1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

Questions Asked from Inverse Trigonometric Functions

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