JEE Main Previous Year Trigonometry Questions and Solutions

Q 1 :

If the domain of the function $\sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + \log_{e} (\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10})$ is $(α, β]$ , then $3 α + 10 β$ is equal to [2024]

95
97
98
100

1

2

3

4

(2)

$\sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + \log_{e} (\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10})$

Since, $- 1 \leq \sin^{- 1} x \leq 1$

$∴$ $- 1 \leq \frac{3 x - 22}{2 x - 19} \leq 1$

$\frac{3 x - 22}{2 x - 19} + 1 \geq 0 and \frac{3 x - 22}{2 x - 19} - 1 \leq 0$

$\Rightarrow \frac{3 x - 22 + 2 x - 19}{2 x - 19} \geq 0 and \frac{3 x - 22 - 2 x + 19}{2 x - 19} \leq 0$

$\Rightarrow \frac{5 x - 41}{2 x - 19} \geq 0 and \frac{x - 3}{2 x - 19} \leq 0$

$x \in (- \infty, \frac{41}{5}] \cup (\frac{19}{2}, \infty) and x \in [3, \frac{19}{2})$

$\Rightarrow x \in [3, \frac{41}{5}]$ ...(i)

and $\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10} > 0$ $;$ $\frac{(3 x - 5) (x - 1)}{(x - 5) (x + 2)} > 0$

$\Rightarrow x \in (- \infty, - 2) \cup (1, \frac{5}{3}) \cup (5, \infty)$ ...(ii)

Taking intersection of individual domains

$x \in (5, \frac{41}{5}]$ $\Rightarrow α = 5 and β = \frac{41}{5}$

$Hence, 3 α + 10 β = 15 + 82 = 97$

Q 2 :

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[- 1, 1]$ such that $\cos^{- 1} x - \sin^{- 1} y = α,$ $\frac{- π}{2} \leq α \leq π .$ Then, the minimum value of $x^{2} + y^{2} + 2 x y$ $\sin α$ is [2024]

$- 1$
$0$
$\frac{1}{2}$
$\frac{- 1}{2}$

1

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3

4

(2)

$\cos^{- 1} x - \frac{π}{2} + \cos^{- 1} y = α$

$\Rightarrow \cos^{- 1} x + \cos^{- 1} y = \frac{π}{2} + α$

$∴$ $\frac{π}{2} + α \in (0, \frac{3 π}{2})$ $[∵ α \in [\frac{- π}{2}, π]]$

$\cos^{- 1} (x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}) = \frac{π}{2} + α$

$\Rightarrow x y - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = - \sin α$

$\Rightarrow x y + \sin α = \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}$

$\Rightarrow x^{2} y^{2} + \sin^{2} α + 2 x y \sin α = 1 - x^{2} - y^{2} + x^{2} y^{2}$

$\Rightarrow x^{2} + y^{2} + 2 x y \sin α = \cos^{2} α$

Thus, required minimum value is 0.

Q 3 :

If the domain of the function $f (x) = \sin^{- 1} (\frac{x - 1}{2 x + 3})$ is $R - (α, β),$ then $12 α β$ is equal to : [2024]

24
32
40
36

1

2

3

4

(2)

We have, $f (x) = \sin^{- 1} (\frac{x - 1}{2 x + 3}),$ since domain of $\sin^{- 1} x is [- 1, 1], \forall x .$

$\Rightarrow - 1 \leq \frac{x - 1}{2 x + 3} \leq 1 \Rightarrow \frac{x - 1}{2 x + 3} + 1 \geq 0 and \frac{x - 1}{2 x + 3} - 1 \leq 0$

$\Rightarrow \frac{x - 1 + 2 x + 3}{2 x + 3} \geq 0 and \frac{x - 1 - 2 x - 3}{2 x + 3} \leq 0$

$\Rightarrow \frac{3 x + 2}{2 x + 3} \geq 0 and \frac{- x - 4}{2 x + 3} \leq 0$

Now, $\frac{3 x + 2}{2 x + 3} \geq 0 \Rightarrow x < - \frac{3}{2} or x \geq - \frac{2}{3}$

$\Rightarrow x \in (- \infty, - \frac{3}{2}) \cup [\frac{- 2}{3}, \infty)$

and $\frac{- x - 4}{2 x + 3} \leq 0 \Rightarrow x \leq - 4 or x > - \frac{3}{2}$

$\Rightarrow x \in (- \infty, - 4] \cup (\frac{- 3}{2}, \infty)$

$Hence, x \in (- \infty, - 4] \cup [\frac{- 2}{3}, \infty)$ $i . e ., x \in R - (- 4, \frac{- 2}{3})$

So, Domain : $R - (- 4, \frac{- 2}{3}) \Rightarrow α = - 4 and β = \frac{- 2}{3}$

$∴$ $12 \times β = 12 \times (- 4) \times (- \frac{2}{3}) = 32$

Q 4 :

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$ is: [2024]

1
more than 2
2
0

1

2

3

4

(1)

Given, $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (\frac{3 x}{1 - 2 x^{2}}) = \frac{π}{4}$ $(∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y}))$

$\Rightarrow \frac{3 x}{1 - 2 x^{2}} = \tan \frac{π}{4} \Rightarrow \frac{3 x}{1 - 2 x^{2}} = 1 \Rightarrow 2 x^{2} + 3 x - 1 = 0$

$\Rightarrow x = \frac{- 3 \pm \sqrt{17}}{4}$ $∴$ $Possible value of x = \frac{- 3 + \sqrt{17}}{4}$

Hence, only 1 positive real value of $x$ satisfies the equation.

Q 5 :

Let $x = \frac{m}{n}$ $(m, n$ are co-prime natural numbers $)$ be a solution of the equation $\cos (2 \sin^{- 1} x) = \frac{1}{9}$ and let $α, β (α > β)$ be the roots of the equation $m x^{2} - n x - m + n = 0$ . Then the point $(α, β)$ lies on the line [2024]

$3 x - 2 y = - 2$
$5 x - 8 y = - 9$
$3 x + 2 y = 2$
$5 x + 8 y = 9$

1

2

3

4

(4)

We have, $\cos (2 \sin^{- 1} x) = \frac{1}{9}$

Put $\sin^{- 1} x = θ$

$\cos 2 θ = \frac{1}{9} \Rightarrow 1 - 2 \sin^{2} θ = \frac{1}{9}$

$\Rightarrow \sin θ = \frac{2}{3}$ $(∵ x > 0 and \sin θ lies in the first quadrant)$

$\Rightarrow x = \frac{2}{3} = \frac{m}{n} .$ So, $m = 2$ and $n = 3$

The given equation becomes, $2 x^{2} - 3 x - 2 + 3 = 0$

$\Rightarrow 2 x^{2} - 3 x + 1 = 0 \Rightarrow 2 x^{2} - 2 x - x + 1 = 0$

$\Rightarrow (2 x - 1) (x - 1) = 0$ $\Rightarrow x = \frac{1}{2} or x = 1$

$\Rightarrow α = 1 and β = \frac{1}{2}, which lies on the line 5 x + 8 y = 9$

Q 6 :

If the domain of the function $f (x) = \cos^{- 1} (\frac{2 - | x |}{4}) + {\log_{e} (3 - x)}^{- 1}$ is $[- α, β) - {γ},$ then $α + β + γ$ is equal to [2024]

9
8
11
12

1

2

3

4

(3)

We have, $f (x) = \cos^{- 1} (\frac{2 - | x |}{4}) + {\log_{e} (3 - x)}^{- 1}$

For $f (x)$ be defined $- 1 \leq \frac{2 - | x |}{4} \leq 1$

$\Rightarrow - 4 \leq 2 - | x | \leq 4 \Rightarrow - 6 \leq - | x | \leq 2$

$\Rightarrow 6 \geq | x | \geq - 2$

Since, $| x | \geq - 2$ so $- 6 \leq x \leq 6$ ...(i)

Also, $3 - x > 0$ and $3 - x \neq 1$

$\Rightarrow x < 3 and x \neq 2$ ...(ii)

Taking intersection of (i) and (ii), we get $x \in [- 6, 3) - {2}$

$\Rightarrow α = 6, β = 3, and γ = 2$

$\Rightarrow α + β + γ = 6 + 3 + 2 = 11$

Q 7 :

For $α, β, γ \neq 0, if \sin^{- 1} α + \sin^{- 1} β + \sin^{- 1} γ = π and (α + β + γ) (α - γ + β) = 3 α β, then γ equals$ [2024]

$\frac{\sqrt{3}}{2}$
$\sqrt{3}$
$\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3} - 1}{2 \sqrt{2}}$

1

2

3

4

(1)

Given, $\sin^{- 1} α + \sin^{- 1} β + \sin^{- 1} γ = π$

and $(α + β + γ) (α + β - γ) = 3 α β$

Let $\sin^{- 1} α = A, \sin^{- 1} β = B, \sin^{- 1} γ = C$

$\Rightarrow A + B + C = π and (α + β + γ) (α + β - γ) = 3 α β$

$α^{2} + β^{2} + 2 α β - γ^{2} = 3 α β, α^{2} + β^{2} - γ^{2} = α β$

$\frac{α^{2} + β^{2} - γ^{2}}{2 α β} = \frac{1}{2}$

$\Rightarrow \cos C = \frac{1}{2} \Rightarrow C = 60 ° \Rightarrow \sin C = \frac{\sqrt{3}}{2} = γ$

So, $γ = \frac{\sqrt{3}}{2}$

Q 8 :

If $a = \sin^{- 1} (\sin (5)) and b = \cos^{- 1} (\cos (5)),$ then $a^{2} + b^{2}$ is equal to [2024]

$25$
$8 π^{2} - 40 π + 50$
$4 π^{2} - 20 π + 50$
$4 π^{2} + 25$

1

2

3

4

(2)

We have, $a = \sin^{- 1} (\sin (5)) and b = \cos^{- 1} (\cos (5))$

$\Rightarrow a = - 2 π + 5$ $(∵ \sin^{- 1} (\sin x) = - 2 π + x, if \frac{3 π}{2} \leq x \leq \frac{5 π}{2})$

$b = 2 π - 5,$ $(∵ \cos^{- 1} (\cos x) = 2 π - x, π \leq x \leq 2 π)$

$∴$ $a^{2} + b^{2} = {(- 2 π + 5)}^{2} + {(2 π - 5)}^{2}$

$= 4 π^{2} + 25 - 20 π + 4 π^{2} + 25 - 20 π = 8 π^{2} - 40 π + 50$

Q 9 :

For $n \in N, if \cot^{- 1} 3 + \cot^{- 1} 4 + \cot^{- 1} 5 + \cot^{- 1} n = \frac{π}{4},$ then $n$ is equal to ______. [2024]

(47)

$\cot^{- 1} 3 + \cot^{- 1} 4 + \cot^{- 1} 5 + \cot^{- 1} n = \frac{π}{4}$

$\Rightarrow \tan^{- 1} \frac{1}{3} + \tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{n} = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{4}}{1 - \frac{1}{12}}) + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{n} = \frac{π}{4}$

$\Rightarrow \tan^{- 1} (\frac{7}{11}) + \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1$

$\Rightarrow \tan^{- 1} (\frac{(\frac{7}{11} + \frac{1}{5})}{1 - \frac{7}{11} \cdot \frac{1}{5}}) + \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1$

$\Rightarrow \tan^{- 1} (\frac{23}{24}) + \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1$

$\Rightarrow \tan^{- 1} \frac{1}{n} = \tan^{- 1} 1 - \tan^{- 1} \frac{23}{24}$

$\Rightarrow \tan^{- 1} \frac{1}{n} = \tan^{- 1} (\frac{1 - \frac{23}{24}}{1 + \frac{23}{24}}) \Rightarrow \frac{1}{n} = \frac{1}{47} \Rightarrow n = 47$

Q 10 :

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5},$ is _____. [2024]

(0)

We have, $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5}$

$\Rightarrow 2 (\frac{π}{2} - \cos^{- 1} x) + 3 \cos^{- 1} x = \frac{2 π}{5}$

$\Rightarrow π + \cos^{- 1} x = \frac{2 π}{5} \Rightarrow \cos^{- 1} x = \frac{- 3 π}{5}$

Which is not possible as $\cos^{- 1} x \in [0, π]$

$∴ No solution exists.$

Q 11 :

If the domain of the function $f (x) = \log_{e} (\frac{2 x + 3}{4 x^{2} + x - 3}) + \cos^{- 1} (\frac{2 x - 1}{x + 2})$ is $(α, β],$ then the value of $5 β - 4 α$ is equal to [2024]

12
10
11
9

1

2

3

4

(1)

Given function,

$f (x) = \log_{e} (\frac{2 x + 3}{4 x^{2} + x - 3}) + \cos^{- 1} (\frac{2 x - 1}{x + 2})$

Here, $\log (\frac{2 x + 3}{4 x^{2} + x - 3})$ will be defined, if

$\frac{2 x + 3}{4 x^{2} + x - 3} > 0 \Rightarrow \frac{2 x + 3}{(4 x - 3) (x + 1)} > 0$

$∴$ $x \in (\frac{- 3}{2}, - 1) \cup (\frac{3}{4}, \infty)$ ...(i)

Also, $\cos^{- 1} (\frac{2 x - 1}{x + 2})$ will be defined, if $- 1 \leq \frac{2 x - 1}{x + 2} \leq 1 \Rightarrow 0 \leq \frac{3 x + 1}{x + 2}$

$∴$ $x \in ([- \infty, - 2] \cup [\frac{- 1}{3}, \infty])$ ...(ii)

Also, $\frac{x - 3}{x + 2} \leq 0$

$∴$ $x \in [- 2, 3]$ ...(iii)

From (i), (ii) and (iii), we get

$β = 3$ and $α = \frac{3}{4} \Rightarrow 5 β - 4 α = 12$

Q 12 :

Considering the principal values of the inverse trigonometric functions, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$ , $- \frac{1}{2} < x < \frac{1}{\sqrt{2}}$ , is equal to [2025]

$\frac{- 5 π}{6} - \sin^{- 1} x$
$\frac{π}{4} + \sin^{- 1} x$
$\frac{5 π}{6} - \sin^{- 1} x$
$\frac{π}{6} + \sin^{- 1} x$

1

2

3

4

(4)

Let $\sin^{- 1} x = θ, \frac{- π}{4} < θ < \frac{π}{4}$

$\Rightarrow x = \sin θ$

Now, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$

$= \sin^{- 1} (\sin θ \cos \frac{π}{6} + \cos θ \sin \frac{π}{6})$

$= \sin^{- 1} (\sin (θ + \frac{π}{6})) = θ + \frac{π}{6} = \sin^{- 1} x + \frac{π}{6}$

Q 13 :

The sum of the infinite series $\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$ is: [2025]

$\frac{π}{2} + \cot^{- 1} (\frac{1}{2})$
$\frac{π}{2} - \tan^{- 1} (\frac{1}{2})$
$\frac{π}{2} + \tan^{- 1} (\frac{1}{2})$
$\frac{π}{2} - \cot^{- 1} (\frac{1}{2})$

1

2

3

4

(2)

The given infinite series is :

$\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$

$\Rightarrow T_{n} = \cot^{- 1} (\frac{4 n^{2} + 3}{4})$

$= \tan^{- 1} (\frac{4}{4 n^{2} + 3}) = \tan^{- 1} (\frac{1}{n^{2} + \frac{3}{4}})$

$= \tan^{- 1} (\frac{1}{n^{2} - \frac{1}{4} + 1}) = \tan^{- 1} (\frac{(n + \frac{1}{2}) - (n - \frac{1}{2})}{1 + (n + \frac{1}{2}) (n - \frac{1}{2})})$

$= \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} (n - \frac{1}{2})$

$T_{1} = \tan^{- 1} \frac{3}{2} - \tan^{- 1} \frac{1}{2}$

$T_{2} = \tan^{- 1} \frac{5}{2} - \tan^{- 1} \frac{3}{2}$ and so on.

$∴ \sum_{r = 1}^{n} T_{r} = \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} \frac{1}{2}$

$as n \to \infty$

$\sum_{r = 1}^{\infty} T_{r} = \frac{π}{2} - \tan^{- 1} \frac{1}{2}$

Q 14 :

The value of $\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$ is equal to [2025]

$π + \frac{3}{2}$
$π - \frac{5}{4}$
$π - \frac{3}{2}$
$π + \frac{5}{2}$

1

2

3

4

(2)

We have,

$\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$

$= \cot^{- 1} (\frac{| \sec (2) | - 1}{\tan (2)}) - \cot^{- 1} (\frac{| \sec (\frac{1}{2}) | + 1}{\tan (\frac{1}{2})})$

$= \cot^{- 1} (\frac{- 1 - \cos 2}{\sin 2}) - \cot^{- 1} (\frac{1 + \cos (\frac{1}{2})}{\sin (\frac{1}{2})})$

$= \cot^{- 1} (\frac{- 2 \cos^{2} (1)}{2 \cos (1) \sin (1)}) - \cot^{- 1} (\frac{2 \cos^{2} (\frac{1}{4})}{2 \cos (\frac{1}{4}) \sin (\frac{1}{4})})$

$= \cot^{- 1} (\cot (- 1)) - \cot^{- 1} (\cot \frac{1}{4})$

$= π - 1 - \frac{1}{4} = π - \frac{5}{4}$ .

Q 15 :

Let the domains of the functions $f (x) = \log_{4} \log_{3}$ $\log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))$ and $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$ be ( $α, β$ ) and [ $γ, δ$ ], respectively. Then $α^{2} + β^{2} + γ^{2} + δ^{2}$ is equal to : [2025]

13
16
15
14

1

2

3

4

(3)

We have,

$f (x) = \log_{4} \log_{3} \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))$

For f(x) to be defined we need,

$\log_{3} \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 0$

$\Rightarrow \log_{7} (8 - \log_{2} (x^{2} + 4 x + 5)) > 1$

$\Rightarrow 8 - \log_{2} (x^{2} + 4 x + 5) > 7$

$\Rightarrow \log_{2} (x^{2} + 4 x + 5) < 1$

$\Rightarrow x^{2} + 4 x + 5 < 2$

$\Rightarrow x^{2} + 4 x + 3 < 0$

$\Rightarrow (x + 1) (x + 3) < 0$

$\Rightarrow x \in (- 3, - 1) i . e ., α = - 3, β = - 1$ ... (i)

Also, $x^{2} + 4 x + 5 > 0, but D = 16 - 20 = - 4 < 0$

Also, we have $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$

$\Rightarrow - 1 \leq \frac{7 x + 10}{x - 2} \leq 1 \Rightarrow - 1 \leq \frac{7 (x - 2) + 24}{x - 2} \leq 1$

$\Rightarrow - 8 \leq \frac{24}{x - 2} \leq - 6 \Rightarrow \frac{1}{- 6} \leq \frac{x - 2}{24} \leq \frac{1}{- 8}$

$\Rightarrow - 4 \leq x - 2 \leq - 3 \Rightarrow - 2 \leq x \leq - 1$

$\Rightarrow x \in [- 2, - 1] i . e ., γ = - 2, δ = - 1$

$∴ α^{2} + β^{2} + γ^{2} + δ^{2} = {(- 3)}^{2} + {(- 1)}^{2} + {(- 2)}^{2} + {(- 1)}^{2} = 15$

Q 16 :

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16 ({(\sec^{- 1} x)}^{2} + {({cosec}^{- 1} x)}^{2})$ is : [2025]

$22 π^{2}$
$24 π^{2}$
$18 π^{2}$
$31 π^{2}$

1

2

3

4

(1)

$f (x) = 16 ({(\sec^{- 1} x)}^{2} + {({cosec}^{- 1} x)}^{2})$

$= 16 [(\sec^{- 1} x + {{cosec}^{- 1} x)}^{2} - 2 \sec^{- 1} x {cosec}^{- 1} x]$

$= 16 [{(\frac{π}{2})}^{2} - 2 (\frac{π}{2} - {cosec}^{- 1} x) {cosec}^{- 1} x]$

$= 16 [\frac{π^{2}}{4} - π {cosec}^{- 1} x + 2 {({cosec}^{- 1} x)}^{2}]$

$= 32 [{({cosec}^{- 1} x)}^{2} - \frac{π}{2} {cosec}^{- 1} x + \frac{π^{2}}{8}]$

$\Rightarrow f (x) = 32 [{({cosec}^{- 1} x - \frac{π}{4})}^{2} + \frac{π^{2}}{16}]$

f(x) is maximum when x = –1

$∴ f_{m a x} = 32 \times 10 \frac{π^{2}}{16}$

f(x) is minimum when $x = \sqrt{2}$

$∴ f_{\min} = 32 \times \frac{π^{2}}{16}$

$∴$ Required sum $= 32 \times 10 \frac{π^{2}}{16} + 32 \times \frac{π^{2}}{16} = 22 π^{2}$

Q 17 :

If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ , then $\cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)$ is equal to [2025]

$x - \tan^{- 1} \frac{5}{12}$
$x + \tan^{- 1} \frac{5}{12}$
$x + \tan^{- 1} \frac{4}{5}$
$x - \tan^{- 1} \frac{4}{3}$

1

2

3

4

(1)

Given, $\frac{π}{2} \leq x \leq \frac{3 π}{4}$

Let $\cos α = \frac{12}{13} [∴ \cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)]$

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

$= x - \tan^{- 1} (\frac{5}{12})$

Q 18 :

If $α > β > γ > 0$ , then the expression $\cot^{- 1} {β + \frac{(1 + β^{2})}{(α - β)}} + \cot^{- 1} {γ + \frac{(1 + γ^{2})}{(β - γ)}} + \cot^{- 1} {α + \frac{(1 + α^{2})}{(γ - α)}}$ is equal to: [2025]

$3 π$
0
$\frac{π}{2} - (α + β + γ)$
$π$

1

2

3

4

(4)

$β + \frac{(1 + β^{2})}{α - β} = \frac{α β - β^{2} + 1 + β^{2}}{α - β} = \frac{α β + 1}{α - β}$

Similarly, $γ + \frac{(1 + γ^{2})}{β - γ} = \frac{β γ + 1}{β - γ} and α + \frac{(1 + α^{2})}{γ - α} = \frac{α γ + 1}{γ - α}$

$∴ \cot^{- 1} (\frac{α β + 1}{α - β}) + \cot^{- 1} (\frac{β γ + 1}{β - γ}) + \cot^{- 1} (\frac{α γ + 1}{γ - α})$

$= \tan^{- 1} (\frac{α - β}{1 + α β}) + \tan^{- 1} (\frac{β - γ}{1 - β γ}) + π - \cot^{- 1} (\frac{α γ + 1}{α - γ})$

$= \tan^{- 1} (\frac{α - β}{1 + α β}) + \tan^{- 1} (\frac{β - γ}{1 + β γ}) - \tan^{- 1} (\frac{- γ + α}{1 + γ α}) + π$

$= \tan^{- 1} (α) - \tan^{- 1} (β) + \tan^{- 1} (β) - \tan^{- 1} (γ) + \tan^{- 1} (γ) - \tan^{- 1} (α) + π = π$

Q 19 :

$\cos (\sin^{- 1} \frac{3}{5} + \sin^{- 1} \frac{5}{13} + \sin^{- 1} \frac{33}{65})$ is equal to : [2025]

1
0
$\frac{32}{65}$
$\frac{33}{65}$

1

2

3

4

(2)

$\cos (\sin^{- 1} \frac{3}{5} + \sin^{- 1} \frac{5}{13} + \sin^{- 1} \frac{33}{65})$

$= \cos (\tan^{- 1} \frac{3}{4} + \tan^{- 1} \frac{5}{12} + \tan^{- 1} \frac{33}{56})$

$= \cos (\tan^{- 1} \frac{(\frac{3}{4} + \frac{5}{12})}{(1 - \frac{3}{4} \cdot \frac{5}{12})} + \tan^{- 1} \frac{33}{56})$

$= \cos (\tan^{- 1} \frac{56}{33} + \tan^{- 1} \frac{33}{56})$

$= \cos [\cot^{- 1} (\frac{33}{56}) + \tan^{- 1} (\frac{33}{56})]$

$= \cos (\frac{π}{2}) = 0$ $(∵ \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2})$

Q 20 :

Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = $\sec^{- 1}$ (2[x] + 1) is : [2025]

$(- \infty, \infty)$
$(- \infty, \infty) - {0}$
$(- \infty, - 1] \cup [0, \infty)$
$(- \infty, - 1] \cup [1, \infty)$

1

2

3

4

(1)

We have, $2 [x] + 1 \leq - 1 or 2 [x] + 1 \geq 1$

$\Rightarrow [x] \leq - 1 or [x] \geq 0$

$\Rightarrow x \in (- \infty, 0) or x \in [0, \infty)$

$\Rightarrow x \in (- \infty, \infty)$

Q 21 :

If $y = \cos (\frac{π}{3} + \cos^{- 1} \frac{x}{2})$ , then ${(x - y)}^{2} + 3 y^{2}$ is equal to [2025]

(3)

We have, $y = \cos (\frac{π}{3} + \cos^{- 1} \frac{x}{2})$

$= \cos \frac{π}{3} \cos (\cos^{- 1} \frac{x}{2}) - \sin \frac{π}{3} \sin (\cos^{- 1} \frac{x}{2})$

$= \frac{1}{2} \times \frac{x}{2} - \frac{\sqrt{3}}{2} \sin (\sin^{- 1} \sqrt{1 - \frac{x^{2}}{4}})$

$= \frac{x}{4} - \frac{\sqrt{3}}{2} \frac{\sqrt{4 - x^{2}}}{2}$

$\Rightarrow 4 y = x - \sqrt{3 (4 - x^{2})} \Rightarrow (x - 4 y) = \sqrt{3 (4 - x^{2})}$

On squaring both sides, we get

$x^{2} + 16 y^{2} - 8 x y = 12 - 3 x^{2}$

$\Rightarrow 4 x^{2} + 16 y^{2} - 8 x y = 12 \Rightarrow x^{2} + 4 y^{2} - 2 x y = 3$

$∴ {(x - y)}^{2} + 3 y^{2} = 3$

Q 22 :

Let the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))$ be $(γ, δ)$ . Then $| 7 (α + β) + 4 (γ + δ) |$ is equal to __________. [2025]

(96)

We have, $f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})$

Since, domain of $\cos^{— 1} x$ is [–1, 1].

$∴ - 1 \leq \frac{4 x + 5}{3 x - 7} \leq 1$

For, $\frac{4 x + 5}{3 x - 7} \geq - 1 \Rightarrow \frac{4 x + 5 + 3 x - 7}{3 x - 7} \geq 0$

$\Rightarrow \frac{7 x - 2}{3 x - 7} \geq 0 \Rightarrow x \in (- \infty, \frac{2}{7}] \cup (\frac{7}{3}, \infty)$ ... (i)

Now, for $\frac{4 x + 5}{3 x - 7} \leq 1 \Rightarrow \frac{4 x + 5 - 3 x + 7}{3 x - 7} \leq 0$

$\Rightarrow \frac{x + 12}{3 x - 7} \leq 0 \Rightarrow x \in [- 12, \frac{7}{3})$ ... (ii)

On combining (i) and (ii), we get domain of f(x) is $[- 12, \frac{2}{7}]$ .

$∴ α = - 12 and β = \frac{2}{7}$

Now, $g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))$

For g(x) to be defined, we have

$2 - 6 \log_{27} (2 x + 5) > 0 \Rightarrow 6 \log_{27} (2 x + 5) < 2$

$\Rightarrow \log_{27} (2 x + 5) < \frac{1}{3} \Rightarrow (2 x + 5) < {(27)}^{\frac{1}{3}}$

$\Rightarrow 2 x + 5 < 3 \Rightarrow 2 x < - 2 \Rightarrow x < - 1$

Also, 2x + 5 > 0 [For $\log_{27} (2 x + 5)$ to be defined]

$\Rightarrow x > \frac{- 5}{2}$

$∴$ Domain of g(x) is $(\frac{- 5}{2}, - 1)$ $∴ γ = \frac{- 5}{2} and δ = - 1$

$∴$ $| 7 (α + β) + 4 (γ + δ) |$ $=$ $| 7 (- 12 + \frac{2}{7}) + 4 (\frac{- 5}{2} - 1) |$

$=$ $| - 82 - 14 |$ $= 96$

Q 23 :

If for some $α, β; α \leq β, α + β = 8$ and $\sec^{2} (\tan^{- 1} α) + {cosec}^{2} (\cot^{- 1} β) = 36$ , then $α^{2} + β$ is __________. [2025]

(14)

Given, $\sec^{2} (\tan^{- 1} α) + {cosec}^{2} (\cot^{- 1} β) = 36$

$∴ 1 + \tan^{2} (\tan^{- 1} α) + 1 + \cot^{2} (\cot^{- 1} β) = 36$

$\Rightarrow α^{2} + β^{2} = 34$

Given, $α + β = 8$ ... (i)

$\Rightarrow α^{2} + β^{2} + 2 α β = 64$

$\Rightarrow 2 α β = 30 \Rightarrow α β = 15$

$∴ β - α = \sqrt{{(α + β)}^{2} - 4 α β}$ [ $∵ β \geq α$ ]

$β - α = 2$ ... (ii)

From (i) and (ii), we get

$α = 3, β = 5$

$∴ α^{2} + β = 9 + 5 = 14$

Q 24 :

Let $S = {x : \cos^{- 1} x = π + \sin^{- 1} x + \sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} {(2 x - 1)}^{2}$ is equal to __________. [2025]

(5)

We have, $\cos^{- 1} x = π + \sin^{- 1} x + \sin^{- 1} (2 x + 1)$

$\Rightarrow 2 \cos^{- 1} x - \sin^{- 1} (2 x + 1) = \frac{3 π}{2}$

$\Rightarrow 2 α - β = \frac{3 π}{2}, where \cos^{- 1} x = α, \sin^{- 1} (2 x + 1) = β$

$\Rightarrow 2 α = \frac{3 π}{2} + β \Rightarrow \cos 2 α = \sin β \Rightarrow 2 \cos^{2} α - 1 = \sin β$

$\Rightarrow 2 x^{2} - 1 = 2 x + 1$ [ $∵ x = \cos α and 2 x + 1 = \sin β$ ]

$\Rightarrow x^{2} - x - 1 = 0$

$\Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$

$\Rightarrow x = \frac{1 - \sqrt{5}}{2}$ ( $∵ x = \frac{1 + \sqrt{5}}{2}$ rejected as $\cos α \leq 1$ )

Now, $\sum_{x \in S} {(2 x - 1)}^{2} = \sum_{x \in S} (4 x^{2} + 1 - 4 x) = 5$

Q 25 :

The range of $f (x) = 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$ is [2023]

$[0, 2 π)$
$[0, π]$
$[0, 2 π]$
$[0, π)$

1

2

3

4

(1)

$Given, 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$

$\frac{x^{2}}{x^{2} + 1} = \frac{x^{2} + 1 - 1}{x^{2} + 1} = 1 - \frac{1}{x^{2} + 1}$

$x^{2} \geq 0; x^{2} + 1 \geq 1$ $∴$ $0 < \frac{1}{x^{2} + 1} \leq 1 \Rightarrow - 1 \leq - \frac{1}{x^{2} + 1} < 0;$

$0 \leq 1 - \frac{1}{x^{2} + 1} < 1; 0 \leq 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1}) < 2 π$

Q 26 :

Let $S$ be the set of all solutions of the equation $\cos^{- 1} (2 x) - 2 \cos^{- 1} (\sqrt{1 - x^{2}}) = π,$ $x \in [- \frac{1}{2}, \frac{1}{2}] .$ Then $\sum_{x \in S} 2 \sin^{- 1} (x^{2} - 1)$ is equal to [2023]

$0$
$\frac{- 2 π}{3}$
$π - 2 \sin^{- 1} (\frac{\sqrt{3}}{4})$
$π - \sin^{- 1} (\frac{\sqrt{3}}{4})$

1

2

3

4

(2)

Q 27 :

Let $S = {x \in R : 0 < x < 1 and 2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})} .$ If $n (S)$ denotes the number of elements in S, then [2023]

$n (S) = 2 and only one element in S is less than \frac{1}{2}$
$n (S) = 1 and the element in S is less than \frac{1}{2}$
$n (S) = 1 and the element in S is more than \frac{1}{2}$
$n (S) = 0$

1

2

3

4

(2)

$0 < x < 1$

$2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ ...(i)

Let $\tan^{- 1} x = θ \in (0, \frac{π}{4}) \Rightarrow x = \tan θ$

$\Rightarrow 2 \tan^{- 1} (\tan (\frac{π}{4} - θ)) = \cos^{- 1} (\cos 2 θ)$

$\Rightarrow 2 (\frac{π}{4} - θ) = 2 θ$

$\Rightarrow 4 θ = \frac{π}{2}$ $∴$ $θ = \frac{π}{8}$ $x = \tan \frac{π}{8} \Rightarrow x = \sqrt{2} - 1 ≃ 0.414$

Q 28 :

$\tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$ is equal to [2023]

$\frac{π}{2}$
$\frac{π}{4}$
$\frac{π}{6}$
$\frac{π}{3}$

1

2

3

4

(4)

$\tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$

$\tan^{- 1} (\frac{1 + \sqrt{3}}{\sqrt{3} (\sqrt{3} + 1)}) + \tan^{- 1} (\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{6 + 3 \sqrt{3}}})$

$\tan^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{3 (2 + \sqrt{3})}})$

$\tan^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$

Q 29 :

If the domain of the function $f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$ is $(α, β],$ then $36$ $| α + β |$ is equal to [2023]

45
63
72
54

1

2

3

4

(1)

We have,

$f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$

Let $f_{1} (x) = \log_{e} (4 x^{2} + 11 x + 6)$

$f_{2} (x) = \sin^{- 1} (4 x + 3)$

and $f_{3} (x) = \cos^{- 1} (\frac{10 x + 6}{3})$

Now, we will find the domain of $f_{1}, f_{2}$ and $f_{3}$

Consider, $f_{1} (x) = \log (4 x^{2} + 11 x + 6)$

Now, $4 x^{2} + 11 x + 6 > 0$

$\Rightarrow x > \frac{- 11 + \sqrt{121 - 96}}{8} and x < \frac{- 11 - \sqrt{121 - 96}}{8}$

$\Rightarrow x > \frac{- 3}{4} and x < - 2$

Consider, $f_{2} (x) = \sin^{- 1} (4 x + 3)$

So, $- 1 \leq 4 x + 3 \leq 1$

$\Rightarrow - 4 \leq 4 x \leq - 2 \Rightarrow - 1 \leq x \leq - \frac{1}{2}$

$∴$ $D (f_{2} (x)) is$ $[- 1, - \frac{1}{2}]$

Consider, $f_{3} (x) = \cos^{- 1} (\frac{10 x + 6}{3})$

So, $- 1 \leq \frac{10 x + 6}{3} \leq 1$

$\Rightarrow - 3 \leq 10 x + 6 \leq 3 \Rightarrow \frac{- 9}{10} \leq x \leq \frac{- 3}{10}$

$\Rightarrow D (f_{3} (x)) is [\frac{- 9}{10}, \frac{- 3}{10}]$

Now, $D (f (x))$ is $D (f_{1} (x)) \cap D (f_{2} (x)) \cap D (f_{3} (x))$

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

So, $α = - \frac{3}{4} and β = \frac{- 1}{2}$

So, $36$ $| α + β |$ $= 36$ $| - \frac{3}{4} - \frac{1}{2} |$ $= 36 \times \frac{5}{4} = 45$

Q 30 :

If $\sin^{- 1} \frac{α}{17} + \cos^{- 1} \frac{4}{5} - \tan^{- 1} \frac{77}{36} = 0,$ $0 < α < 13,$ then $\sin^{- 1} (\sin α) + \cos^{- 1} (\cos α)$ is equal to [2023]

$16 - 5 π$
$16$
$0$
$π$

1

2

3

4

(4)

We have,

$\sin^{- 1} \frac{α}{17} + \cos^{- 1} \frac{4}{5} - \tan^{- 1} \frac{77}{36} = 0; 0 < α < 13$

$\Rightarrow \sin^{- 1} \frac{α}{17} = \tan^{- 1} \frac{77}{36} - \cos^{- 1} \frac{4}{5} = \tan^{- 1} \frac{77}{36} - \sin^{- 1} \frac{3}{5}$

$\Rightarrow \frac{α}{17} = \sin (\tan^{- 1} \frac{77}{36} - \sin^{- 1} \frac{3}{5})$

$= \sin (\tan^{- 1} \frac{77}{36}) \cdot \cos (\sin^{- 1} \frac{3}{5}) - \cos (\tan^{- 1} \frac{77}{36}) \sin (\sin^{- 1} \frac{3}{5})$

$\Rightarrow \frac{α}{17} = \sin (\sin^{- 1} \frac{77}{85}) \cdot \cos (\cos^{- 1} \frac{4}{5}) - \cos (\cos^{- 1} \frac{36}{85}) (\frac{3}{5})$

$= \frac{77}{85} \times \frac{4}{5} - \frac{36}{85} \times \frac{3}{5} \Rightarrow α = 8$

$Now, \sin^{- 1} (\sin α) + \cos^{- 1} (\cos α) = \sin^{- 1} (\sin 8) + \cos^{- 1} (\cos 8)$

$= 3 π - 8 + 8 - 2 π [∵ 3 π - 8 \in [- \frac{π}{2}, \frac{π}{2}]] = π$

Topic Question Set

Q 1 :

If the domain of the function $\sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + \log_{e} (\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10})$ is $(α, β]$ , then $3 α + 10 β$ is equal to [2024]

Q 2 :

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[- 1, 1]$ such that $\cos^{- 1} x - \sin^{- 1} y = α,$ $\frac{- π}{2} \leq α \leq π .$ Then, the minimum value of $x^{2} + y^{2} + 2 x y$ $\sin α$ is [2024]

Q 3 :

If the domain of the function $f (x) = \sin^{- 1} (\frac{x - 1}{2 x + 3})$ is $R - (α, β),$ then $12 α β$ is equal to : [2024]

Q 4 :

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$ is: [2024]

Q 5 :

Let $x = \frac{m}{n}$ $(m, n$ are co-prime natural numbers $)$ be a solution of the equation $\cos (2 \sin^{- 1} x) = \frac{1}{9}$ and let $α, β (α > β)$ be the roots of the equation $m x^{2} - n x - m + n = 0$ . Then the point $(α, β)$ lies on the line [2024]

Q 6 :

If the domain of the function $f (x) = \cos^{- 1} (\frac{2 - | x |}{4}) + {\log_{e} (3 - x)}^{- 1}$ is $[- α, β) - {γ},$ then $α + β + γ$ is equal to [2024]

Q 7 :

For $α, β, γ \neq 0, if \sin^{- 1} α + \sin^{- 1} β + \sin^{- 1} γ = π and (α + β + γ) (α - γ + β) = 3 α β, then γ equals$ [2024]

Q 8 :

If $a = \sin^{- 1} (\sin (5)) and b = \cos^{- 1} (\cos (5)),$ then $a^{2} + b^{2}$ is equal to [2024]

Q 9 :

For $n \in N, if \cot^{- 1} 3 + \cot^{- 1} 4 + \cot^{- 1} 5 + \cot^{- 1} n = \frac{π}{4},$ then $n$ is equal to ______. [2024]

Q 10 :

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5},$ is _____. [2024]

Q 11 :

If the domain of the function $f (x) = \log_{e} (\frac{2 x + 3}{4 x^{2} + x - 3}) + \cos^{- 1} (\frac{2 x - 1}{x + 2})$ is $(α, β],$ then the value of $5 β - 4 α$ is equal to [2024]

Q 12 :

Considering the principal values of the inverse trigonometric functions, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$ , $- \frac{1}{2} < x < \frac{1}{\sqrt{2}}$ , is equal to [2025]

Q 13 :

The sum of the infinite series $\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$ is: [2025]

Q 14 :

The value of $\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$ is equal to [2025]

Q 15 :

Let the domains of the functions $f (x) = \log_{4} \log_{3}$ $\log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))$ and $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$ be ( $α, β$ ) and [ $γ, δ$ ], respectively. Then $α^{2} + β^{2} + γ^{2} + δ^{2}$ is equal to : [2025]

Q 16 :

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16 ({(\sec^{- 1} x)}^{2} + {({cosec}^{- 1} x)}^{2})$ is : [2025]

Q 17 :

If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ , then $\cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)$ is equal to [2025]

Q 18 :

If $α > β > γ > 0$ , then the expression $\cot^{- 1} {β + \frac{(1 + β^{2})}{(α - β)}} + \cot^{- 1} {γ + \frac{(1 + γ^{2})}{(β - γ)}} + \cot^{- 1} {α + \frac{(1 + α^{2})}{(γ - α)}}$ is equal to: [2025]

Q 19 :

$\cos (\sin^{- 1} \frac{3}{5} + \sin^{- 1} \frac{5}{13} + \sin^{- 1} \frac{33}{65})$ is equal to : [2025]

Q 20 :

Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = $\sec^{- 1}$ (2[x] + 1) is : [2025]

Q 21 :

If $y = \cos (\frac{π}{3} + \cos^{- 1} \frac{x}{2})$ , then ${(x - y)}^{2} + 3 y^{2}$ is equal to [2025]

Q 22 :

Let the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))$ be $(γ, δ)$ . Then $| 7 (α + β) + 4 (γ + δ) |$ is equal to __________. [2025]

Q 23 :

If for some $α, β; α \leq β, α + β = 8$ and $\sec^{2} (\tan^{- 1} α) + {cosec}^{2} (\cot^{- 1} β) = 36$ , then $α^{2} + β$ is __________. [2025]

Q 24 :

Let $S = {x : \cos^{- 1} x = π + \sin^{- 1} x + \sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} {(2 x - 1)}^{2}$ is equal to __________. [2025]

Q 25 :

The range of $f (x) = 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$ is [2023]

Q 26 :

Let $S$ be the set of all solutions of the equation $\cos^{- 1} (2 x) - 2 \cos^{- 1} (\sqrt{1 - x^{2}}) = π,$ $x \in [- \frac{1}{2}, \frac{1}{2}] .$ Then $\sum_{x \in S} 2 \sin^{- 1} (x^{2} - 1)$ is equal to [2023]

(2)

Q 27 :

Let $S = {x \in R : 0 < x < 1 and 2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})} .$ If $n (S)$ denotes the number of elements in S, then [2023]

Q 28 :

$\tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$ is equal to [2023]

Q 29 :

If the domain of the function $f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$ is $(α, β],$ then $36$ $| α + β |$ is equal to [2023]

Q 30 :

If $\sin^{- 1} \frac{α}{17} + \cos^{- 1} \frac{4}{5} - \tan^{- 1} \frac{77}{36} = 0,$ $0 < α < 13,$ then $\sin^{- 1} (\sin α) + \cos^{- 1} (\cos α)$ is equal to [2023]

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Topic Question Set

Q 1 : If the domain of the function sin-1(3x-222x-19)+loge(3x2-8x+5x2-3x-10) is (α,β], then 3α+10β is equal to [2024]

Q 2 : Given that the inverse trigonometric function assumes principal values only. Let x,y be any two real numbers in [-1,1] such that cos-1x-sin-1y=α, -π2≤α≤π. Then, the minimum value of x2+y2+2xy sinα is [2024]

Q 3 : If the domain of the function f(x)=sin-1(x-12x+3) is R-(α,β), then 12αβ is equal to : [2024]

Q 4 : Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying tan-1(x)+tan-1(2x)=π4 is: [2024]

Q 5 : Let x=mn (m,n are co-prime natural numbers) be a solution of the equation cos(2sin-1x)=19 and let α,β(α>β) be the roots of the equation mx2-nx-m+n=0. Then the point (α,β) lies on the line [2024]

Q 6 : If the domain of the function f(x)=cos-1(2-|x|4)+{loge(3-x)}-1 is [-α,β)-{γ}, then α+β+γ is equal to [2024]

Q 7 : For α,β,γ≠0, if sin-1α+sin-1β+sin-1γ=π and (α+β+γ)(α-γ+β)=3αβ, then γ equals [2024]

Q 8 : If a=sin-1(sin(5)) and b=cos-1(cos(5)), then a2+b2 is equal to [2024]

Q 9 : For n∈N, if cot-13+cot-14+cot-15+cot-1n=π4, then n is equal to ______. [2024]

Q 10 : Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2sin-1x+3cos-1x=2π5, is _____. [2024]

Q 11 : If the domain of the function f(x)=loge(2x+34x2+x-3)+cos-1(2x-1x+2) is (α,β], then the value of 5β-4α is equal to [2024]

Q 12 : Considering the principal values of the inverse trigonometric functions, sin–1(32x+121–x2), –12<x<12, is equal to [2025]

Q 13 : The sum of the infinite series cot–1(74)+cot–1(194)+cot–1(394)+cot–1(674)+... is: [2025]

Q 14 : The value of cot–1(1+tan2(2)–1tan(2))–cot–1(1+tan2(12)+1tan(12)) is equal to [2025]

Q 15 : Let the domains of the functions f(x)=log4log3 log7(8–log2(x2+4x+5)) and g(x)=sin–1(7x+10x–2) be (α,β) and [γ,δ], respectively. Then α2+β2+γ2+δ2 is equal to : [2025]

Q 16 : Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of 16((sec–1x)2+(cosec–1x)2) is : [2025]

Q 17 : If π2≤x≤3π4, then cos–1(1213cosx+513sinx) is equal to [2025]

Q 18 : If α>β>γ>0, then the expression cot–1{β+(1+β2)(α–β)}+cot–1{γ+(1+γ2)(β–γ)}+cot–1{α+(1+α2)(γ–α)} is equal to: [2025]

Q 19 : cos(sin–135+sin–1513+sin–13365) is equal to : [2025]

Q 20 : Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = sec–1(2[x] + 1) is : [2025]

Q 21 : If y=cos(π3+cos–1x2), then (x–y)2+3y2 is equal to [2025]

Q 22 : Let the domain of the function f(x)=cos–1(4x+53x–7) be [α,β] and the domain of g(x)=log2(2–6 log27(2x+5)) be (γ,δ). Then |7(α+β)+4(γ+δ)| is equal to __________. [2025]

Q 23 : If for some α,β;α≤β,α+β=8 and sec2(tan–1α)+cosec2(cot–1β)=36, then α2+β is __________. [2025]

Q 24 : Let S={x : cos–1x=π+sin–1x+sin–1(2x+1)}. Then ∑x∈S(2x–1)2 is equal to __________. [2025]

Q 25 : The range of f(x)=4sin-1(x2x2+1) is [2023]

Q 26 : Let S be the set of all solutions of the equation cos-1(2x)-2cos-1(1-x2)=π, x∈[-12, 12]. Then ∑x∈S2sin-1(x2-1) is equal to [2023]

(2)

Q 27 : Let S={ x∈R:0<x<1 and 2tan-1(1-x1+x)=cos-1(1-x21+x2)}. If n(S) denotes the number of elements in S, then [2023]

Q 28 : tan-1(1+33+3)+sec-1(8+436+33) is equal to [2023]

Q 29 : If the domain of the function f(x)=loge(4x2+11x+6)+sin-1(4x+3)+cos-1(10x+63) is (α,β], then 36|α+β| is equal to [2023]

Q 30 : If sin-1α17+cos-145-tan-17736=0, 0<α<13, then sin-1(sinα)+cos-1(cosα) is equal to [2023]

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Q 1 :

If the domain of the function $\sin^{- 1} (\frac{3 x - 22}{2 x - 19}) + \log_{e} (\frac{3 x^{2} - 8 x + 5}{x^{2} - 3 x - 10})$ is $(α, β]$ , then $3 α + 10 β$ is equal to [2024]

Q 2 :

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[- 1, 1]$ such that $\cos^{- 1} x - \sin^{- 1} y = α,$ $\frac{- π}{2} \leq α \leq π .$ Then, the minimum value of $x^{2} + y^{2} + 2 x y$ $\sin α$ is [2024]

Q 3 :

If the domain of the function $f (x) = \sin^{- 1} (\frac{x - 1}{2 x + 3})$ is $R - (α, β),$ then $12 α β$ is equal to : [2024]

Q 4 :

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan^{- 1} (x) + \tan^{- 1} (2 x) = \frac{π}{4}$ is: [2024]

Q 5 :

Let $x = \frac{m}{n}$ $(m, n$ are co-prime natural numbers $)$ be a solution of the equation $\cos (2 \sin^{- 1} x) = \frac{1}{9}$ and let $α, β (α > β)$ be the roots of the equation $m x^{2} - n x - m + n = 0$ . Then the point $(α, β)$ lies on the line [2024]

Q 6 :

If the domain of the function $f (x) = \cos^{- 1} (\frac{2 - | x |}{4}) + {\log_{e} (3 - x)}^{- 1}$ is $[- α, β) - {γ},$ then $α + β + γ$ is equal to [2024]

Q 7 :

For $α, β, γ \neq 0, if \sin^{- 1} α + \sin^{- 1} β + \sin^{- 1} γ = π and (α + β + γ) (α - γ + β) = 3 α β, then γ equals$ [2024]

Q 8 :

If $a = \sin^{- 1} (\sin (5)) and b = \cos^{- 1} (\cos (5)),$ then $a^{2} + b^{2}$ is equal to [2024]

Q 9 :

For $n \in N, if \cot^{- 1} 3 + \cot^{- 1} 4 + \cot^{- 1} 5 + \cot^{- 1} n = \frac{π}{4},$ then $n$ is equal to ______. [2024]

Q 10 :

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5},$ is _____. [2024]

Q 11 :

If the domain of the function $f (x) = \log_{e} (\frac{2 x + 3}{4 x^{2} + x - 3}) + \cos^{- 1} (\frac{2 x - 1}{x + 2})$ is $(α, β],$ then the value of $5 β - 4 α$ is equal to [2024]

Q 12 :

Considering the principal values of the inverse trigonometric functions, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$ , $- \frac{1}{2} < x < \frac{1}{\sqrt{2}}$ , is equal to [2025]

Q 13 :

The sum of the infinite series $\cot^{- 1} (\frac{7}{4}) + \cot^{- 1} (\frac{19}{4}) + \cot^{- 1} (\frac{39}{4}) + \cot^{- 1} (\frac{67}{4}) + . . .$ is: [2025]

Q 14 :

The value of $\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$ is equal to [2025]

Q 15 :

Let the domains of the functions $f (x) = \log_{4} \log_{3}$ $\log_{7} (8 - \log_{2} (x^{2} + 4 x + 5))$ and $g (x) = \sin^{- 1} (\frac{7 x + 10}{x - 2})$ be ( $α, β$ ) and [ $γ, δ$ ], respectively. Then $α^{2} + β^{2} + γ^{2} + δ^{2}$ is equal to : [2025]

Q 16 :

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16 ({(\sec^{- 1} x)}^{2} + {({cosec}^{- 1} x)}^{2})$ is : [2025]

Q 17 :

If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ , then $\cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)$ is equal to [2025]

Q 18 :

If $α > β > γ > 0$ , then the expression $\cot^{- 1} {β + \frac{(1 + β^{2})}{(α - β)}} + \cot^{- 1} {γ + \frac{(1 + γ^{2})}{(β - γ)}} + \cot^{- 1} {α + \frac{(1 + α^{2})}{(γ - α)}}$ is equal to: [2025]

Q 19 :

$\cos (\sin^{- 1} \frac{3}{5} + \sin^{- 1} \frac{5}{13} + \sin^{- 1} \frac{33}{65})$ is equal to : [2025]

Q 20 :

Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = $\sec^{- 1}$ (2[x] + 1) is : [2025]

Q 21 :

If $y = \cos (\frac{π}{3} + \cos^{- 1} \frac{x}{2})$ , then ${(x - y)}^{2} + 3 y^{2}$ is equal to [2025]

Q 22 :

Let the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 5}{3 x - 7})$ be $[α, β]$ and the domain of $g (x) = \log_{2} (2 - 6 \log_{27} (2 x + 5))$ be $(γ, δ)$ . Then $| 7 (α + β) + 4 (γ + δ) |$ is equal to __________. [2025]

Q 23 :

If for some $α, β; α \leq β, α + β = 8$ and $\sec^{2} (\tan^{- 1} α) + {cosec}^{2} (\cot^{- 1} β) = 36$ , then $α^{2} + β$ is __________. [2025]

Q 24 :

Let $S = {x : \cos^{- 1} x = π + \sin^{- 1} x + \sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} {(2 x - 1)}^{2}$ is equal to __________. [2025]

Q 25 :

The range of $f (x) = 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$ is [2023]

Q 26 :

Let $S$ be the set of all solutions of the equation $\cos^{- 1} (2 x) - 2 \cos^{- 1} (\sqrt{1 - x^{2}}) = π,$ $x \in [- \frac{1}{2}, \frac{1}{2}] .$ Then $\sum_{x \in S} 2 \sin^{- 1} (x^{2} - 1)$ is equal to [2023]

Q 27 :

Let $S = {x \in R : 0 < x < 1 and 2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})} .$ If $n (S)$ denotes the number of elements in S, then [2023]

Q 28 :

$\tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$ is equal to [2023]

Q 29 :

If the domain of the function $f (x) = \log_{e} (4 x^{2} + 11 x + 6) + \sin^{- 1} (4 x + 3) + \cos^{- 1} (\frac{10 x + 6}{3})$ is $(α, β],$ then $36$ $| α + β |$ is equal to [2023]

Q 30 :

If $\sin^{- 1} \frac{α}{17} + \cos^{- 1} \frac{4}{5} - \tan^{- 1} \frac{77}{36} = 0,$ $0 < α < 13,$ then $\sin^{- 1} (\sin α) + \cos^{- 1} (\cos α)$ is equal to [2023]