1

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

If  x = sin$-$1(sin10) and y = cos$-$1(cos10), then y $-$ x is equal to :
A
0
B
10
C
7$\pi$
D
$\pi$

## Explanation

x = sin$-$1 sin 10 = 3$\pi$ $-$ 10

y = cos$-$1cos 10 = 4$\pi$ $-$ 10

y $-$ x = (4$\pi$ $-$ 10) $-$ (3$\pi$ $-$ 10) = $\pi$
2

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

The value of $\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right)$ is -
A
${{22} \over {23}}$
B
${{23} \over {22}}$
C
${{21} \over {19}}$
D
${{19} \over {21}}$

## Explanation

$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)$

$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)$

$\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)}$

$\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}$

(Where tanA $=$ 20, tanB $=$ 1)   ${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$
3

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

All x satisfying the inequality (cot–1 x)2– 7(cot–1 x) + 10 > 0, lie in the interval :
A
(cot 2, $\infty$)
B
(–$\infty$, cot 5) $\cup$ (cot 2, $\infty$)
C
(cot 5, cot 4)
D
(– $\infty$, cot 5) $\cup$ (cot 4, cot 2)

## Explanation

cot$-$1 x > 5,    cot$-$1 x < 2

$\Rightarrow$  x < cot5,   x > cot2
4

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

Considering only the principal values of inverse functions, the set
A = { x $\ge$ 0: tan$-$1(2x) + tan$-$1(3x) = ${\pi \over 4}$}
A
contains two elements
B
contains more than two elements
C
is an empty set
D
is a singleton

## Explanation

tan$-$1(2x) + tan$-$1(3x) = $\pi$/4

$\Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$ = 1

$\Rightarrow$  6x2 + 5x $-$ 1 = 0

x = $-$1 or x = ${1 \over 6}$

x = ${1 \over 6}$

$\because$  x > 0

NEET