Inverse Trigonometric Functions — Page 3 | JEE Mathematics

Q21

Let

\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)

and

\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)

where the inverse trigonometric functions take principal values. Then, the equation whose roots are

\alpha

and

\beta

is :

A

15{x^2} - 8x - 7 = 0

B

5{x^2} - 12x + 7 = 0

C

25{x^2} - 18x - 7 = 0

D

25{x^2} - 32x + 7 = 0

Correct Answer

Option C

Solution

Given,

\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)

We know,

2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)

$\therefore$

2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times {1 \over 5} - 1} \right) = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right)

$\therefore$

\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( { - {3 \over 5}} \right)} \right.} \right)

= \tan \left( {{{5\pi } \over {16}}\sin \left( {\pi - {{\cos }^{ - 1}}\left( { {3 \over 5}} \right)} \right.} \right)

= \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right.} \right)

= \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right)} \right.} \right)

= \tan \left( {{{5\pi } \over {16}} \times {4 \over 5}} \right)

= \tan \left( {{\pi \over 4}} \right)

= 1

$\therefore$

\alpha = 1

Also given,

\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)

= \cos \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)

= \cos \left( {2{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)

= \cos \left( {{{\cos }^{ - 1}}\left( {2\left. {{{\left( {{3 \over 5}} \right)}^2} - 1} \right)} \right.} \right)

= \cos \left( {{{\cos }^{-1}}\left( {{{18} \over {25}} - 1} \right.} \right)

= {{18} \over {25}} - 1

= {{18 - 25} \over {25}}

= - {7 \over {25}}

$\therefore$

\beta = - {7 \over {25}}

$\therefore$ The quadratic equation with roots $\alpha$ and $\beta$ is

{x^2} - (\alpha + \beta )x + \alpha \beta = 0

\Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0

\Rightarrow 25{x^2} - 18x - 7 = 0

Q22

The domain of the function

f(x)=\sin ^{-1}\left(\dfrac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)

is :

A

[1, \infty)

B

[-1,2]

C

[-1, \infty)

D

(-\infty, 2]

Correct Answer

Option C

Solution

$f(x)=\sin ^{-1}\left(\dfrac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$

-1 \leq \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1

\begin{aligned} & \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 \\\\ & x^{2}-3 x+2 \leq x^{2}+2 x+7 \\\\ & 5 x \geq-5 \\\\ & x \geq-1 \end{aligned}

And

\frac{x^{2}-3 x+2}{x^{2}+2 x+7} \geq-1

x^{2}-3 x+2 \geq-x^{2}-2 x-7

$2 x^{2}-x+9 \geq 0$

x \in R

(i) $\cap$ (ii) Domain $\in[-1, \infty)$

Q23

\tan \left(2 \tan ^{-1} \dfrac{1}{5}+\sec ^{-1} \dfrac{\sqrt{5}}{2}+2 \tan ^{-1} \dfrac{1}{8}\right)

is equal to :

A 1

B 2

C

\dfrac{1}{4}

D

\dfrac{5}{4}

Correct Answer

Option B

Solution

\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)

= \tan \left( {2{{\tan }^{ - 1}}\left( {{{{1 \over 5} + {1 \over 8}} \over {1 - {1 \over 5}\,.\,{1 \over 8}}}} \right) + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2}} \right)

= \tan \left[ {2{{\tan }^{ - 1}}{1 \over 3} + {{\tan }^{ - 1}}{1 \over 2}} \right]

= \tan \left[ {{{\tan }^{ - 1}}{{{2 \over 3}} \over {1 - {1 \over 9}}} + {{\tan }^{ - 1}}{1 \over 2}} \right]

= \tan \left[ {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{1 \over 2}} \right]

= \tan \left[ {{{\tan }^{ - 1}}{{{3 \over 4} + {1 \over 2}} \over {1 - {3 \over 8}}}} \right] = \tan \left[ {{{\tan }^{ - 1}}{{{5 \over 4}} \over {{5 \over 8}}}} \right]

= \tan [{\tan ^{ - 1}}2] = 2

Q24

If 0 < a, b < 1, and tan

-

1a + tan

-

1b =

{\pi \over 4}

, then the value of

(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....

is :

A

{\log _e}

2

B e

C

{\log _e}\left( {{e \over 2}} \right)

D e2 = 1

Correct Answer

Option A

Solution

tan $-$ 1a + tan $-$ 1b =

{\pi \over 4}

0 < a, b < 1

\Rightarrow {{a + b} \over {1 - ab}} = 1

a + b = 1 $-$ ab (a + 1)(b + 1) = 2 Now

\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3} + ....} \right]

= {\log _e}(1 + a) + {\log _e}(1 + b)

( $\because$ expansion of loge(1 + x))

= {\log _e}[(1 + a)(1 + b)]

= {\log _e}2

Q25

The domain of the function

{{\mathop{\rm cosec}\nolimits} ^{ - 1}}\left( {{{1 + x} \over x}} \right)

is :

A

\left( { - 1, - {1 \over 2}} \right] \cup (0,\infty )

B

\left[ { - {1 \over 2},0} \right) \cup [1,\infty )

C

\left( { - {1 \over 2},\infty } \right) - \{ 0\}

D

\left[ { - {1 \over 2},\infty } \right) - \{ 0\}

Correct Answer

Option D

Solution

{{1 + x} \over x} \in ( - \infty , - 1] \cup [1,\infty )

{1 \over x} \in ( - \infty , - 2] \cup [0,\infty )

x \in \left[ { - {1 \over 2},0} \right) \cup (0,\infty )

x \in \left[ { - {1 \over 2},0} \right) \cup \{ 0\}

Q26

Let (a, b)

\subset(0,2 \pi)

be the largest interval for which

\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi)

, holds. If

\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0

and

\alpha-\beta=b-a

, then

\alpha

is equal to :

A

\dfrac{\pi}{16}

B

\dfrac{\pi}{48}

C

\dfrac{\pi}{8}

D

\dfrac{\pi}{12}

Correct Answer

Option D

Solution

$\sin ^{-1} \sin \theta-\left(\dfrac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0$ $\Rightarrow \sin ^{-1} \sin \theta>\dfrac{\pi}{4}$ $\Rightarrow \sin \theta>\dfrac{1}{\sqrt{2}}$ So, $\theta \in\left(\dfrac{\pi}{4}, \dfrac{3 \pi}{4}\right)$ $\theta \in\left(\dfrac{\pi}{4}, \dfrac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b})$ $b-a=\dfrac{\pi}{2}=\alpha-\beta$ $\Rightarrow \beta=\alpha-\dfrac{\pi}{2}$ $\Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0$ $x=3,9 \alpha+3 \beta+\dfrac{\pi}{2}+0=0$

\begin{aligned} & \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\ & \Rightarrow 12 \alpha-\pi=0 \\\\ & \alpha=\frac{\pi}{12} \end{aligned}

Q27

If

a=\sin ^{-1}(\sin (5))

and

b=\cos ^{-1}(\cos (5))

, then

a^2+b^2

is equal to

A 25

B

4 \pi^2+25

C

8 \pi^2-40 \pi+50

D

4 \pi^2-20 \pi+50

Correct Answer

Option C

Solution

\begin{aligned} & a=\sin ^{-1}(\sin 5)=5-2 \pi \\ & \text{ and } b=\cos ^{-1}(\cos 5)=2 \pi-5 \\ & \therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 \\ & =8 \pi^2-40 \pi+50 \end{aligned}

Q28

The sum of the infinite series

\cot ^{-1}\left(\dfrac{7}{4}\right)+\cot ^{-1}\left(\dfrac{19}{4}\right)+\cot ^{-1}\left(\dfrac{39}{4}\right)+\cot ^{-1}\left(\dfrac{67}{4}\right)+\ldots

. is :

A

\dfrac{\pi}{2}+\cot ^{-1}\left(\dfrac{1}{2}\right)

B

\dfrac{\pi}{2}-\cot ^{-1}\left(\dfrac{1}{2}\right)

C

\dfrac{\pi}{2}-\tan ^{-1}\left(\dfrac{1}{2}\right)

D

\dfrac{\pi}{2}+\tan ^{-1}\left(\dfrac{1}{2}\right)

Correct Answer

Option C

Solution

\begin{aligned} & \cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \\ & T_r=\cot ^{-1}\left(\frac{4 r^2+3}{4}\right) \\ & T_r=\tan ^{-1}\left(\frac{1}{\left(\frac{3}{4}+r^2\right)}\right) \end{aligned}

\begin{aligned} & T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+r^2-1 / 4}\right) \\ & T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right) \\ & T_r=\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right) \\ & T_1=\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

\begin{aligned} & T_2=\tan ^{-1}\left(\frac{5}{2}\right)-\tan ^{-1}\left(\frac{3}{2}\right) \\ & \vdots \qquad\qquad \vdots \qquad \qquad\qquad \vdots \\ & T_n=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \Sigma T_r=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \Sigma T_r=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

Q29

{\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)

is equal to :

A

{\pi \over 2}

B

{\pi \over 3}

C

{\pi \over 6}

D

{\pi \over 4}

Correct Answer

Option B

Solution

{\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)

= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \over 2}}}

= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {\sqrt 3 (\sqrt 3 + 1)}}} \right) + {\sec ^{ - 1}}{\left( {{{4({1^2} + {{(\sqrt 3 )}^2} + 2\,.\,1\,.\,\sqrt 3 } \over {{3^2}{{(\sqrt 3 )}^2} + 2\,.\,3\,.\,\sqrt 3 }}} \right)^{{1 \over 2}}}

= {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{4{{(\sqrt 3 + 1)}^2}} \over {{{(3 + \sqrt 3 )}^2}}}} \right)^{{1 \over 2}}}

= {\pi \over 6} + {\sec ^{ - 1}}\left( {{{2(\sqrt 3 + 1)} \over {\sqrt 3 (\sqrt 3 + 1)}}} \right)

= {\pi \over 6} + {\sec ^{ - 1}}\left( {{2 \over {\sqrt 3 }}} \right)

= {\pi \over 6} + {\cos ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right)

= {\pi \over 6} + {\pi \over 6}

= {\pi \over 3}

Q30

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of

16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)

is :

A

24 \pi^2

B

18 \pi^2

C

22 \pi^2

D

31 \pi^2

Correct Answer

Option C

Solution

Let $f(x) = 16\left[\left(\sec^{-1} x\right)^2 + \left(\operatorname{cosec}^{-1} x\right)^2\right]$ .

We can express $f(x)$ as: $f(x) = 16\left[\left(\sec^{-1} x + \operatorname{cosec}^{-1} x\right)^2 - 2\left(\sec^{-1} x\right)\left(\dfrac{\pi}{2} - \sec^{-1} x\right)\right]$ This simplifies to: $f(x) = 16\left[\dfrac{\pi^2}{4} - \pi \sec^{-1} x + 2 \left(\sec^{-1} x\right)^2\right], \quad \text{where } \sec^{-1} x \in [0, \pi] - \left\{\dfrac{\pi}{2}\right\}$ Further simplification gives: $f(x) = 16\left[2\left(\sec^{-1} x - \dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{4} - \dfrac{\pi^2}{8}\right]$ For the maximum value when $\sec^{-1} x = \pi$ : $\max = 16\left[2\pi^2 - \pi^2 + \dfrac{\pi^2}{4}\right] = 20\pi^2$ For the minimum value when $\sec^{-1} x = \dfrac{\pi}{4}$ : $\min = 16\left[\dfrac{2 \times \pi^2}{16} - \dfrac{\pi^2}{4} + \dfrac{\pi^2}{4}\right] = 2\pi^2$ Therefore, the sum of the maximum and minimum values is: $\text{Sum} = 22\pi^2$