If the domain of the function $f(x)=\log _{e}\left(4 x^{2}+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right)$ is $(\alpha, \beta]$, then $36|\alpha+\beta|$ is equal to :

To find the domain of the function, we need to consider the individual functions and their respective domains. We have: $f_1(x) = \ln(4x^2 + 11x + 6)$ $f_2(x) = \sin^{-1}(4x + 3)$ $f_3(x) = \cos^{-1}\left(\frac{10x + 6}{3}\right)$ For $f_1(x)$: $$ 4x^2 + 11x + 6 > 0 $$ Factoring the quadratic expression: $$ (4x + 3)(x + 2) > 0 $$ From this inequality, we have: $$ x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right) $$ For $f_2(x)$: $$ -1 \le 4x + 3 \le 1 $$ From these inequalities, w

Inverse Trigonometric Functions — Page 6 | JEE Mathematics

Q51

If

\dfrac{\pi}{2} \leq x \leq \dfrac{3 \pi}{4}

, then

\cos ^{-1}\left(\dfrac{12}{13} \cos x+\dfrac{5}{13} \sin x\right)

is equal to

A

x+\tan ^{-1} \dfrac{5}{12}

B

x-\tan ^{-1} \dfrac{4}{3}

C

x+\tan ^{-1} \dfrac{4}{5}

D

x-\tan ^{-1} \dfrac{5}{12}

Correct Answer

Option D

Solution

\begin{aligned} & \frac{\pi}{2} \leq x \leq \frac{3 \pi}{4} \\ & \cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{12} \sin x\right) \\ & \cos ^{-1}(\cos x \cos \alpha+\sin x \sin \alpha) \\ & \cos ^{-1}(\cos (x-\alpha)) \\ & \Rightarrow x-\alpha \text{ because } x-\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ & \Rightarrow x-\tan ^{-1} \frac{5}{12} \end{aligned}

Q52

If

{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha

,where –1

\le

x

\le

1, – 2

\le

y

\le

2, x

\le

{y \over 2}

, then for all x, y, 4x2 – 4xy cos

\alpha

+ y2 is equal to :

A 4 sin2

\alpha

B 2 sin2

\alpha

C 4 sin2

\alpha

- 2x2y2

D 4 cos2

\alpha

+ 2x2y2

Correct Answer

Option A

Solution

{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha

\Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha

\Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha

\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}}

squaring both sides

{x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha

$\therefore$ 4x2 – 4xy cos $\alpha$ + y2 = 4

{\sin ^2}\alpha

Q53

Let [x] denote the greatest integer less than or equal to x. Then the domain of

f(x) = \sec^{-1}(2[x] + 1)

is:

A

(-\infty, \infty)

B

(-\infty, \infty)- \{0\}

C

(-\infty, -1] \cup [0, \infty)

D

(-\infty, -1] \cup [1, \infty)

Correct Answer

Option A

Solution

\begin{aligned} & 2[\mathrm{x}]+1 \leq-1 \text{ or } 2[\mathrm{x}]+1 \geq 1 \\ & \Rightarrow[\mathrm{x}] \leq-1 \cup[\mathrm{x}] \geq 0 \\ & \Rightarrow \mathrm{x} \in(-\infty, 0) \cup \mathrm{x} \in[0, \infty) \\ & \Rightarrow \mathrm{x} \in(-\infty, \infty) \end{aligned}

Q54

The value of

\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)

is :

A

{{26} \over {25}}

B

{{25} \over {26}}

C

{{50} \over {51}}

D

{{52} \over {51}}

Correct Answer

Option A

Solution

\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)

= \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)

= \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right)

= \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1)

= \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right)

= \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right)

= {{26} \over {25}}

Q55

The sum of possible values of x for tan

-

1(x + 1) + cot

-

1

\left( {{1 \over {x - 1}}} \right)

= tan

-

1

\left( {{8 \over {31}}} \right)

is :

A

-

{{{32} \over 4}}

B

-

{{{33} \over 4}}

C

-

{{{31} \over 4}}

D

-

{{{30} \over 4}}

Correct Answer

Option A

Solution

tan $-$ 1(x + 1) + cot $-$ 1

\left( {{1 \over {x - 1}}} \right)

= tan $-$ 1

\left( {{8 \over {31}}} \right)

$\Rightarrow$ tan $-$ 1(x + 1) + tan $-$ 1(x - 1) = tan $-$ 1

\left( {{8 \over {31}}} \right)

$\Rightarrow$

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

= tan $-$ 1

\left( {{8 \over {31}}} \right)

$\Rightarrow$

{{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}

\Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}

\Rightarrow 4{x^2} + 31x - 8 = 0

\Rightarrow x = - 8,{1 \over 4}

but at

x = {1 \over 4}

LHS > {\pi \over 2}

and

RHS < {\pi \over 2}

So, only solution is x = $-$ 8 = $-$

{{{32} \over 4}}

Q56

cosec

\left[ {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right]

is equal to :

A

{{75} \over {56}}

B

{{65} \over {56}}

C

{{56} \over {33}}

D

{{65} \over {33}}

Correct Answer

Option B

Solution

\cos ec\left( {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)

\cos ec\left( {2{{\tan }^{ - 1}}\left( {{1 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)

= \cos ec\left( {{{\tan }^{ - 1}}\left( {{{2\left( {{1 \over 5}} \right)} \over {1 - {{\left( {{1 \over 5}} \right)}^2}}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)

= \cos ec\left( {{{\tan }^{ - 1}}\left( {{5 \over {12}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)

Let

{\tan ^{ - 1}}(5/12) = \theta \Rightarrow \sin \theta = {5 \over {13}},\cos \theta = {{12} \over {13}}

and

{\cos ^{ - 1}}\left( {{4 \over 5}} \right) = \phi \Rightarrow \cos \phi = {4 \over 5}

and

\sin \phi = {3 \over 5}

= \cos ec(\theta + \phi )

= {1 \over {\sin \theta \cos \phi + \cos \theta \sin \phi }}

= {1 \over {{5 \over {13}}.{4 \over 5} + {{12} \over {13}}.{3 \over 5}}} = {{65} \over {56}}

Q57

If

\alpha>\beta>\gamma>0

, then the expression

\cot ^{-1}\left\{\beta+\dfrac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\dfrac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\dfrac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}

is equal to :

A

3 \pi

B

\dfrac{\pi}{2}-(\alpha+\beta+\gamma)

C

\pi

D 0

Correct Answer

Option C

Solution

\begin{aligned} & \Rightarrow \cot ^{-1}\left(\frac{\alpha \beta+1}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{\beta \gamma+1}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{\alpha \gamma+1}{\gamma-\alpha}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{\alpha-\beta}{1+\alpha \beta}\right)+\tan ^{-1}\left(\frac{\beta-\gamma}{1+\beta \gamma}\right)+\pi+\tan ^{-1}\left(\frac{\gamma-\alpha}{1+\gamma \alpha}\right) \\ & \Rightarrow\left(\tan ^{-1} \alpha-\tan ^{-1} \beta\right)+\left(\tan ^{-1} \beta-\tan ^{-1} \gamma\right)+\left(\pi+\tan ^{-1} \gamma-\tan ^{-1} \alpha\right) \\ & \Rightarrow \pi \end{aligned}

Q58

The domain of the function

f(x)=\sin ^{-1}\left[2 x^{2}-3\right]+\log _{2}\left(\log _{\dfrac{1}{2}}\left(x^{2}-5 x+5\right)\right)

, where [t] is the greatest integer function, is :

A

\left(-\sqrt{\dfrac{5}{2}}, \dfrac{5-\sqrt{5}}{2}\right)

B

\left(\dfrac{5-\sqrt{5}}{2}, \dfrac{5+\sqrt{5}}{2}\right)

C

\left(1, \dfrac{5-\sqrt{5}}{2}\right)

D

\left[1, \dfrac{5+\sqrt{5}}{2}\right)

Correct Answer

Option C

Solution

- 1 \le 2{x^2} - 3 or

2 \le 2{x^2} or

1 \le {x^2}

x \in \left( { - \sqrt {{5 \over 2}} , - 1} \right] \cup \left[ {1,\sqrt {{5 \over 2}} } \right)

{\log _{{1 \over 2}}}({x^2} - 5x + 5) > 0

0

{x^2} - 5x + 5 > 0

&

{x^2} - 5x + 4

x \in \left( { - \infty ,{{5 - \sqrt 5 } \over 2}} \right) \cup \left( {{{5 + \sqrt 5 } \over 2},\infty } \right)

&

x \in ( - \infty ,1) \cup (4,\infty )

Taking intersection

x \in \left( {1,{{5 - \sqrt 5 } \over 2}} \right)$$

Q59

If the domain of the function

f(x)=\log _{e}\left(4 x^{2}+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1}\left(\dfrac{10 x+6}{3}\right)

is

(\alpha, \beta]

, then

36|\alpha+\beta|

is equal to :

A 72

B 54

C 45

D 63

Correct Answer

Option C

Solution

To find the domain of the function, we need to consider the individual functions and their respective domains.

We have: $f_1(x) = \ln(4x^2 + 11x + 6)$ $f_2(x) = \sin^{-1}(4x + 3)$ $f_3(x) = \cos^{-1}\left(\dfrac{10x + 6}{3}\right)$ For $f_1(x)$ :

4x^2 + 11x + 6 > 0

Factoring the quadratic expression:

(4x + 3)(x + 2) > 0

From this inequality, we have:

x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)

For $f_2(x)$ :

-1 \le 4x + 3 \le 1

From these inequalities, we get:

x \in \left[-1, -\frac{1}{2}\right]

For $f_3(x)$ :

-1 \le \frac{10x + 6}{3} \le 1

From these inequalities, we get:

x \in \left[-\frac{9}{10}, -\frac{3}{10}\right]

Now, we need to find the intersection of the domains of the three functions:

\left(-\infty, -2\right) \cup \left(-\frac{3}{4}, \infty\right) \cap \left[-1, -\frac{1}{2}\right] \cap \left[-\frac{9}{10}, -\frac{3}{10}\right]

To find the intersection, let's analyze the intervals: The interval $(-\infty, -2) \cup \left(-\dfrac{3}{4}, \infty\right)$ contains all $x$ values less than $-2$ and greater than $-\dfrac{3}{4}$ .

The interval $\left[-1, -\dfrac{1}{2}\right]$ contains all $x$ values between $-1$ and $-\dfrac{1}{2}$ .

The interval $\left[-\dfrac{9}{10}, -\dfrac{3}{10}\right]$ contains all $x$ values between $-\dfrac{9}{10}$ and $-\dfrac{3}{10}$ .

Looking at the intervals, we can see that the intersection is:

x \in \left(-\frac{3}{4}, -\frac{1}{2}\right]

Thus, the domain of the function is $(\alpha, \beta] = \left(-\dfrac{3}{4}, -\dfrac{1}{2}\right]$ .

Now, we need to find the value of $36|\alpha + \beta|$ :

36\left|-\frac{3}{4} - \frac{1}{2}\right| = 36\left|-\frac{5}{4}\right| =45

Q60

{\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))

is equal to : (The inverse trigonometric functions take the principal values)

A 3

\pi

-

11

B 4

\pi

-

9

C 4

\pi

-

11

D 3

\pi

+ 1

Correct Answer

Option C

Solution

{\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))

= (2\pi - 5) + (6 - 2\pi ) - (12 - 4\pi )

= 4\pi - 11

.