1

### 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)   ${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$
2

### 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
3

### 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
4

### JEE Main 2019 (Online) 8th April Morning Slot

If $\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$, $\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$ where $0 < \alpha ,\beta < {\pi \over 2}$ , then $\alpha$ - $\beta$ is equal to
A
${\tan ^{ - 1}}\left( {{9 \over {14 }}} \right)$
B
${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$
C
${\cos ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$
D
${\tan ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$

## Explanation

Here $\cos \alpha = {3 \over 5}$

$\therefore$ $\tan \alpha = {4 \over 3}$

and $\tan \beta = {1 \over 3}$

We know,

$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$

= ${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$ = ${9 \over {13}}$

$\therefore$ $\left( {\alpha - \beta } \right)$ = ${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$ = ${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$