Inverse Trigonometric Functions — Page 2 | JEE Mathematics

Q11

If $$0

A

4 \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)

B

4 x \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)

C

2 x \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)

D

4 \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)

Correct Answer

Option B

Solution

Let

{{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta } = k \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = k(\alpha + \beta )

\Rightarrow \alpha + \beta = {\pi \over {2k}}

Now,

{{2\pi \,\alpha } \over {\alpha + \beta }} = {{2\pi \,\alpha } \over {{\pi \over {2k}}}} = 4k\alpha = 4{\sin ^{ - 1}}x

Here

\sin \left( {{{2\pi \,\alpha } \over {\alpha + \beta }}} \right) = \sin (4{\sin ^{ - 1}}x)

Let

{\sin ^{ - 1}}x = \theta

$\because$

x \in \left( {0,{1 \over {\sqrt 2 }}} \right) \Rightarrow \theta \in \left( {0,{\pi \over 4}} \right)

\Rightarrow x = \sin \theta

\Rightarrow \cos \theta = \sqrt {1 - {x^2}}

\Rightarrow \sin 2\theta = 2x\,.\,\sqrt {1 - {x^2}}

\Rightarrow \cos 2\theta = \sqrt {1 - 4{x^2}(1 - {x^2})} = \sqrt {{{(2{x^2} - 1)}^2}} = 1 - 2{x^2}

$\because$

\left( {\cos 2\theta > 0\,\mathrm{as}\,2\theta \in \left( {0,{\pi \over 2}} \right)} \right)

\Rightarrow \sin 4\theta = 2\,.\,2x\sqrt {1 - {x^2}} (1 - 2{x^2})

= 4x\sqrt {1 - {x^2}} (1 - 2{x^2})

Q12

Let m and M respectively be the minimum and the maximum values of

f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]

. Then m + M is equal to :

A

1 + \sqrt 2 + \pi

B

\left( {1 + \sqrt 2 } \right)\pi

C

\pi + \sqrt 2

D

1 + \pi

Correct Answer

Option A

Solution

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

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

= {\pi \over 2} + \sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2x + {1 \over {\sqrt 2 }}\cos 2x} \right)

= {\pi \over 2} + \sqrt 2 \left( {\cos {\pi \over 4}\sin 2x + \sin {\pi \over 4}\cos 2x} \right)

= {\pi \over 2} + \sqrt 2 \,.\,\sin \left( {2x + {\pi \over 4}} \right)

f(x) is maximum when

\sin \left( {2x + {\pi \over 4}} \right)

is maximum means

x = {\pi \over 8}

or

\sin \left( {2 \times {\pi \over 8} + {\pi \over 4}} \right) = \sin {\pi \over 2} = 1

$\therefore$

{\left[ {f(x)} \right]_{\max }} = {\pi \over 2} + \sqrt 2 \,.\,1 = {\pi \over 2} + \sqrt 2 = M

f(x) is minimum when

\sin \left( {2x + {\pi \over 4}} \right)

is minimum means

x = 0

or

\sin \left( {2 \times 0 + {\pi \over 4}} \right) = {1 \over {\sqrt 2 }}

$\therefore$

{\left[ {f(x)} \right]_{\min }} = {\pi \over 2} + \sqrt 2 \,.\,{1 \over {\sqrt 2 }} = {\pi \over 2} + 1 = m

$\therefore$

m + M = {\pi \over 2} + \sqrt 2 + {\pi \over 2} + 1 = \pi + \sqrt 2 + 1

Q13

If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0

A 16

B

\pi

C 16

-

5

\pi

D 0

Correct Answer

Option B

Solution

$\sin ^{-1}\left(\dfrac{\alpha}{17}\right)=-\cos ^{4}\left(\dfrac{4}{5}\right)+\tan ^{-1}\left(\dfrac{77}{36}\right)$ Let $\cos ^{-1}\left(\dfrac{4}{5}\right)=p$ and $\tan ^{-1}\left(\dfrac{77}{36}\right)=q$ $\Rightarrow \sin \left(\sin ^{-1} \dfrac{\alpha}{17}\right)=\sin (q-p)$ $=\sin q \cdot \cos p-\cos q \cdot \sin p$ $\Rightarrow \dfrac{\alpha}{17}=\dfrac{77}{85} \cdot \dfrac{4}{5}-\dfrac{36}{85} \cdot \dfrac{3}{5}$ $\Rightarrow \alpha=\dfrac{200}{25}=8$ $\sin ^{-1} \sin 8+\cos ^{-1} \cos 8$ $= -8+3 \pi+8-2 \pi$ $=\pi$

Q14

Let

S = \left\{ {x \in R:0 If

\mathrm{n(S)}

denotes the number of elements in

\mathrm{S}$$ then :

A

\mathrm{n}(\mathrm{S})=0

B

\mathrm{n}(\mathrm{S})=1

and only one element in

\mathrm{S}

is less than

\dfrac{1}{2}

.

C

\mathrm{n}(\mathrm{S})=1

and the elements in

\mathrm{S}

is more than

\dfrac{1}{2}

.

D

\mathrm{n}(\mathrm{S})=1

and the element in

\mathrm{S}

is less than

\dfrac{1}{2}

.

Correct Answer

Option D

Solution

{\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}

$\begin{aligned} & \text{ Put } x=\tan \theta \quad \theta \in\left(0, \dfrac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\dfrac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left(\dfrac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\\\ & 2 \tan ^{-1}\left[\tan \left(\dfrac{\pi}{4}-\theta\right)\right]=\cos ^{-1}[\cos (2 \theta)] \\\\ & \Rightarrow 2\left(\dfrac{\pi}{4}-\theta\right)=2 \theta \Rightarrow \theta=\dfrac{\pi}{8} \\\\ & \Rightarrow x=\tan \dfrac{\pi}{8}=\sqrt{2}-1 \simeq 0.414\end{aligned}$

Q15

The number of real roots of the equation

{\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}

is :

A 1

B 2

C 4

D 0

Correct Answer

Option D

Solution

{\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}

For equation to be defined, x2 + x $\ge$ 0 $\Rightarrow$ x2 + x + 1 $\ge$ 1 $\therefore$ Only possibility that the equation is defined x2 + x = 0 $\Rightarrow$ x = 0; x = $-$ 1 None of these values satisfy $\therefore$ No of roots = 0

Q16

If S is the sum of the first 10 terms of the series

{\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....

then tan(S) is equal to :

A

{10 \over {11}}

B

{5 \over {11}}

C -

{6 \over {5}}

D

{5 \over {6}}

Correct Answer

Option D

Solution

S =

{\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....

=

{\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}} \right) + ...

$\therefore$ Tr =

{\tan ^{ - 1}}\left( {{1 \over {1 + r \times \left( {r + 1} \right)}}} \right)

= tan–1(r + 1) – tan–1r $\therefore$ T1 = tan–12 – tan–11 T2 = tan–13 – tan–12 T3 = tan–14 – tan–13 . . .

T10 = tan-111 – tan–110 $\therefore$ S = tan–111 – tan–11 =

{\tan ^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)

$\therefore$ tan(S) =

\tan \left( {{{\tan }^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)} \right)

=

{{{11 - 1} \over {1 + 11}}}

=

{{10} \over {12}} = {5 \over 6}

Q17

If the domain of the function

f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }}

is the interval (

\alpha

,

\beta

], then

\alpha

+

\beta

is equal to :

A

{3 \over 2}

B 2

C

{1 \over 2}

D 1

Correct Answer

Option A

Solution

O \le {x^2} - x + 1 \le 1

\Rightarrow {x^2} - x \le 0

\Rightarrow x \in [0,1]

Also,

0 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi \over 2}

\Rightarrow 0 < {{2x - 1} \over 2} \le 1

\Rightarrow 0 < 2x - 1 \le 2

1 < 2x \le 3

{1 \over 2} < x \le {3 \over 2}

Taking intersection

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

\Rightarrow \alpha = {1 \over 2},\beta = 1

\Rightarrow \alpha + \beta = {3 \over 2}

Q18

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy

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

is equal to :

A 2

B 0

C 3

D 1

Correct Answer

Option C

Solution

{\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x

{\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x

{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x

x = 0

or

3\sqrt {25 - 16{x^2}} + 4\sqrt {25 - 9{x^2}} = 25

4\sqrt {25 - 9{x^2}} = 25 - 3\sqrt {25 - 16{x^2}}

Squaring we get

16(25 - 9{x^2}) = 625 - 9(25 - 16{x^2}) - 150\sqrt {25 - 16{x^2}}

400 = 625 + 225 - 150\sqrt {25 - 16{x^2}}

\sqrt {25 - 16{x^2}} = 3 \Rightarrow 25 - 16{x^2} = 9

\Rightarrow {x^2} = 1

Put x = 0, 1, $-$ 1 in the original equation We see that all values satisfy the original equation. Number of solution = 3

Q19

The value of

{\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)

is equal to :

A

\pi - {\sin ^{ - 1}}\left( {{{63} \over {65}}} \right)

B

{\pi \over 2} - {\sin ^{ - 1}}\left( {{{56} \over {65}}} \right)

C

{\pi \over 2} - {\cos ^{ - 1}}\left( {{9 \over {65}}} \right)

D

\pi - {\cos ^{ - 1}}\left( {{{33} \over {65}}} \right)

Correct Answer

Option B

Solution

{\sin ^{ - 1}}{{12} \over {13}} - {\sin ^{ - 1}}{3 \over 5} = {\sin ^{ - 1}}\left( {{{12} \over {13}}.{4 \over 5}.{3 \over 5}.{5 \over {13}}} \right)

\Rightarrow {\sin ^{ - 1}}{{33} \over {65}} = {\pi \over 2} - {\cos ^{ - 1}}{{33} \over {65}}

\Rightarrow {\pi \over 2} - {\sin ^{ - 1}}{{56} \over {65}}

Q20

2

\pi

-

\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)

is equal to :

A

{{7\pi } \over 4}

B

{{5\pi } \over 4}

C

{{3\pi } \over 2}

D

{\pi \over 2}

Correct Answer

Option C

Solution

2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)

= 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)

= 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}

= 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)

=

2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)

= 2\pi - {\pi \over 2}

= {{3\pi } \over 2}