JEE Main Previous Year Trigonometry Questions and Solutions

Q 1 :

Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

$\frac{6 - \sqrt{6}}{(3 \sqrt{6} - 2)}$
$\frac{4}{(3 \sqrt{6} - 2)}$
$\frac{4}{(3 \sqrt{6} + 2)}$
$\frac{6 + \sqrt{6}}{(3 \sqrt{6} + 2)}$

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4

(2)

We have, $4 \cos θ - 3 \sin θ = 1$

$\Rightarrow 4 \cos θ - 1 = 3 \sin θ \Rightarrow {(4 \cos θ - 1)}^{2} = {(3 \sin θ)}^{2}$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 \sin^{2} θ$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 (1 - \cos^{2} θ)$

$\Rightarrow 25 \cos^{2} θ - 8 \cos θ - 8 = 0$

$\Rightarrow \cos θ = \frac{8 \pm \sqrt{64 + 800}}{50}$

$= \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12 \sqrt{6}}{50} = \frac{4 \pm 6 \sqrt{6}}{25}$

Since, $θ \in [0, π / 4]$ so $\cos θ = \frac{4 + 6 \sqrt{6}}{25}$

$= \frac{(4 - 6 \sqrt{6}) (4 + 6 \sqrt{6})}{25 (4 - 6 \sqrt{6})} = \frac{- 200}{25 \times 2 (2 - 3 \sqrt{6})} = \frac{4}{3 \sqrt{6} - 2}$

Q 2 :

Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

$18 π$
$9 π$
$6 π$
$15 π$

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3

4

(3)

$| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8}$

$\Rightarrow$ $| \cos θ \cos (\frac{π}{3} - θ) \cos (\frac{π}{3} + θ) |$ $\leq \frac{1}{8}$

$\Rightarrow \frac{1}{4} | \cos 3 θ | \leq \frac{1}{8} \Rightarrow | \cos 3 θ | \leq \frac{1}{2} \Rightarrow \frac{- 1}{2} \leq \cos 3 θ \leq \frac{1}{2}$

$cos 3 θ is max if \cos 3 θ = \frac{1}{2}$

$\Rightarrow 3 θ = 2 n π \pm \frac{π}{3} \Rightarrow θ = \frac{2 n π}{3} \pm \frac{π}{9}$

when $θ \in [0, 2 π]$

$∴$ $θ = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}, \frac{11 π}{9}, \frac{13 π}{9}, \frac{17 π}{9}$

$Hence, \sum θ_{i} = 6 π$

Q 3 :

The number of solutions of the equation $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0;$ $x \in [- 2 π, 2 π]$ is : [2024]

0
3
1
2

1

2

3

4

(1)

We have, $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0$

$\Rightarrow 4 (1 - \cos^{2} x) - 4 \cos^{3} x + 9 - 4 \cos x = 0$

$\Rightarrow 4 \cos^{3} x + 4 \cos^{2} x + 4 \cos x = 13$

Ist clear that L.H.S. $\leq 12$ and never 13

$∴$ Solution does not exists.

Q 4 :

If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

$\frac{\sqrt{10} - 10}{12}$
$\frac{10 - \sqrt{10}}{6}$
$\frac{\sqrt{10} - 10}{6}$
$\frac{10 - \sqrt{10}}{12}$

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3

4

(1)

We have, $4 \cos θ + 5 \sin θ = 1$

$\Rightarrow 4 (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + 5 (\frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) = 1$

$\Rightarrow 5 \tan^{2} (θ / 2) - 10 \tan (θ / 2) - 3 = 0$

$\Rightarrow \tan (θ / 2) = \frac{10 \pm \sqrt{100 - 4 (5) (- 3)}}{2 \times 5} = \frac{10 \pm \sqrt{160}}{10} = \frac{5 \pm \sqrt{40}}{5}$

$∵$ $α \in (\frac{- π}{2}, \frac{π}{2})$ So, $\frac{α}{2} \in (\frac{- π}{4}, \frac{π}{4})$

$\Rightarrow \tan (\frac{α}{2}) \in (- 1, 1)$ $∴$ $\tan \frac{α}{2} = \frac{5 - \sqrt{40}}{5}$

$Hence, \tan α = \frac{2 \tan \frac{α}{2}}{1 - \tan^{2} \frac{α}{2}} = \frac{2 (\frac{5 - \sqrt{40}}{5})}{1 - {(1 - \frac{\sqrt{40}}{5})}^{2}}$

$= \frac{2 (\frac{5 - \sqrt{40}}{5})}{1 - (1 + \frac{40}{25} - \frac{2 \sqrt{40}}{5})}$

$= \frac{2 (\frac{5 - \sqrt{40}}{5})}{\frac{2 \sqrt{40}}{5} - \frac{8}{5}} = \frac{5 - \sqrt{40}}{\sqrt{40} - 4} \times \frac{\sqrt{40} + 4}{\sqrt{40} + 4}$

$= \frac{5 \sqrt{40} + 20 - 40 - 4 \sqrt{40}}{40 - 16} = \frac{\sqrt{40} - 20}{24} = \frac{\sqrt{10} - 10}{12}$

Q 5 :

The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

0
- 1
1
3

1

2

3

4

(2)

We have, $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{\cos 2 x (3 + \cos^{2} 2 x)}{(\cos^{2} x - \sin^{2} x) ((\cos^{2} x + \sin^{2} x)^{2} - \cos^{2} x \sin^{2} x)} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{(3 + \cos^{2} 2 x)}{(1 - \cos^{2} x \sin^{2} x)} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{3 + {(\cos^{2} x - \sin^{2} x)}^{2}}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{3 + {(\cos^{2} x + \sin^{2} x)}^{2} - 4 \cos^{2} x \sin^{2} x}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{4 - 4 \cos^{2} x \sin^{2} x}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow 4 = x^{3} - x^{2} + 6 \Rightarrow x^{3} - x^{2} + 2 = 0$

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

$\Rightarrow x = - 1 is the only real solution.$

Q 6 :

For $α, β \in (0, \frac{π}{2}),$ let $3 \sin (α + β) = 2 \sin (α - β)$ and a real number $k$ be such that $\tan α = k$ $\tan β .$ Then, the value of $k$ is equal to [2024]

$- \frac{2}{3}$
$\frac{2}{3}$
$5$
$- 5$

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4

(4)

We have, $3 \sin (α + β) = 2 \sin (α - β)$

$\Rightarrow \frac{\sin (α + β)}{\sin (α - β)} = \frac{2}{3} \Rightarrow \frac{\sin (α + β) + \sin (α - β)}{\sin (α + β) - \sin (α - β)} = \frac{5}{- 1}$ (Using componendo and dividendo)

$\Rightarrow \frac{2 \sin α \cos β}{2 \cos α \sin β} = - 5 \Rightarrow \frac{\tan α}{\tan β} = - 5$

$\Rightarrow \tan α = - 5 \tan β$ $∴$ $k = - 5$

Q 7 :

The number of solutions of the equation $e^{\sin x} - 2 e^{- \sin x} = 2,$ is [2024]

0
2
1
more than 2

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4

(1)

We have, $e^{\sin x} - 2 e^{- \sin x} = 2$ ...(i)

Let $e^{\sin x} = t$

$∴$ $(i) becomes$ $t - \frac{2}{t} = 2$

$\Rightarrow t^{2} - 2 = 2 t \Rightarrow t^{2} - 2 t - 2 = 0$

$\Rightarrow t = \frac{2 \pm \sqrt{4 - 4 (1) (- 2)}}{2} \Rightarrow t = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$

Now, $e^{\sin x} = 1 + \sqrt{3} and e^{\sin x} = 1 - \sqrt{3} (Not possible)$

$\Rightarrow \sin x = \ln (1 + \sqrt{3}) > 1 (Not possible)$

$∴$ $There is no solution.$

Q 8 :

Let $S = {\sin^{2} 2 θ : (\sin^{4} θ + \cos^{4} θ) x^{2} + (\sin 2 θ) x + (\sin^{6} θ + \cos^{6} θ) = 0 has real roots} .$ If $α$ and $β$ be the smallest and largest elements of the set $S,$ respectively, then $3 ({(α - 2)}^{2} + {(β - 1)}^{2})$ equals _________ . [2024]

(4)

For real roots $D \geq 0$

$\Rightarrow {(\sin 2 θ)}^{2} - 4 (\sin^{4} θ + \cos^{4} θ) (\sin^{6} θ + \cos^{6} θ) \geq 0$

$\Rightarrow \sin^{2} 2 θ \geq 4 (\sin^{4} θ + \cos^{4} θ) (\sin^{6} θ + \cos^{6} θ)$

Put $\sin^{2} 2 θ = t$

$\Rightarrow t \geq 4 (1 - \frac{t}{2}) (1 - \frac{3 t}{4}) \Rightarrow 2 t \geq (2 - t) (4 - 3 t)$

$\Rightarrow 3 t^{2} - 12 t + 8 \leq 0$ $\Rightarrow t^{2} - 4 t + \frac{8}{3} \leq 0$

$\Rightarrow {(t - 2)}^{2} + \frac{8}{3} - 4 \leq 0$ $\Rightarrow {(t - 2)}^{2} \leq \frac{4}{3}$

$\Rightarrow - \frac{2}{\sqrt{3}} \leq t - 2 \leq \frac{2}{\sqrt{3}}$ $\Rightarrow 2 - \frac{2}{\sqrt{3}} \leq t \leq 2 + \frac{2}{\sqrt{3}}$

$∵$ $t \in [0, 1]$ $\Rightarrow 2 - \frac{2}{\sqrt{3}} \leq t \leq 1$

Since, $α$ and $β$ be the smallest and largest elements of set $S$

$∴$ $α = 2 - \frac{2}{\sqrt{3}}, β = 1$

$Hence, 3 [{(α - 2)}^{2} + {(β - 1)}^{2}] = 3 [{(2 - \frac{2}{\sqrt{3}} - 2)}^{2} + {(1 - 1)}^{2}]$ $= 3 [\frac{4}{3} + 0] = 4 .$

Q 9 :

Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[p, q]$ and $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °,$ then $p q r$ is equal to _____ . [2024]

(48)

$\cos 2 x + a \sin x = 2 a - 7$

$1 - 2 \sin^{2} x + a \sin x = 2 a - 7$

Let $t = \sin x$ $∴$ $2 t^{2} - a t + 2 a - 8 = 0$

$\Rightarrow (t - 2) (2 t + 4 - a) = 0 \Rightarrow t = 2 or t = \frac{a - 4}{2}$

$∴$ $\sin x = 2 (not possible) or 2 \sin x = a - 4$

$\Rightarrow a = 2 \sin x + 4 \in [2, 6]$ $[∵$ $\sin x \in [- 1, 1]]$

$So, p = 2, q = 6$

Now, $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °$

$= (\tan 9 ° + \cot 9 °) - (\tan 27 ° + \cot 27 °)$

$= 2 c o s e c 18 ° - 2 c o s e c 54 °$ $[∵$ $\tan θ + \cot θ = 2 c o s e c 2 θ]$

$= 2 [\frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1}] = 4 \Rightarrow r = 4$

$∴$ $p q r = 2 \times 6 \times 4 = 48$

Q 10 :

If $2 \tan^{2} θ - 5 \sec θ = 1$ has exactly 7 solutions in the interval $[0, \frac{n π}{2}],$ for the least value of $n \in N,$ then $\sum_{k = 1}^{n} \frac{k}{2^{k}}$ is equal to [2024]

$\frac{1}{2^{15}} (2^{14} - 14)$
$1 - \frac{15}{2^{13}}$
$\frac{1}{2^{13}} (2^{14} - 15)$
$\frac{1}{2^{14}} (2^{15} - 15)$

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(3)

Given, $2 \tan^{2} θ - 5 \sec θ = 1$

$\Rightarrow 2 (\sec^{2} θ - 1) - 5 \sec θ = 1$

$\Rightarrow 2 \sec^{2} θ - 5 \sec θ - 3 = 0$

$\Rightarrow (2 \sec θ + 1) (\sec θ - 3) = 0 \Rightarrow \sec θ = - \frac{1}{2}, 3$

$\Rightarrow \cos θ = - 2, \frac{1}{3}$ $(\cos θ \in [- 1, 1])$

$The least value of n \in N, for which exactly 7 solutions are possible is 13 .$

So, $\sum_{k = 1}^{13} \frac{k}{2^{k}} = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \dots + \frac{13}{2^{13}}$

Let $S = \frac{1}{2} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \dots + \frac{13}{2^{13}}$

$\frac{S}{2} = \frac{1}{2^{2}} + \frac{2}{2^{3}} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}$

$\Rightarrow \frac{S}{2} = \frac{1}{2} [\frac{1 - \frac{1}{2^{13}}}{1 - \frac{1}{2}}] - \frac{13}{2^{14}} \Rightarrow S = 2 (\frac{2^{13} - 1}{2^{13}}) - \frac{13}{2^{13}}$

$\Rightarrow S = \frac{2^{14} - 15}{2^{13}}$

Q 11 :

If $2 \sin^{3} x + \sin 2 x \cos x + 4 \sin x - 4 = 0$ has exactly 3 solutions in the interval $[0, \frac{n π}{2}], n \in N,$ then the roots of the equation $x^{2} + n x + (n - 3) = 0$ belong to [2024]

$(0, \infty)$
$(- \infty, 0)$
$(- \frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2})$
$Z$

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2

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(2)

$2 \sin^{3} x + \sin 2 x \cos x + 4 \sin x - 4 = 0$

$\Rightarrow 2 \sin^{3} x + 2 \sin x \cos^{2} x + 4 \sin x - 4 = 0$

$\Rightarrow 2 \sin x (\sin^{2} x + \cos^{2} x) + 4 \sin x = 4 \Rightarrow \sin x = \frac{4}{6} = \frac{2}{3}$

Clearly, from the graph, the given equation has exactly 3 solutions in $[0, \frac{5 π}{2}] \Rightarrow x = 5$

So, we have equation, $x^{2} + 5 x + 2 = 0$

$\Rightarrow x = \frac{- 5 + \sqrt{25 - 8}}{2} \Rightarrow x = \frac{- 5 + \sqrt{17}}{2} \in (- \infty, 0)$

Q 12 :

The number of solutions of $\sin^{2} x + (2 + 2 x - x^{2}) \sin x - 3 {(x - 1)}^{2} = 0,$ where $- π \leq x \leq π,$ is _________. [2024]

(2)

We have $\sin^{2} x + (2 + 2 x - x^{2}) \sin x - 3 {(x - 1)}^{2} = 0$ where $- π \leq x \leq π$

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

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

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

$\Rightarrow \sin x + 3 \neq 0 \Rightarrow \sin x - {(x - 1)}^{2} = 0 \Rightarrow \sin x = {(x - 1)}^{2}$

There are two points of intersection.

So, number of solutions are 2.

Q 13 :

If $θ \in [- 2 π, 2 π]$ , then the number of solutions of $2 \sqrt{2} \cos^{2} θ + (2 - \sqrt{6}) \cos θ - \sqrt{3} = 0$ , is equal to : [2025]

8
12
10
6

1

2

3

4

(1)

We have, $2 \sqrt{2} \cos^{2} θ + 2 \cos θ - \sqrt{6} \cos θ - \sqrt{3} = 0$

$\Rightarrow (2 \cos θ - \sqrt{3}) (\sqrt{2} \cos θ + 1) = 0 \Rightarrow \cos θ = [\frac{\sqrt{3}}{2}, \frac{- 1}{\sqrt{2}}]$

$\Rightarrow \frac{- 11 π}{6}, \frac{- π}{6}, \frac{π}{6}, \frac{11 π}{6}, \frac{5 π}{4}, \frac{3 π}{4}, \frac{- 5 π}{4}, \frac{- 3 π}{4}$

Hence, number of solutions = 8.

Q 14 :

If $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , then the number of solutions of $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ , is equal to : [2025]

7
10
8
6

1

2

3

4

(4)

We have, $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$

Let $cosec θ = x$

$∴ \sqrt{3} x^{2} - 2 (\sqrt{3} - 1) x - 4 = 0$

$x = \frac{2 (\sqrt{3} - 1) \pm \sqrt{4 {(\sqrt{3} - 1)}^{2} + 16 \sqrt{3}}}{2 \sqrt{3}}$

$= \frac{2 (\sqrt{3} - 1) \pm \sqrt{16 - 8 \sqrt{3} + 16 \sqrt{3}}}{2 \sqrt{3}}$

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

$= \frac{2 (\sqrt{3} - 1) \pm 2 (\sqrt{3} + 1)}{2 \sqrt{3}}$

$= \frac{(\sqrt{3} - 1) \pm (\sqrt{3} + 1)}{\sqrt{3}}$

$∴ x_{1} = \frac{(\sqrt{3} - 1) + (\sqrt{3} + 1)}{\sqrt{3}} = 2$

and $x_{2} = \frac{(\sqrt{3} - 1) - (\sqrt{3} + 1)}{\sqrt{3}} = \frac{- 2}{\sqrt{3}}$

Now, Put x = $cosec θ$

When $cosec θ = 2 \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}, \frac{5 π}{6}, \frac{- 7 π}{6}$

When $cosec θ = \frac{- 2}{\sqrt{3}} \Rightarrow \sin θ = - \frac{\sqrt{3}}{2}$

$\Rightarrow θ = \frac{4 π}{3}, \frac{5 π}{3}, \frac{- π}{3}, \frac{- 2 π}{3}$

Now, $\frac{5 π}{3} \notin [\frac{- 7 π}{6}, \frac{4 π}{3}]$

$∴$ Required number of solutions are = 6.

Q 15 :

The number of solutions of the equation $2 x + 3 \tan x = π, x \in [- 2 π, 2 π] - {\pm \frac{π}{2}, \pm \frac{3 π}{2}}$ is [2025]

4
5
6
3

1

2

3

4

(2)

We have, 2x + 3 tan x = $π$

$\Rightarrow \tan x = \frac{π}{3} - \frac{2 x}{3}$

$∴$ Number of solutions = 5.

Q 16 :

The number of solutions of the equation $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$ is [2025]

5
3
6
4

1

2

3

4

(1)

Given, $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$

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

$\Rightarrow 2 \sqrt{3} \sin^{2} x + 4 \sin x - \sqrt{3} s in x - 2 = 0$

$\Rightarrow (2 \sin x - 1) (\sqrt{3} \sin x + 2) = 0$

$\Rightarrow \sin x = \frac{1}{2} [∵ \sin x \neq \frac{- 2}{\sqrt{3}}]$

$\Rightarrow x = \frac{- 11 π}{6}, \frac{- 7 π}{6}, \frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}$

$∴$ Number of solutions = 5.

Q 17 :

If $10 \sin^{4} θ + 15 \cos^{4} θ = 6$ , then the value of $\frac{27 {cosec}^{6} θ + 8 \sec^{6} θ}{16 \sec^{8} θ}$ is [2025]

$\frac{3}{4}$
$\frac{2}{5}$
$\frac{3}{5}$
$\frac{1}{5}$

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2

3

4

(2)

We have, $10 \sin^{4} θ + 15 \cos^{4} θ = 6$

$\Rightarrow 10 \sin^{4} θ + 15 + 15 \sin^{4} θ - 30 \sin^{2} θ = 6$

Let $\sin^{2} θ = p$

$\Rightarrow 10 p^{2} + 15 + 15 p^{2} - 30 p = 6$

$\Rightarrow 25 p^{2} - 30 p = - 9 \Rightarrow {(5 p - 3)}^{2} = 0$

$∴ \sin^{2} θ = 3 / 5 and \cos^{2} θ = 2 / 5$

So, $\frac{27 {cosec}^{6} θ + 8 \sec^{6} θ}{16 \sec^{8} θ} = \frac{27 \times \frac{125}{27} + 8 \times \frac{125}{8}}{16 \times \frac{625}{16}}$

$= \frac{125 + 125}{625} = \frac{2}{5}$ .

Q 18 :

Let the product of $ω_{1} = (8 + i) \sin θ + (7 + 4 i) \cos θ$ and $ω_{2} = (1 + 8 i) \sin θ + (4 + 7 i) \cos θ$ be $α + i β$ , $i = \sqrt{- 1}$ . Let p and q be the maximum and the minimum values of $α + β$ respectively. Then p + q is equal to : [2025]

140
130
160
150

1

2

3

4

(2)

We have $ω_{1} = (8 \sin θ + 7 \cos θ) + i (\sin θ + 4 \cos θ)$ and $ω_{2} = (\sin θ + 4 \cos θ) + i (8 \sin θ + 7 \cos θ)$

Now, $ω_{1} ω_{2} = 8 \sin^{2} θ + 39 \sin θ \cos θ + 28 \cos^{2} θ - 8 \sin^{2} θ - 39 \cos θ \sin θ - 28 \cos^{2} θ$

$+ i (\sin^{2} θ + 16 \cos^{2} θ + 8 \sin θ \cos θ + 64 \sin^{2} θ + 49 \cos^{2} θ + 112 \sin θ \cos θ)$

Now, $α + i β = i (65 + 120 \sin θ \cos θ) = i (65 + 60 \sin 2 θ)$

(when sin 2 $θ$ = 1)

$p = | α + β |_{m a x} = 65 + 60 = 125$

(when sin 2 $θ$ = –1)

$q = | α + β |_{\min} = 65 - 60 = 5$

$∴$ p + q = 125 + 5 =130.

Q 19 :

The number of solutions of the equation $\cos 2 θ \cos \frac{θ}{2} + \cos \frac{5 θ}{2} = 2 \cos^{3} \frac{5 θ}{2} in [- \frac{π}{2}, \frac{π}{2}]$ is: [2025]

5
9
7
6

1

2

3

4

(3)

We have, $\cos 2 θ \cos \frac{θ}{2} + \cos \frac{5 θ}{2} = 2 \cos^{3} \frac{5 θ}{2}$

$\Rightarrow \frac{1}{2} (2 \cos 2 θ \cos \frac{θ}{2}) + \cos \frac{5 θ}{2} = \frac{1}{2} (\cos \frac{15 θ}{2} + 3 \cos \frac{5 θ}{2})$

$\Rightarrow \frac{1}{2} (\cos \frac{5 θ}{2} + \cos \frac{3 θ}{2}) = \cos \frac{5 θ}{2} \cos 5 θ$

$\Rightarrow \cos \frac{5 θ}{2} + \cos \frac{3 θ}{2} = \cos \frac{15 θ}{2} + \cos \frac{5 θ}{2}$

$\Rightarrow \cos \frac{3 θ}{2} = \cos \frac{15 θ}{2} \Rightarrow \cos \frac{15 θ}{2} - \cos \frac{3 θ}{2} = 0$

$\Rightarrow 2 \sin 3 θ \sin \frac{9 θ}{2} = 0 \Rightarrow 3 θ = n π or \frac{9 θ}{2} = m π$

$\Rightarrow θ = \frac{n π}{3} or θ = \frac{2 m π}{9} for m, n \in Z$

$∴ θ = {- \frac{π}{3}, \frac{π}{3}, 0} and θ = {- \frac{4 π}{9}, \frac{- 2 π}{9}, 0, \frac{2 π}{9}, \frac{4 π}{9}}$

Required number of solutions = 7.

Q 20 :

The sum of all values of $θ \in [0, 2 π]$ satisfying $2 \sin^{2} θ = \cos 2 θ$ and $2 \cos^{2} θ = 3 \sin θ$ is: [2025]

$π / 2$
$π$
$5 π / 6$
$4 π$

1

2

3

4

(2)

$2 \sin^{2} θ = \cos 2 θ$

$\Rightarrow 2 \sin^{2} θ = 1 - 2 \sin^{2} θ \Rightarrow 4 \sin^{2} θ = 1$

$\Rightarrow \sin^{2} θ = \frac{1}{4} \Rightarrow \sin θ = \pm \frac{1}{2}$ ... (i)

Also, $2 \cos^{2} θ = 3 \sin θ$

$\Rightarrow 2 (1 - \sin^{2} θ) = 3 \sin θ$

$\Rightarrow 2 \sin^{2} θ + 3 \sin θ - 2 = 0$

$\Rightarrow 2 \sin^{2} θ + 4 \sin θ - \sin θ - 2 = 0$

$\Rightarrow (2 \sin θ - 1) (\sin θ + 2) = 0$

$\Rightarrow \sin θ = \frac{1}{2} (\sin θ = - 2 (not possible))$ ... (ii)

Thus, $θ = \frac{π}{6}, \frac{5 π}{6} (θ \in [0, 2 π])$ [Using (i) & (ii)]

$∴$ Required sum = $π$

Q 21 :

Let $α_{θ}$ and $β θ$ be the distinct roots of $2 x^{2} + (\cos θ) x - 1 = 0, θ \in (0, 2 π)$ . If m and M are the minimum and the maximum values of $α_{θ}^{4} + β_{θ}^{4}$ , then 16(M + m) equals: [2025]

24
25
17
27

1

2

3

4

(2)

$α_{θ}^{4} + β_{θ}^{4} = {(α_{θ}^{2} + β_{θ}^{2})}^{2} - 2 α_{θ}^{2} β_{θ}^{2}$

$= {[{(α_{θ} + β_{θ})}^{2} - 2 α_{θ} β_{θ}]}^{2} - 2 {(α_{θ} β_{θ})}^{2}$

$= {[\frac{\cos^{2} θ}{4} + 1]}^{2} - 2 (\frac{1}{4}) [∵ α_{θ} + β_{θ} = \frac{- \cos θ}{2} and α_{θ} β_{θ} = - \frac{1}{2}]$

$= {(\frac{\cos^{2} θ}{4} + 1)}^{2} - \frac{1}{2}$

Since, $0 \leq \cos^{2} θ \leq 1$

$∴ M = \frac{25}{16} - \frac{1}{2} = \frac{17}{16}$ and $m = \frac{1}{2}$ .

Hence, 16(M + m) = 25

Q 22 :

Let $A = {x \in (0, π) - {\frac{π}{2}} : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2}$ and $B = {x \geq 0 : \sqrt{x} (\sqrt{x} - 4) - 3 | \sqrt{x} - 2 | + 6 = 0}$ .

Then $n (A \cup B)$ is equal to : [2025]

4
2
8
6

1

2

3

4

(3)

For $A : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2$

$\Rightarrow \log_{(2 / π)} (| \sin x \cdot \cos x |) = 2$

$\Rightarrow | \sin x \cdot \cos x | = {(\frac{2}{π})}^{2}$

$\Rightarrow | 2 \sin x \cdot \cos x | = \frac{8}{π^{2}} \Rightarrow | \sin 2 x | = \frac{8}{π^{2}}$

The region is given by

Since, line $y = \frac{8}{π^{2}}$ cuts the graph of $y =$ $| \sin 2 x |$ four times. Thus, there are four solutions for A.

For $B : \sqrt{x} (\sqrt{x} - 4) - 3 | \sqrt{x} - 2 | + 6 = 0$

When $x < 4, \sqrt{x} (\sqrt{x} - 4) + 3 (\sqrt{x} - 2) + 6 = 0$

$\Rightarrow x - 4 \sqrt{x} + 3 \sqrt{x} - 6 + 6 = 0 \Rightarrow x - \sqrt{x} = 0$

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

$\Rightarrow x = 0 or 1$

When $x > 4, \sqrt{x} (\sqrt{x} - 4) - 3 (\sqrt{x} - 2) + 6 = 0$

$\Rightarrow x - 4 \sqrt{x} - 3 \sqrt{x} + 6 + 6 = 0 \Rightarrow x - 7 \sqrt{x} + 12 = 0$

$\Rightarrow (\sqrt{x} - 3) (\sqrt{x} - 4) = 0 \Rightarrow \sqrt{x} = 3 or 4$

$\Rightarrow x = 9 or 16$

Thus, there are four solutions for B as well.

Now, $n (A \cup B) = n (A) + n (B) = 4 + 4 = 8$

Q 23 :

$Let S = {x \in (- \frac{π}{2}, \frac{π}{2}) : 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10}$ and $β = \sum_{x \in S} \tan^{2} (\frac{x}{3}),$ then $\frac{1}{6} {(β - 14)}^{2}$ is equal to [2023]

16
8
64
32

1

2

3

4

(4)

$We have, 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10$

Putting $9^{\tan^{2} x} = t$ , we get

$∴$ $\frac{9}{t} + t = 10 \Rightarrow t^{2} - 10 t + 9 = 0$

$\Rightarrow t^{2} - 9 t - t + 9 = 0 \Rightarrow (t - 1) (t - 9) = 0$ $\Rightarrow t = 1, 9$

$∴$ $9^{\tan^{2} x} = 1 or 9^{\tan^{2} x} = 9$

$\Rightarrow \tan^{2} x = 0 or \tan^{2} x = 1$ $\Rightarrow \tan x = 0 or \tan x = + 1$

$∴$ $x = 0, \frac{π}{4}, - \frac{π}{4}$

$∴$ $x \in (\frac{- π}{2}, \frac{π}{2})$ $...(i)$

Now, $β = \sum \tan^{2} (\frac{x}{3}) = \tan^{2} 0 + \tan^{2} \frac{π}{12} + \tan^{2} (- \frac{π}{12})$

$= 0 + 2 {(\tan 15 °)}^{2}$

$= 2 {(2 - \sqrt{3})}^{2} = 2 (7 - 4 \sqrt{3})$

$β = 14 - 8 \sqrt{3}$ $...(ii)$

$∴$ $\frac{1}{6} {(β - 14)}^{2} = \frac{1}{6} {(14 - 8 \sqrt{3} - 14)}^{2}$ $[from (ii)]$

$= \frac{192}{6} = 32$

Q 24 :

The number of elements in the set $S = {θ \in [0, 2 π] : 3 \cos^{4} θ - 5 \cos^{2} θ - 2 \sin^{6} θ + 2 = 0}$ is [2023]

10
8
9
12

1

2

3

4

(3)

$3 \cos^{4} θ - 5 \cos^{2} θ - 2 \sin^{6} θ + 2 = 0$

$\Rightarrow 3 \cos^{4} θ - 3 \cos^{2} θ - 2 \cos^{2} θ - 2 \sin^{6} θ + 2 = 0$

$\Rightarrow 3 \cos^{4} θ - 3 \cos^{2} θ + 2 \sin^{2} θ - 2 \sin^{6} θ = 0$

$\Rightarrow 3 \cos^{2} θ (\cos^{2} θ - 1) - 2 \sin^{2} θ (\sin^{4} θ - 1) = 0$

$\Rightarrow - 3 \cos^{2} θ \sin^{2} θ + 2 \sin^{2} θ (1 + \sin^{2} θ) \cos^{2} θ = 0$

$\Rightarrow \sin^{2} θ \cos^{2} θ (2 + 2 \sin^{2} θ - 3) = 0$

$\Rightarrow \sin^{2} θ \cos^{2} θ (2 \sin^{2} θ - 1) = 0$

Case I: $\sin^{2} θ = 0 \to 3 solutions: θ = {0, π, 2 π}$

Case II: $\cos^{2} θ = 0 \to 2 solutions: θ = {\frac{π}{2}, \frac{3 π}{2}}$

Case III: $\sin^{2} θ = \frac{1}{2} \to 4 solutions: θ = {\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}}$

$Number of solutions = 9$

Q 25 :

Let $f (θ) = 3 (\sin^{4} (\frac{3 π}{2} - θ) + \sin^{4} (3 π + θ)) - 2 (1 - \sin^{2} 2 θ)$ and $S = {θ \in [0, π] : f' (θ) = - \frac{\sqrt{3}}{2}} .$ If $4 β = \sum_{θ \in S} θ,$ then $f (β)$ is equal to [2023]

$\frac{11}{8}$
$\frac{3}{2}$
$\frac{9}{8}$
$\frac{5}{4}$

1

2

3

4

(4)

$Given f (θ) = 3 (\sin^{4} (\frac{3 π}{2} - θ) + \sin^{4} (3 π + θ))$ $- 2 (1 - \sin^{2} 2 θ)$

$= 3 (\cos^{4} θ + \sin^{4} θ)$ $- 2 \cos^{2} 2 θ$

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

$= 3 - \frac{3}{2} (1 - \cos^{2} 2 θ) - 2 \cos^{2} 2 θ = \frac{3}{2} - \frac{1}{2} \cos^{2} 2 θ$

$= \frac{3}{2} - \frac{1}{2} (\frac{1 + \cos 4 θ}{2}) = \frac{5}{4} - \frac{\cos 4 θ}{4}$

$Now, f' (θ) = \sin 4 θ = \frac{- \sqrt{3}}{2}$

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

$\Rightarrow θ = (\frac{π}{4} + \frac{π}{12}), (\frac{π}{2} - \frac{π}{12}), (\frac{3 π}{4} + \frac{π}{12}), (π - \frac{π}{12})$

$\Rightarrow 4 β = \frac{π}{4} + \frac{π}{2} + \frac{3 π}{4} = \frac{5 π}{2}$

$\Rightarrow β = \frac{5 π}{8} \Rightarrow f (β) = \frac{5}{4} - \frac{\cos \frac{5 π}{2}}{4} = \frac{5}{4}$

Q 26 :

The set of all values of $λ$ for which the equation $\cos^{2} 2 x - 2 \sin^{4} x - 2 \cos^{2} x = λ$ has a real solution $x,$ is [2023]

$[- 2, - 1]$
$[- 2, - \frac{3}{2}]$
$[- 1, - \frac{1}{2}]$
$[- \frac{3}{2}, - 1]$

1

2

3

4

(4)

$The equation is given \cos^{2} 2 x - 2 \sin^{4} x - 2 \cos^{2} x = λ$

$then we can write the given equation as$

$λ = \cos^{2} 2 x - 2 \sin^{4} x - 2 \cos^{2} x$

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

$\Rightarrow λ = 4 \cos^{4} x - 4 \cos^{2} x + 1 - 2 (1 - 2 \cos^{2} x + \cos^{4} x)$ $- 2 \cos^{2} x$

$\Rightarrow λ = 2 \cos^{4} x - 2 \cos^{2} x + 1 - 2 \Rightarrow λ = 2 \cos^{4} x - 2 \cos^{2} x - 1$

$λ = 2 [\cos^{4} x - \cos^{2} x - \frac{1}{2}], λ = 2 [{(\cos^{2} x - \frac{1}{2})}^{2} - \frac{3}{4}]$

$So λ_{\max} = 2 [\frac{1}{4} - \frac{3}{4}] = 2 \times (\frac{- 2}{4}) = - 1 (maximum value)$

$and λ_{\min} = 2 [0 - \frac{3}{4}] = - \frac{3}{2} (minimum value)$

$So range of the value of λ is [\frac{- 3}{2}, - 1] .$

$Hence option (4) is correct answer.$

Q 27 :

If $\tan 15 ° + \frac{1}{\tan 75 °} + \frac{1}{\tan 105 °} + \tan 195 ° = 2 a$ , then the value of $(a + \frac{1}{a})$ is [2023]

$4$
$2$
$4 - 2 \sqrt{3}$
$5 - \frac{3}{2} \sqrt{3}$

1

2

3

4

(1)

$\tan 15 ° + \frac{1}{\tan 75 °} + \frac{1}{\tan 105 °} + \tan 195 ° = 2 a$

We know that $\tan 15 ° = 2 - \sqrt{3}$

$\frac{1}{\tan 75 °} = \tan 15 ° = 2 - \sqrt{3},$ $\frac{1}{\tan 105 °} = - \cot 75 ° = \sqrt{3} - 2$

and $\tan 195 ° = \tan 15 ° = 2 - \sqrt{3}$

So, $2 (2 - \sqrt{3}) = 2 a \Rightarrow a = 2 - \sqrt{3}$

$\Rightarrow a + \frac{1}{a} = 2 - \sqrt{3} + \frac{1}{2 - \sqrt{3}} = 4$

Q 28 :

If the solution of the equation $\log_{\cos x} (\cot x) + 4 \log_{\sin x} (\tan x) = 1,$ $x \in (0, \frac{π}{2}),$ is $\sin^{- 1} (\frac{α + \sqrt{β}}{2}),$ where $α, β$ are integers, then $α + β$ is equal to [2023]

6
4
5
3

1

2

3

4

(2)

$We have \log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1$

$\Rightarrow \log_{\cos x} \cos x - \log_{\cos x} \sin x + 4 [\log_{\sin x} \sin x - \log_{\sin x} \cos x] = 1$

$\Rightarrow 1 - \log_{\cos x} \sin x + 4 - 4 \log_{\sin x} \cos x = 1$

$\Rightarrow - \log_{\cos x} \sin x + 4 - 4 \log_{\sin x} \cos x = 0$

Put $\log_{\cos x} \sin x = y$ $∴$ $- y + 4 - \frac{4}{y} = 0$

$\Rightarrow y^{2} - 4 y + 4 = 0 \Rightarrow {(y - 2)}^{2} = 0$ $∴$ $y = 2 = \log_{\cos x} \sin x$

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

$∴$ $\sin x = \frac{- 1 \pm \sqrt{1 - 4 \times 1 \times (- 1)}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$

Since $x \in (0, π / 2)$ ,

$∴$ $\sin x = \frac{- 1 + \sqrt{5}}{2} \Rightarrow x = \sin^{- 1} (\frac{- 1 + \sqrt{5}}{2})$

$∴$ $α = - 1, β = 5,$ So, $α + β = 4$

Q 29 :

Let $S = {θ \in [0, 2 π) : \tan (π \cos θ) + \tan (π \sin θ) = 0}$ . Then $\sum_{θ \in S} \sin^{2} (θ + \frac{π}{4})$ is equal to ________ . [2023]

(2)

$Given, S = {θ \in [0, 2 π) : \tan (π \cos θ) + \tan (π \sin θ) = 0}$

$Now, \tan (π \cos θ) + \tan (π \sin θ) = 0$

$\Rightarrow \tan (π \cos θ) = - \tan (π \sin θ)$

$\Rightarrow \tan (π \cos θ) = \tan (- π \sin θ)$

$∴$ $π \cos θ = n π - π \sin θ$

$∴$ $\sin θ + \cos θ = n,$ where, $n \in I$

Possible values are $n = 0, 1, - 1$ because

$- \sqrt{2} \leq \sin θ + \cos θ \leq \sqrt{2}$

$Now, it gives θ \in {0, \frac{π}{2}, \frac{3 π}{4}, \frac{7 π}{4}, \frac{3 π}{2}, π}$

$So, \sum_{θ \in S} \sin^{2} (θ + \frac{π}{4}) = 2 (0) + 4 (\frac{1}{2}) = 2$

Q 30 :

If $m$ and $n$ respectively are the numbers of positive and negative values of $θ$ in the interval $[- π, π]$ that satisfy the equation $\cos 2 θ \cos \frac{θ}{2} = \cos 3 θ \cos \frac{9 θ}{2},$ then $m n$ is equal to _______. [2023]

(25)

$We have, \cos 2 θ \cdot \cos \frac{θ}{2} = \cos 3 θ \cdot \cos \frac{9 θ}{2}$

$\Rightarrow 2 \cos 2 θ \cdot \cos \frac{θ}{2} = 2 \cos \frac{9 θ}{2} \cdot \cos 3 θ$

$\Rightarrow \cos \frac{5 θ}{2} + \cos \frac{3 θ}{2} = \cos \frac{15 θ}{2} + \cos \frac{3 θ}{2}$

$\Rightarrow \cos \frac{15 θ}{2} = \cos \frac{5 θ}{2} \Rightarrow \frac{15 θ}{2} = 2 k π \pm \frac{5 θ}{2}$

$\Rightarrow 5 θ = 2 k π or 10 θ = 2 k π \Rightarrow θ = \frac{2 k π}{5} or θ = \frac{k π}{5}$

$\Rightarrow θ = \frac{k π}{5}$

$Also, - π \leq \frac{k π}{5} \leq π \Rightarrow - 5 \leq k \leq 5$ $∴$ $m = 5, n = 5$

Topic Question Set

Q 1 :

Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

Q 2 :

Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

Q 3 :

The number of solutions of the equation $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0;$ $x \in [- 2 π, 2 π]$ is : [2024]

Q 4 :

If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

Q 5 :

The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

Q 6 :

For $α, β \in (0, \frac{π}{2}),$ let $3 \sin (α + β) = 2 \sin (α - β)$ and a real number $k$ be such that $\tan α = k$ $\tan β .$ Then, the value of $k$ is equal to [2024]

Q 7 :

The number of solutions of the equation $e^{\sin x} - 2 e^{- \sin x} = 2,$ is [2024]

Q 8 :

Let $S = {\sin^{2} 2 θ : (\sin^{4} θ + \cos^{4} θ) x^{2} + (\sin 2 θ) x + (\sin^{6} θ + \cos^{6} θ) = 0 has real roots} .$ If $α$ and $β$ be the smallest and largest elements of the set $S,$ respectively, then $3 ({(α - 2)}^{2} + {(β - 1)}^{2})$ equals _________ . [2024]

Q 9 :

Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[p, q]$ and $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °,$ then $p q r$ is equal to _____ . [2024]

Q 10 :

If $2 \tan^{2} θ - 5 \sec θ = 1$ has exactly 7 solutions in the interval $[0, \frac{n π}{2}],$ for the least value of $n \in N,$ then $\sum_{k = 1}^{n} \frac{k}{2^{k}}$ is equal to [2024]

Q 11 :

If $2 \sin^{3} x + \sin 2 x \cos x + 4 \sin x - 4 = 0$ has exactly 3 solutions in the interval $[0, \frac{n π}{2}], n \in N,$ then the roots of the equation $x^{2} + n x + (n - 3) = 0$ belong to [2024]

Q 12 :

The number of solutions of $\sin^{2} x + (2 + 2 x - x^{2}) \sin x - 3 {(x - 1)}^{2} = 0,$ where $- π \leq x \leq π,$ is _________. [2024]

Q 13 :

If $θ \in [- 2 π, 2 π]$ , then the number of solutions of $2 \sqrt{2} \cos^{2} θ + (2 - \sqrt{6}) \cos θ - \sqrt{3} = 0$ , is equal to : [2025]

Q 14 :

If $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , then the number of solutions of $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ , is equal to : [2025]

Q 15 :

The number of solutions of the equation $2 x + 3 \tan x = π, x \in [- 2 π, 2 π] - {\pm \frac{π}{2}, \pm \frac{3 π}{2}}$ is [2025]

Q 16 :

The number of solutions of the equation $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$ is [2025]

Q 17 :

If $10 \sin^{4} θ + 15 \cos^{4} θ = 6$ , then the value of $\frac{27 {cosec}^{6} θ + 8 \sec^{6} θ}{16 \sec^{8} θ}$ is [2025]

Q 18 :

Let the product of $ω_{1} = (8 + i) \sin θ + (7 + 4 i) \cos θ$ and $ω_{2} = (1 + 8 i) \sin θ + (4 + 7 i) \cos θ$ be $α + i β$ , $i = \sqrt{- 1}$ . Let p and q be the maximum and the minimum values of $α + β$ respectively. Then p + q is equal to : [2025]

Q 19 :

The number of solutions of the equation $\cos 2 θ \cos \frac{θ}{2} + \cos \frac{5 θ}{2} = 2 \cos^{3} \frac{5 θ}{2} in [- \frac{π}{2}, \frac{π}{2}]$ is: [2025]

Q 20 :

The sum of all values of $θ \in [0, 2 π]$ satisfying $2 \sin^{2} θ = \cos 2 θ$ and $2 \cos^{2} θ = 3 \sin θ$ is: [2025]

Q 21 :

Let $α_{θ}$ and $β θ$ be the distinct roots of $2 x^{2} + (\cos θ) x - 1 = 0, θ \in (0, 2 π)$ . If m and M are the minimum and the maximum values of $α_{θ}^{4} + β_{θ}^{4}$ , then 16(M + m) equals: [2025]

Q 22 :

Let $A = {x \in (0, π) - {\frac{π}{2}} : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2}$ and $B = {x \geq 0 : \sqrt{x} (\sqrt{x} - 4) - 3 | \sqrt{x} - 2 | + 6 = 0}$ .

Then $n (A \cup B)$ is equal to : [2025]

Q 23 :

$Let S = {x \in (- \frac{π}{2}, \frac{π}{2}) : 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10}$ and $β = \sum_{x \in S} \tan^{2} (\frac{x}{3}),$ then $\frac{1}{6} {(β - 14)}^{2}$ is equal to [2023]

Q 24 :

The number of elements in the set $S = {θ \in [0, 2 π] : 3 \cos^{4} θ - 5 \cos^{2} θ - 2 \sin^{6} θ + 2 = 0}$ is [2023]

Q 25 :

Let $f (θ) = 3 (\sin^{4} (\frac{3 π}{2} - θ) + \sin^{4} (3 π + θ)) - 2 (1 - \sin^{2} 2 θ)$ and $S = {θ \in [0, π] : f' (θ) = - \frac{\sqrt{3}}{2}} .$ If $4 β = \sum_{θ \in S} θ,$ then $f (β)$ is equal to [2023]

Q 26 :

The set of all values of $λ$ for which the equation $\cos^{2} 2 x - 2 \sin^{4} x - 2 \cos^{2} x = λ$ has a real solution $x,$ is [2023]

Q 27 :

If $\tan 15 ° + \frac{1}{\tan 75 °} + \frac{1}{\tan 105 °} + \tan 195 ° = 2 a$ , then the value of $(a + \frac{1}{a})$ is [2023]

Q 28 :

If the solution of the equation $\log_{\cos x} (\cot x) + 4 \log_{\sin x} (\tan x) = 1,$ $x \in (0, \frac{π}{2}),$ is $\sin^{- 1} (\frac{α + \sqrt{β}}{2}),$ where $α, β$ are integers, then $α + β$ is equal to [2023]

Q 29 :

Let $S = {θ \in [0, 2 π) : \tan (π \cos θ) + \tan (π \sin θ) = 0}$ . Then $\sum_{θ \in S} \sin^{2} (θ + \frac{π}{4})$ is equal to ________ . [2023]

Q 30 :

If $m$ and $n$ respectively are the numbers of positive and negative values of $θ$ in the interval $[- π, π]$ that satisfy the equation $\cos 2 θ \cos \frac{θ}{2} = \cos 3 θ \cos \frac{9 θ}{2},$ then $m n$ is equal to _______. [2023]

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Topic Question Set

Q 1 : Suppose θ∈[0,π4] is a solution of 4cosθ-3sinθ=1. Then cosθ is equal to : [2024]

Q 2 : Let |cosθcos(60-θ)cos(60+θ)|≤18,θ∈[0,2π]. Then, the sum of all θ∈[0,2π], where cos3θ attains its maximum value, is: [2024]

Q 3 : The number of solutions of the equation 4sin2x-4cos3x+9-4cosx=0; x∈[-2π,2π] is : [2024]

Q 4 : If α,-π2<α<π2 is the solution of 4cosθ+5sinθ=1, then the value of tanα is [2024]

Q 5 : The sum of the solutions x∈R of the equation 3cos2x+cos32xcos6x-sin6x=x3-x2+6 is [2024]

Q 6 : For α,β∈(0,π2), let 3sin(α+β)=2sin(α-β) and a real number k be such that tanα=k tanβ. Then, the value of k is equal to [2024]

Q 7 : The number of solutions of the equation esinx-2e-sinx=2, is [2024]

Q 8 : Let S={sin22θ:(sin4θ+cos4θ)x2+(sin2θ)x+(sin6θ+cos6θ)=0 has real roots}. If α and β be the smallest and largest elements of the set S, respectively, then 3((α-2)2+(β-1)2) equals _________ . [2024]

Q 9 : Let the set of all a∈R such that the equation cos2x+asinx=2a-7 has a solution be [p,q] and r=tan9°-tan27°-1cot63°+tan81°, then pqr is equal to _____ . [2024]

Q 10 : If 2tan2θ-5secθ=1 has exactly 7 solutions in the interval [0,nπ2], for the least value of n∈N, then ∑k=1nk2k is equal to [2024]

Q 11 : If 2sin3x+sin2xcosx+4sinx-4=0 has exactly 3 solutions in the interval [0,nπ2], n∈N, then the roots of the equation x2+nx+(n-3)=0 belong to [2024]

Q 12 : The number of solutions of sin2x+(2+2x-x2)sinx-3(x-1)2=0, where -π≤x≤π, is _________. [2024]

Q 13 : If θ∈[–2π,2π], then the number of solutions of 22cos2θ+(2–6)cosθ–3=0, is equal to : [2025]

Q 14 : If θ∈[–7π6,4π3], then the number of solutions of 3cosec2θ–2(3–1)cosecθ–4=0, is equal to : [2025]

Q 15 : The number of solutions of the equation 2x+3tanx=π, x∈[–2π,2π]–{±π2,±3π2} is [2025]

Q 16 : The number of solutions of the equation (4–3)sinx–23cos2x=–41+3,x∈[–2π,5π2] is [2025]

Q 17 : If 10 sin4θ+15 cos4θ=6, then the value of 27 cosec6θ+8 sec6θ16 sec8θ is [2025]

Q 18 : Let the product of ω1=(8+i) sin θ+(7+4i) cos θ and ω2=(1+8i) sin θ+(4+7i) cos θ be α+iβ, i=–1. Let p and q be the maximum and the minimum values of α+β respectively. Then p + q is equal to : [2025]

Q 19 : The number of solutions of the equation cos2θ cosθ2+cos5θ2=2cos35θ2in [–π2,π2] is: [2025]

Q 20 : The sum of all values of θ∈[0,2π] satisfying 2sin2θ=cos2θ and 2cos2θ=3sinθ is: [2025]

Q 21 : Let αθ and βθ be the distinct roots of 2x2+(cosθ)x–1=0, θ∈(0,2π). If m and M are the minimum and the maximum values of αθ4+βθ4, then 16(M + m) equals: [2025]

Q 22 : Let A={x∈(0,π)–{π2} : log(2/π)|sinx|+log(2/π)|cosx|=2} and B={x≥0 : x(x–4)–3|x–2|+6=0}. Then n(A∪B) is equal to : [2025]

Q 23 : Let S={x∈(-π2,π2):9 1-tan2x+9 tan2x=10} and β=∑x∈Stan2(x3), then 16(β-14)2 is equal to [2023]

Q 24 : The number of elements in the set S={θ∈[0,2π]:3cos4θ-5cos2θ-2sin6θ+2=0} is [2023]

Q 25 : Let f(θ)=3(sin4(3π2-θ)+sin4(3π+θ))-2(1-sin22θ) and S={θ∈[0,π]:f'(θ)=-32}. If 4β=∑θ∈Sθ, then f(β) is equal to [2023]

Q 26 : The set of all values of λ for which the equation cos22x-2sin4x-2cos2x=λ has a real solution x, is [2023]

Q 27 : If tan15°+1tan75°+1tan105°+tan195°=2a, then the value of (a+1a) is [2023]

Q 28 : If the solution of the equation logcosx(cotx)+4logsinx(tanx)=1, x∈(0,π2), is sin-1(α+β2), where α,β are integers, then α+β is equal to [2023]

Q 29 : Let S={θ∈[0,2π):tan(πcosθ)+tan(πsinθ)=0}. Then ∑θ∈Ssin2(θ+π4) is equal to ________ . [2023]

Q 30 : If m and n respectively are the numbers of positive and negative values of θ in the interval [-π,π] that satisfy the equation cos2θcosθ2=cos3θcos9θ2, then mn is equal to _______. [2023]

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Q 1 :

Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

Q 2 :

Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

Q 3 :

The number of solutions of the equation $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0;$ $x \in [- 2 π, 2 π]$ is : [2024]

Q 4 :

If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

Q 5 :

The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

Q 6 :

For $α, β \in (0, \frac{π}{2}),$ let $3 \sin (α + β) = 2 \sin (α - β)$ and a real number $k$ be such that $\tan α = k$ $\tan β .$ Then, the value of $k$ is equal to [2024]

Q 7 :

The number of solutions of the equation $e^{\sin x} - 2 e^{- \sin x} = 2,$ is [2024]

Q 8 :

Let $S = {\sin^{2} 2 θ : (\sin^{4} θ + \cos^{4} θ) x^{2} + (\sin 2 θ) x + (\sin^{6} θ + \cos^{6} θ) = 0 has real roots} .$ If $α$ and $β$ be the smallest and largest elements of the set $S,$ respectively, then $3 ({(α - 2)}^{2} + {(β - 1)}^{2})$ equals _________ . [2024]

Q 9 :

Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[p, q]$ and $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °,$ then $p q r$ is equal to _____ . [2024]

Q 10 :

If $2 \tan^{2} θ - 5 \sec θ = 1$ has exactly 7 solutions in the interval $[0, \frac{n π}{2}],$ for the least value of $n \in N,$ then $\sum_{k = 1}^{n} \frac{k}{2^{k}}$ is equal to [2024]

Q 11 :

If $2 \sin^{3} x + \sin 2 x \cos x + 4 \sin x - 4 = 0$ has exactly 3 solutions in the interval $[0, \frac{n π}{2}], n \in N,$ then the roots of the equation $x^{2} + n x + (n - 3) = 0$ belong to [2024]

Q 12 :

The number of solutions of $\sin^{2} x + (2 + 2 x - x^{2}) \sin x - 3 {(x - 1)}^{2} = 0,$ where $- π \leq x \leq π,$ is _________. [2024]

Q 13 :

If $θ \in [- 2 π, 2 π]$ , then the number of solutions of $2 \sqrt{2} \cos^{2} θ + (2 - \sqrt{6}) \cos θ - \sqrt{3} = 0$ , is equal to : [2025]

Q 14 :

If $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , then the number of solutions of $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ , is equal to : [2025]

Q 15 :

The number of solutions of the equation $2 x + 3 \tan x = π, x \in [- 2 π, 2 π] - {\pm \frac{π}{2}, \pm \frac{3 π}{2}}$ is [2025]

Q 16 :

The number of solutions of the equation $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$ is [2025]

Q 17 :

If $10 \sin^{4} θ + 15 \cos^{4} θ = 6$ , then the value of $\frac{27 {cosec}^{6} θ + 8 \sec^{6} θ}{16 \sec^{8} θ}$ is [2025]

Q 18 :

Let the product of $ω_{1} = (8 + i) \sin θ + (7 + 4 i) \cos θ$ and $ω_{2} = (1 + 8 i) \sin θ + (4 + 7 i) \cos θ$ be $α + i β$ , $i = \sqrt{- 1}$ . Let p and q be the maximum and the minimum values of $α + β$ respectively. Then p + q is equal to : [2025]

Q 19 :

The number of solutions of the equation $\cos 2 θ \cos \frac{θ}{2} + \cos \frac{5 θ}{2} = 2 \cos^{3} \frac{5 θ}{2} in [- \frac{π}{2}, \frac{π}{2}]$ is: [2025]

Q 20 :

The sum of all values of $θ \in [0, 2 π]$ satisfying $2 \sin^{2} θ = \cos 2 θ$ and $2 \cos^{2} θ = 3 \sin θ$ is: [2025]

Q 21 :

Let $α_{θ}$ and $β θ$ be the distinct roots of $2 x^{2} + (\cos θ) x - 1 = 0, θ \in (0, 2 π)$ . If m and M are the minimum and the maximum values of $α_{θ}^{4} + β_{θ}^{4}$ , then 16(M + m) equals: [2025]

Q 22 :

Let $A = {x \in (0, π) - {\frac{π}{2}} : \log_{(2 / π)} | \sin x | + \log_{(2 / π)} | \cos x | = 2}$ and $B = {x \geq 0 : \sqrt{x} (\sqrt{x} - 4) - 3 | \sqrt{x} - 2 | + 6 = 0}$ .

Then $n (A \cup B)$ is equal to : [2025]

Q 23 :

$Let S = {x \in (- \frac{π}{2}, \frac{π}{2}) : 9^{1 - \tan^{2} x} + 9^{\tan^{2} x} = 10}$ and $β = \sum_{x \in S} \tan^{2} (\frac{x}{3}),$ then $\frac{1}{6} {(β - 14)}^{2}$ is equal to [2023]

Q 24 :

The number of elements in the set $S = {θ \in [0, 2 π] : 3 \cos^{4} θ - 5 \cos^{2} θ - 2 \sin^{6} θ + 2 = 0}$ is [2023]

Q 25 :

Let $f (θ) = 3 (\sin^{4} (\frac{3 π}{2} - θ) + \sin^{4} (3 π + θ)) - 2 (1 - \sin^{2} 2 θ)$ and $S = {θ \in [0, π] : f' (θ) = - \frac{\sqrt{3}}{2}} .$ If $4 β = \sum_{θ \in S} θ,$ then $f (β)$ is equal to [2023]

Q 26 :

The set of all values of $λ$ for which the equation $\cos^{2} 2 x - 2 \sin^{4} x - 2 \cos^{2} x = λ$ has a real solution $x,$ is [2023]

Q 27 :

If $\tan 15 ° + \frac{1}{\tan 75 °} + \frac{1}{\tan 105 °} + \tan 195 ° = 2 a$ , then the value of $(a + \frac{1}{a})$ is [2023]

Q 28 :

If the solution of the equation $\log_{\cos x} (\cot x) + 4 \log_{\sin x} (\tan x) = 1,$ $x \in (0, \frac{π}{2}),$ is $\sin^{- 1} (\frac{α + \sqrt{β}}{2}),$ where $α, β$ are integers, then $α + β$ is equal to [2023]

Q 29 :

Let $S = {θ \in [0, 2 π) : \tan (π \cos θ) + \tan (π \sin θ) = 0}$ . Then $\sum_{θ \in S} \sin^{2} (θ + \frac{π}{4})$ is equal to ________ . [2023]

Q 30 :

If $m$ and $n$ respectively are the numbers of positive and negative values of $θ$ in the interval $[- π, π]$ that satisfy the equation $\cos 2 θ \cos \frac{θ}{2} = \cos 3 θ \cos \frac{9 θ}{2},$ then $m n$ is equal to _______. [2023]