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1

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
For $$y = {\sin ^{ - 1}}\left\{ {{{5x + 12\sqrt {1 - {x^2}} } \over {13}}} \right\};\left| x \right| \le 1$$, if $$a(1 - {x^2}){y_2} + bx{y_1} = 0$$ then (a, b) =
A
(2, 1)
B
(1, $$-$$1)
C
($$-$$1, 1)
D
(1, 2)

Explanation

Given,

$$y = {\sin ^{ - 1}}\left\{ {{{5x + 12\sqrt {1 - {x^2}} } \over {13}}} \right\}$$

$$ = {\sin ^{ - 1}}\left\{ {{{5x} \over {13}} + {{12\sqrt {1 - {x^2}} } \over {13}}} \right\}$$

Let $$ = \sin {\theta _1} = {5 \over {13}}$$ and cos$$\theta$$2 = x

$$\therefore$$ $$y = {\sin ^{ - 1}}(\sin {\theta _1}\,.\,\cos {\theta _2} + \cos {\theta _1}\,.\,\sin {\theta _2})$$

$$ = {\sin ^{ - 1}}\{ \sin ({\theta _1} + {\theta _2})\} $$

$$ = {\theta _1} + {\theta _2} = {\sin ^{ - 1}}{5 \over {13}} + {\cos ^{ - 1}}x$$

$$ \Rightarrow {y_1} = - {1 \over {\sqrt {1 - {x^2}} }}$$

and $${y_2} = {{ - x} \over {(1 - {x^2})\sqrt {1 - {x^2}} }}$$

$$ \Rightarrow {y_2}(1 - {x^2}) = x{y_1}$$

$$ \Rightarrow {y_2}(1 - {x^2}) - x{y_1} = 0$$

$$\therefore$$ a = 1, b = $$-$$1

$$\therefore$$ (a, b) = (1, $$-$$1)
$$y = {\sin ^{ - 1}}\left\{ {{{5x + 12\sqrt {1 - {x^2}} } \over {13}}} \right\};\left| x \right| \le 1$$ -এর ক্ষেত্রে, $$a(1 - {x^2}){y_2} + bx{y_1} = 0$$ হলে (a, b) =
A
(2, 1)
B
(1, $$-$$1)
C
($$-$$1, 1)
D
(1, 2)

Explanation

দেওয়া,

$$y = {\sin ^{ - 1}}\left\{ {{{5x + 12\sqrt {1 - {x^2}} } \over {13}}} \right\}$$

$$ = {\sin ^{ - 1}}\left\{ {{{5x} \over {13}} + {{12\sqrt {1 - {x^2}} } \over {13}}} \right\}$$

ধরা যাক $$ = \sin {\theta _1} = {5 \over {13}}$$ and cos$$\theta$$2 = x

$$\therefore$$ $$y = {\sin ^{ - 1}}(\sin {\theta _1}\,.\,\cos {\theta _2} + \cos {\theta _1}\,.\,\sin {\theta _2})$$

$$ = {\sin ^{ - 1}}\{ \sin ({\theta _1} + {\theta _2})\} $$

$$ = {\theta _1} + {\theta _2} = {\sin ^{ - 1}}{5 \over {13}} + {\cos ^{ - 1}}x$$

$$ \Rightarrow {y_1} = - {1 \over {\sqrt {1 - {x^2}} }}$$

এবং $${y_2} = {{ - x} \over {(1 - {x^2})\sqrt {1 - {x^2}} }}$$

$$ \Rightarrow {y_2}(1 - {x^2}) = x{y_1}$$

$$ \Rightarrow {y_2}(1 - {x^2}) - x{y_1} = 0$$

$$\therefore$$ a = 1, b = $$-$$1

$$\therefore$$ (a, b) = (1, $$-$$1)
2

WB JEE 2018

MCQ (Single Correct Answer)
English
Bengali
If $$0 \le A \le {\pi \over 4}$$, then $${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$
A
$${\pi \over 4}$$
B
$$\pi$$
C
0
D
$${\pi \over 2}$$

Explanation

We have, $$0 \le A \le {\pi \over 4}$$

$${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$

$$ = {\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}\left( {{{\cot A + {{\cot }^3}A} \over {1 - {{\cot }^4}A}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{1 \over 2}.{{2\tan A} \over {1 - {{\tan }^2}A}}} \right) + {\tan ^{ - 1}}\left( {{{\tan A} \over {{{\tan }^2}A - 1}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{{\tan A} \over {1 - {{\tan }^2}A}}} \right) - {\tan ^{ - 1}}\left( {{{\tan A} \over {1 - {{\tan }^2}A}}} \right)$$

$$ = 0$$

$$0 \le A \le {\pi \over 4}$$ হলে, $${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$ এর মান -

A
$${\pi \over 4}$$
B
$$\pi$$
C
0
D
$${\pi \over 2}$$

Explanation

$${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$

$$ = {\tan ^{ - 1}}\left( {{{\tan A} \over {1 - {{\tan }^2}A}}} \right) + \pi + {\tan ^{ - 1}}\left( {{{\cot A + {{\cot }^3}A} \over {1 - \cot A.{{\cot }^3}A}}} \right)$$ [$$\therefore$$ $$0 \le A \le {\pi \over 4}$$ $$\therefore$$ $$\cos A \ge \sin A$$ $$\therefore$$ $$\cot A \ge 1$$]

$$ = {\tan ^{ - 1}}\left( {{{\tan A} \over {1 - {{\tan }^2}A}}} \right) + \pi + {\tan ^{ - 1}}\left( {{{\cot A} \over {1 - {{\cot }^2}A}}} \right)$$

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

$$ = \pi $$

3

WB JEE 2017

MCQ (Single Correct Answer)
English
Bengali
The possible values of x, which satisfy the trigonometric equation

$${\tan ^{ - 1}}\left( {{{x - 1} \over {x - 2}}} \right) + {\tan ^{ - 1}}\left( {{{x + 1} \over {x + 2}}} \right) = {\pi \over 4}$$ are
A
$$ \pm {1 \over {\sqrt 2 }}$$
B
$$ \pm $$ $${\sqrt 2 }$$
C
$$ \pm $$ $${1 \over 2}$$
D
$$ \pm $$ 2

Explanation

We have,

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

$$ \Rightarrow {\tan ^{ - 1}}\left[ {{{{{x - 1} \over {x - 2}} + {{x + 1} \over {x + 2}}} \over {1 - {{x - 1} \over {x - 2}}.{{x + 1} \over {x + 2}}}}} \right] = {\pi \over 4}$$

$$ \Rightarrow {{(x - 1)(x + 2) + (x - 2)(x + 1)} \over {(x - 2)(x + 2) - (x - 1)(x + 1)}} = \tan {\pi \over 4}$$

$$ \Rightarrow {{{x^2} + x - 2 + {x^2} - x - 2} \over {{x^2} - 4 - {x^2} + 1}} = 1$$

$$ \Rightarrow {{2{x^2} - 4} \over { - 3}} = 1$$

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

$$ \Rightarrow 2{x^2} = 1$$

$$ \Rightarrow {x^2} = {1 \over 2}$$

$$ \Rightarrow x = \pm {1 \over {\sqrt 2 }}$$

ত্রিকোণমিতিক সমীকরণ $${\tan ^{ - 1}}\left( {{{x - 1} \over {x - 2}}} \right) + {\tan ^{ - 1}}\left( {{{x + 1} \over {x + 2}}} \right) = {\pi \over 4}$$ -এর সম্ভাব্য সমাধান হল

A
$$ \pm \,{1 \over {\sqrt 2 }}$$
B
$$ \pm \,\sqrt 2 $$
C
$$ \pm \,{1 \over 2}$$
D
$$ \pm \,2$$

Explanation

$${\tan ^{ - 1}}{{{{x - 1} \over {x - 2}} + {{x + 1} \over {x + 2}}} \over {1 - {{x - 1} \over {x - 2}}\,.\,{{x + 1} \over {x + 2}}}} = {\pi \over 4}$$

বা, $${{(x - 1)(x + 2) + (x - 2)(x + 1)} \over {({x^2} - 4) - ({x^2} - 1)}} = 1$$

বা, $$2{x^2} - 4 = - 3$$

বা, $${x^2} = {1 \over 2} \Rightarrow x = \, \pm \,{1 \over {\sqrt 2 }}$$

4

WB JEE 2011

MCQ (Single Correct Answer)

The solution set of the inequation $${\cos ^{ - 1}}x < {\sin ^{ - 1}}x$$ is

A
[$$-$$1, 1]
B
$$\left[ {{1 \over {\sqrt 2 }},1} \right]$$
C
[0, 1]
D
$$\left( {{1 \over {\sqrt 2 }},1} \right]$$

Explanation

$${\pi \over 2} - {\sin ^{ - 1}}x < {\sin ^{ - 1}}x \Rightarrow {\sin ^{ - 1}}x > {\pi \over 4}$$

But, $${{ - \pi } \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$$

$$\therefore$$ $${\pi \over 4} < {\sin ^{ - 1}}x \le {\pi \over 2}$$

$$\therefore$$ $$x \in \left( {{1 \over {\sqrt 2 }},1} \right]$$

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