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

2

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]

0%

(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%

(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 $\cos^{- 1} x \in [0, π]$

$∴ No solution exists.$

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]

0%

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%

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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]

0%

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]

0%

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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]