1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

The value of tan-1 $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$$ $$\left| x \right| < {1 \over 2},x \ne 0,$$ is equal to :
A
$${\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$
B
$${\pi \over 4} + {\cos ^{ - 1}}\,{x^2}$$
C
$${\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}$$
D
$${\pi \over 4} - {\cos ^{ - 1}}\,{x^2}$$

Explanation

Given,

tan-1 $$\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$$

Let    x2 = cos $$\theta $$

= tan-1 $$\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]$$

= tan-1 $$\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]$$

= tan-1 $$\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]$$

= tan-1 $$\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]$$

= tan-1 $$\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]$$

= tan-1 $$\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]$$

= tan-1 $$\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)$$

= $${\pi \over 4} + {\theta \over 2}$$

= $${\pi \over 4} + {1 \over 2}$$ cos-1 x2
2
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

A value of x satisfying the equation sin[cot−1 (1+ x)] = cos [tan−1 x], is :
A
$$ - {1 \over 2}$$
B
$$-$$ 1
C
0
D
$$ {1 \over 2}$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

If $${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$ (x > $$3 \over 4$$), then x is equal to
A
$${{\sqrt {145} } \over {10}}$$
B
$${{\sqrt {145} } \over {11}}$$
C
$${{\sqrt {145} } \over {12}}$$
D
$${{\sqrt {146} } \over {12}}$$

Explanation

Given,

$${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$$

$$ \Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$$

$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]$$

$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}$$

$$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}$$

$$ \Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}$$

$$ \Rightarrow 64 = 9\left( {16{x^2} - 9} \right)$$

$$ \Rightarrow 16{x^2} - 9 = {{64} \over 9}$$

$$ \Rightarrow 16{x^2} = {{64} \over 9} + 9$$

$$ \Rightarrow 16{x^2} = - {{145} \over 9}$$

$$ \Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}$$

$$ \Rightarrow x = \pm {{\sqrt {145} } \over {12}}$$

as given that $$x > {3 \over 4}$$

$$ \therefore $$  x $$ \ne $$ $$ - {{\sqrt {145} } \over {12}}$$

$$ \therefore $$  x $$ = {{\sqrt {145} } \over {12}}$$
4
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) = $$\pi $$

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