$