JEE Advanced Trigonometric Functions PYQ – Identities & Max/Min Values (With Solutions PDF)

Q 1 :

Let $f : ℝ \to ℝ$ be a function defined by

$f (x)$ $=$ ${\begin{matrix} x^{2} \sin (\frac{π}{x^{2}}), & if x \neq 0 \\ 0, & if x = 0 \end{matrix}$

Then which of the following statements is TRUE? [2024]

$f (x) = 0$ has infinitely many solutions in the interval $[\frac{1}{10^{10}}, \infty)$ .
$f (x) = 0$ has no solutions in the interval $[\frac{1}{π}, \infty)$
The set of solutions of $f (x) = 0$ in the interval $(0, \frac{1}{10^{10}})$ is finite.
$f (x) = 0$ has more than 25 solutions in the interval $(\frac{1}{π^{2}}, \frac{1}{π})$ .

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(4)

Given, $f (x)$ $=$ ${\begin{matrix} x^{2} \sin (\frac{π}{x^{2}}), & if x \neq 0 \\ 0, & if x = 0 \end{matrix}$

$f (x) = 0 \Rightarrow \sin (\frac{π}{x^{2}}) = 0$

$\Rightarrow \frac{π}{x^{2}} = n π \Rightarrow x^{2} = \frac{1}{n} \Rightarrow x = \frac{1}{\sqrt{n}}$

$(a) If x \in [\frac{1}{10^{10}}, \infty)$

$\frac{1}{\sqrt{n}} \in [\frac{1}{10^{10}}, \infty)$

$\sqrt{n} \in (0, 10^{10}]$

$n \in (0, {(10^{10})}^{2}]$

$Finite values of n are possible, so has finite solution.$

$(b) If x \in [\frac{1}{π}, \infty) \Rightarrow \frac{1}{\sqrt{n}} \in [\frac{1}{π}, \infty)$

$\sqrt{n} \in (0, π] \Rightarrow n \in (0, π^{2}]$ $\Rightarrow n = 1, 2, 3, \dots, 9$

$(c) If x \in (0, \frac{1}{10^{10}}) \Rightarrow \sqrt{n} \in (10^{10}, \infty)$

$n is infinite$

$(d) If x \in (\frac{1}{π^{2}}, \frac{1}{π}) \Rightarrow \sqrt{n} \in (π, π^{2})$

$n \in (π^{2}, π^{4})$ $\Rightarrow n \in (9.8, 97.2, \dots)$

$More than 25 solutions$

Q 2 :

Let $\frac{π}{2} < x < π$ be such that $\cot x = \frac{- 5}{\sqrt{11}}$ . Then

$(\sin \frac{11 x}{2}) (\sin 6 x - \cos 6 x) + (\cos \frac{11 x}{2}) (\sin 6 x + \cos 6 x)$ is equal to [2024]

$\frac{\sqrt{11} - 1}{2 \sqrt{3}}$
$\frac{\sqrt{11} + 1}{2 \sqrt{3}}$
$\frac{\sqrt{11} + 1}{3 \sqrt{2}}$
$\frac{\sqrt{11} - 1}{3 \sqrt{2}}$

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(2)

Let $E = (\sin 6 x \cos \frac{11 x}{2} - \cos 6 x \sin \frac{11 x}{2}) + (\cos 6 x \cos \frac{11 x}{2} + \sin 6 x \sin \frac{11 x}{2})$

$= (\sin (6 x - \frac{11 x}{2}) + \cos (6 x - \frac{11 x}{2}))$

$= \sin \frac{x}{2} + \cos \frac{x}{2}$

$Now, E^{2} = 1 + \sin x \dots (i)$

$∴$ $\cot x = \frac{- 5}{\sqrt{11}}$

$∴$ $E^{2} = 1 + \frac{\sqrt{11}}{6}$

$∴$ $E = \sqrt{\frac{6 + \sqrt{11}}{6}} = \sqrt{\frac{12 + 2 \sqrt{11}}{12}} = \frac{\sqrt{11} + 1}{2 \sqrt{3}}$

Q 3 :

The value of $\sum_{k = 1}^{13} \frac{1}{\sin (\frac{π}{4} + \frac{(k - 1) π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}$ is equal to [2016]

$3 - \sqrt{3}$
$2 (3 - \sqrt{3})$
$2 (\sqrt{3} - 1)$
$2 (2 - \sqrt{3})$

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(3)

$\sum_{k = 1}^{13} \frac{1}{\sin (\frac{π}{4} + (k - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}$

$= \sum_{k = 1}^{13} \frac{1}{\sin \frac{π}{6}} [\frac{\sin {\frac{π}{4} + \frac{k π}{6} - (\frac{π}{4} + (k - 1) \frac{π}{6})}}{\sin (\frac{π}{4} + (k - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}]$

$= \sum_{k = 1}^{13} 2 [\cot (\frac{π}{4} + (k - 1) \frac{π}{6}) - \cot (\frac{π}{4} + \frac{k π}{6})]$

$= 2 [{\cot \frac{π}{4} - \cot (\frac{π}{4} + \frac{π}{6})} + {\cot (\frac{π}{4} + \frac{π}{6}) - \cot (\frac{π}{4} + \frac{2 π}{6})} + \dots + {\cot (\frac{π}{4} + \frac{12 π}{6}) - \cot (\frac{π}{4} + \frac{13 π}{6})}]$

$= 2 [\cot \frac{π}{4} - \cot (\frac{π}{4} + \frac{13 π}{6})]$ $= 2 [1 - \cot \frac{5 π}{12}]$

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

Q 4 :

Let $θ \in (0, \frac{π}{4})$ and $t_{1} = {(\tan θ)}^{\tan θ}, t_{2} = {(\tan θ)}^{\cot θ}, t_{3} = {(\cot θ)}^{\tan θ}, t_{4} = {(\cot θ)}^{\cot θ},$ then [2006]

$t_{1} > t_{2} > t_{3} > t_{4}$
$t_{4} > t_{3} > t_{1} > t_{2}$
$t_{3} > t_{1} > t_{2} > t_{4}$
$t_{2} > t_{3} > t_{1} > t_{4}$

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(2)

$Given: θ \in (0, \frac{π}{4}) \Rightarrow \tan θ < 1 and \cot θ > 1$

$Let \tan θ = 1 - x and \cot θ = 1 + y, where x, y > 0 and are very small, then$

$∴$ $t_{1} = {(1 - x)}^{1 - x}, t_{2} = {(1 - x)}^{1 + y}, t_{3} = {(1 + y)}^{1 - x}, t_{4} = {(1 + y)}^{1 + y}$

$Clearly, t_{4} > t_{3} and t_{1} > t_{2} also, t_{3} > t_{1}$

$∴$ $t_{4} > t_{3} > t_{1} > t_{2}$

Q 5 :

The values of $θ \in (0, 2 π)$ for which $2 \sin^{2} θ - 5 \sin θ + 2 > 0$ , are [2006]

$(0, \frac{π}{6}) \cup (\frac{5 π}{6}, 2 π)$
$(\frac{π}{8}, \frac{5 π}{6})$
$(0, \frac{π}{8}) \cup (\frac{π}{6}, \frac{5 π}{6})$
$(\frac{41 π}{48}, π)$

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(1)

$2 \sin^{2} θ - 5 \sin θ + 2 > 0$

$\Rightarrow (\sin θ - 2) (2 \sin θ - 1) > 0$

$∵$ $- 1 \leq \sin θ \leq 1 \Rightarrow (\sin θ - 2) < 0, so (2 \sin θ - 1) < 0$

$∴$ $\sin θ < \frac{1}{2}$

$From graph, we get x \in (0, \frac{π}{6}) \cup (\frac{5 π}{6}, 2 π)$

Q 6 :

If $α + β = \frac{π}{2}$ and $β + γ = α$ , then $\tan α$ equals [2001]

$2 (\tan β + \tan γ)$
$\tan β + \tan γ$
$\tan β + 2 \tan γ$
$2 \tan β + \tan γ$

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(3)

$Given: α + β = \frac{π}{2} \Rightarrow α = \frac{π}{2} - β$

$\Rightarrow \tan α = \tan (\frac{π}{2} - β) = \cot β = \frac{1}{\tan β}$

$\Rightarrow \tan α \tan β = 1 \Rightarrow 1 + \tan α \tan β = 2$

$Now, \tan (α - β) = \frac{\tan α - \tan β}{1 + \tan α \tan β}$

$\Rightarrow \tan γ = \frac{\tan α - \tan β}{2}$

$\Rightarrow 2 \tan γ = \tan α - \tan β$ $\Rightarrow \tan α = 2 \tan γ + \tan β$

Q 7 :

The maximum value of $(\cos α_{1}) . (\cos α_{2}) \dots (\cos α_{n}),$ under the restrictions

$0 \leq α_{1}, α_{2}, \dots, α_{n} \leq \frac{π}{2} and (\cot α_{1}) . (\cot α_{2}) \dots (\cot α_{n}) = 1$ is [2001]

$\frac{1}{2^{n / 2}}$
$\frac{1}{2^{n}}$
$\frac{1}{2 n}$
$1$

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(1)

$Given: (\cot α_{1}) . (\cot α_{2}) \dots (\cot α_{n}) = 1$

$\Rightarrow (\cos α_{1}) (\cos α_{2}) \dots (\cos α_{n}) = (\sin α_{1}) (\sin α_{2}) \dots (\sin α_{n}) \dots (i)$

$Let y = (\cos α_{1}) (\cos α_{2}) \dots (\cos α_{n}) (to be max.)$

$\Rightarrow y^{2} = (\cos^{2} α_{1}) (\cos^{2} α_{2}) \dots (\cos^{2} α_{n})$

$= \cos α_{1} \sin α_{1} \cos α_{2} \sin α_{2} \dots \cos α_{n} \sin α_{n} (From (i))$

$= \frac{1}{2^{n}} [\sin 2 α_{1} \sin 2 α_{2} \dots \sin 2 α_{n}]$

$Now, 0 \leq α_{1}, α_{2}, \dots, α_{n} \leq \frac{π}{2}$

$\Rightarrow 0 \leq 2 α_{1}, 2 α_{2}, \dots, 2 α_{n} \leq π$

$\Rightarrow 0 \leq \sin 2 α_{1}, \sin 2 α_{2}, \dots, \sin 2 α_{n} \leq 1$

$∴$ $y^{2} \leq \frac{1}{2^{n}} \cdot 1 \Rightarrow y \leq \frac{1}{2^{n / 2}}$

$∴$ $Max. value of y i.e. (\cos α_{1}) . (\cos α_{2}) \dots (\cos α_{n}) = \frac{1}{2^{n / 2}}$

Q 8 :

Let $f (θ) = \sin θ (\sin θ + \sin 3 θ)$ . Then $f (θ)$ is [2000]

$\geq 0$ only when $θ \geq 0$
$\leq 0$ for all real $θ$
$\geq 0$ for all real $θ$
$\leq 0$ only when $θ \leq 0$

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(3)

$f (θ) = \sin θ (\sin θ + \sin 3 θ)$

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

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

$= 4 \sin^{2} θ (1 - \sin^{2} θ)$

$= 4 \sin^{2} θ \cos^{2} θ = {(2 \sin θ \cos θ)}^{2} = {(\sin 2 θ)}^{2} \geq 0, which is true for all θ .$

Q 9 :

Let $α$ and $β$ be real numbers such that $- \frac{π}{4} < β < 0 < α < \frac{π}{4}$ . If $\sin (α + β) = \frac{1}{3}$ and $\cos (α - β) = \frac{2}{3}$ ,

then the greatest integer less than or equal to ${(\frac{\sin α}{\cos β} + \frac{\cos β}{\sin α} + \frac{\cos α}{\sin β} + \frac{\sin β}{\cos α})}^{2}$ is _______. [2022]

(1)

Rearrange the given expression

${(\frac{\sin α}{\cos β} + \frac{\cos α}{\sin β} + \frac{\cos β}{\sin α} + \frac{\sin β}{\cos α})}^{2}$

$= {(\frac{\cos (α - β)}{\sin β \cos β} + \frac{\cos (α - β)}{\sin α . \cos α})}^{2}$

$= {(\frac{4}{3} {\frac{1}{\sin 2 β} + \frac{1}{\sin 2 α}})}^{2}$ $[∵ \cos (α - β) = \frac{2}{3}]$

$= \frac{16}{9} {[\frac{\sin 2 α + \sin 2 β}{\sin 2 α \cdot \sin 2 β}]}^{2}$

$= \frac{64}{9} {(\frac{2 \sin (α + β) . \cos (α - β)}{2 \sin 2 α \cdot \sin 2 β})}^{2}$

$= \frac{64}{9} {(\frac{2 \cdot \frac{1}{3} \cdot \frac{2}{3}}{\cos (2 α - 2 β) - \cos (2 α + 2 β)})}^{2}$

$= \frac{64}{9} {(\frac{\frac{4}{9}}{2 \cos^{2} (α - β) - 1 - 1 + 2 \sin^{2} (α + β)})}^{2}$

$= \frac{64}{9} {(\frac{\frac{4}{9}}{\frac{8}{9} - 2 + \frac{2}{9}})}^{2}$ $= \frac{64}{9} {(- \frac{1}{2})}^{2}$ $= [\frac{16}{9}] = [1.7] = 1$

Q 10 :

The maximum value of the expression $\frac{1}{\sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ}$ is _______. [2010]

(2)

Let $f (θ) = \frac{1}{g (θ)},$

where $g (θ) = \sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ$

$Clearly f is maximum when g is minimum$

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

$= 3 + 2 \cos 2 θ + \frac{3}{2} \sin 2 θ$ $\geq 3 + (- \sqrt{4 + \frac{9}{4}})$

$∴$ $g_{\min} = 3 - \frac{5}{2} = \frac{1}{2} \Rightarrow f_{\max} = 2$

Q 11 :

Let $α = \frac{1}{\sin 60 ° \sin 61 °} + \frac{1}{\sin 62 ° \sin 63 °} + \dots + \frac{1}{\sin 118 ° \sin 119 °} .$ Then the value of ${(\frac{c o s e c 1 °}{α})}^{2}$ is ________. [2025]

(3)

$α = \sum_{r = 30}^{59} \frac{1}{\sin (2 r) ° \sin (2 r + 1) °}$

$\Rightarrow \frac{α}{cosec 1 °} = \sum_{r = 30}^{59} \frac{\sin [(2 r) ° - (2 r + 1) °]}{\sin (2 r) ° \sin (2 r + 1) °}$

$= \sum_{r = 30}^{59} (\cot (2 r) ° - \cot (2 r + 1) °)$

$= \cot 60 ° - \cot 61 ° + \cot 62 ° - \cot 63 ° + \dots + \cot 118 ° - \cot 119 °$

$\Rightarrow \frac{α}{cosec 1 °} = \cot 60 ° = \frac{1}{\sqrt{3}}$

$\Rightarrow {(\frac{cosec 1 °}{α})}^{2} = 3$

Q 12 :

Let $f : [0, 2] \to ℝ$ be the function defined by $f (x) = (3 - \sin (2 π x)) \sin (π x - \frac{π}{4}) - \sin (3 π x + \frac{π}{4}) .$

If $α, β \in [0, 2]$ are such that ${x \in [0, 2] : f (x) \geq 0} = [α, β],$ then the value of $β - α$ is _______. [2020]

(1)

Let $π x - \frac{π}{4} = θ \in [\frac{- π}{4}, \frac{7 π}{4}]$

$∵$ $f (x) \geq 0$

$So, (3 - \sin (\frac{π}{2} + 2 θ)) \sin θ \geq \sin (π + 3 θ)$

$\Rightarrow (3 - \cos 2 θ) \sin θ \geq - \sin 3 θ$

$\Rightarrow \sin θ [3 - 4 \sin^{2} θ + 3 - \cos 2 θ] \geq 0$

$\Rightarrow \sin θ [6 - 2 (1 - \cos 2 θ) - \cos 2 θ] \geq 0$

$\Rightarrow \sin θ (4 + \cos 2 θ) \geq 0$ $\Rightarrow \sin θ \geq 0$

$\Rightarrow θ \in [0, π]$ $\Rightarrow 0 \leq π x - \frac{π}{4} \leq π \Rightarrow x \in [\frac{1}{4}, \frac{5}{4}]$

$\Rightarrow [α, β] = [\frac{1}{4}, \frac{5}{4}]$ $∴$ $β - α = \frac{5}{4} - \frac{1}{4} = 1$

Q 13 :

Let $f (x) = x \sin (π x), x > 0$ . Then for all natural numbers $n, f' (x)$ vanishes at [2013]

A unique point in the interval $(n, n + \frac{1}{2})$
A unique point in the interval $(n + \frac{1}{2}, n + 1)$
A unique point in the interval $(n, n + 1)$
Two points in the interval $(n, n + 1)$

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Select one or more options

(2, 3)

Given: $f (x) = x \sin π x, x > 0$

$\Rightarrow f' (x) = \sin π x + x π \cos π x$

$Now, f' (x) = 0 \Rightarrow \tan π x = - π x$

$From graph of y = \tan π x and y = - π x, it is clear that they intersect each other at a unique point in the intervals$

$(n, n + 1) and (n + \frac{1}{2}, n + 1)$

Q 14 :

Let $θ, φ \in [0, 2 π]$ be such that $2 \cos θ (1 - \sin φ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) \cos φ - 1,$ $tan (2 π - θ) > 0$

and $- 1 < \sin θ < - \frac{\sqrt{3}}{2}$ , then $φ$ cannot satisfy [2012]

$0 < φ < \frac{π}{2}$
$\frac{π}{2} < φ < \frac{4 π}{3}$
$\frac{4 π}{3} < φ < \frac{3 π}{2}$
$\frac{3 π}{2} < φ < 2 π$

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Select one or more options

(1, 3, 4)

$As \tan (2 π - θ) > 0 and - 1 < \sin θ < - \frac{\sqrt{3}}{2}, θ \in [0, 2 π]$

$Hence \frac{3 π}{2} < θ < \frac{5 π}{3}$

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

$\Rightarrow 2 \cos θ (1 - \sin φ) = 2 \sin θ \cos φ - 1$

$\Rightarrow 2 \cos θ + 1 = 2 \sin (θ + φ)$

$As θ \in (\frac{3 π}{2}, \frac{5 π}{3}), 1 < 2 \sin (θ + φ) < 2$

$As θ + φ \in (\frac{π}{6}, \frac{5 π}{6}) or (θ + φ) \in (\frac{13 π}{6}, \frac{17 π}{6})$

$We have φ \in (- \frac{3 π}{2}, - \frac{2 π}{3}) \cup (\frac{2 π}{3}, \frac{7 π}{6})$

Q 15 :

If $\frac{\sin^{4} x}{2} + \frac{\cos^{4} x}{3} = \frac{1}{5}$ , then [2009]

$\tan^{2} x = \frac{2}{3}$
$\frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{1}{125}$
$\tan^{2} x = \frac{1}{3}$
$\frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{2}{125}$

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Select one or more options

(1, 2)

Given:

$\frac{\sin^{4} x}{2} + \frac{\cos^{4} x}{3} = \frac{1}{5} \Rightarrow 3 \sin^{4} x + 2 \cos^{4} x = \frac{6}{5}$

$\Rightarrow \sin^{4} x + 2 [\sin^{4} x + \cos^{4} x] = \frac{6}{5}$

$\Rightarrow \sin^{4} x + 2 [1 - 2 \sin^{2} x \cos^{2} x] = \frac{6}{5}$

$\Rightarrow \sin^{4} x + 2 - 4 \sin^{2} x (1 - \sin^{2} x) = \frac{6}{5}$

$\Rightarrow 5 \sin^{4} x - 4 \sin^{2} x + 2 - \frac{6}{5} = 0$

$\Rightarrow 25 \sin^{4} x - 20 \sin^{2} x + 4 = 0$

$\Rightarrow {(5 \sin^{2} x - 2)}^{2} = 0 \Rightarrow \sin^{2} x = \frac{2}{5}$

$\Rightarrow \cos^{2} x = \frac{3}{5} and \tan^{2} x = \frac{2}{3}$

$Also \frac{\sin^{8} x}{8} + \frac{\cos^{8} x}{27} = \frac{2}{625} + \frac{3}{625} = \frac{5}{625} = \frac{1}{125}$

Topic Question Set

Q 1 :

Let $f : ℝ \to ℝ$ be a function defined by

$f (x)$ $=$ ${\begin{matrix} x^{2} \sin (\frac{π}{x^{2}}), & if x \neq 0 \\ 0, & if x = 0 \end{matrix}$

Then which of the following statements is TRUE? [2024]

Q 2 :

Let $\frac{π}{2} < x < π$ be such that $\cot x = \frac{- 5}{\sqrt{11}}$ . Then

$(\sin \frac{11 x}{2}) (\sin 6 x - \cos 6 x) + (\cos \frac{11 x}{2}) (\sin 6 x + \cos 6 x)$ is equal to [2024]

Q 3 :

The value of $\sum_{k = 1}^{13} \frac{1}{\sin (\frac{π}{4} + \frac{(k - 1) π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}$ is equal to [2016]

Q 4 :

Let $θ \in (0, \frac{π}{4})$ and $t_{1} = {(\tan θ)}^{\tan θ}, t_{2} = {(\tan θ)}^{\cot θ}, t_{3} = {(\cot θ)}^{\tan θ}, t_{4} = {(\cot θ)}^{\cot θ},$ then [2006]

Q 5 :

The values of $θ \in (0, 2 π)$ for which $2 \sin^{2} θ - 5 \sin θ + 2 > 0$ , are [2006]

Q 6 :

If $α + β = \frac{π}{2}$ and $β + γ = α$ , then $\tan α$ equals [2001]

Q 7 :

The maximum value of $(\cos α_{1}) . (\cos α_{2}) \dots (\cos α_{n}),$ under the restrictions

$0 \leq α_{1}, α_{2}, \dots, α_{n} \leq \frac{π}{2} and (\cot α_{1}) . (\cot α_{2}) \dots (\cot α_{n}) = 1$ is [2001]

Q 8 :

Let $f (θ) = \sin θ (\sin θ + \sin 3 θ)$ . Then $f (θ)$ is [2000]

Q 10 :

The maximum value of the expression $\frac{1}{\sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ}$ is _______. [2010]

Q 11 :

Let $α = \frac{1}{\sin 60 ° \sin 61 °} + \frac{1}{\sin 62 ° \sin 63 °} + \dots + \frac{1}{\sin 118 ° \sin 119 °} .$ Then the value of ${(\frac{c o s e c 1 °}{α})}^{2}$ is ________. [2025]

Q 12 :

Let $f : [0, 2] \to ℝ$ be the function defined by $f (x) = (3 - \sin (2 π x)) \sin (π x - \frac{π}{4}) - \sin (3 π x + \frac{π}{4}) .$

If $α, β \in [0, 2]$ are such that ${x \in [0, 2] : f (x) \geq 0} = [α, β],$ then the value of $β - α$ is _______. [2020]

Q 13 :

Let $f (x) = x \sin (π x), x > 0$ . Then for all natural numbers $n, f' (x)$ vanishes at [2013]

Q 14 :

Let $θ, φ \in [0, 2 π]$ be such that $2 \cos θ (1 - \sin φ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) \cos φ - 1,$ $tan (2 π - θ) > 0$

and $- 1 < \sin θ < - \frac{\sqrt{3}}{2}$ , then $φ$ cannot satisfy [2012]

Q 15 :

If $\frac{\sin^{4} x}{2} + \frac{\cos^{4} x}{3} = \frac{1}{5}$ , then [2009]

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Topic Question Set

Q 1 : Let f:ℝ→ℝ be a function defined by f(x)={x2sin(πx2),if x≠00,if x=0 Then which of the following statements is TRUE? [2024]

Q 2 : Let π2<x<π be such that cotx=-511. Then (sin11x2)(sin6x-cos6x)+(cos11x2)(sin6x+cos6x) is equal to [2024]

Q 3 : The value of ∑k=1131sin(π4+(k-1)π6)sin(π4+kπ6) is equal to [2016]

Q 4 : Let θ∈(0,π4) and t1=(tanθ)tanθ, t2=(tanθ)cotθ, t3=(cotθ)tanθ, t4=(cotθ)cotθ, then [2006]

Q 5 : The values of θ∈(0,2π) for which 2sin2θ-5sinθ+2>0, are [2006]

Q 6 : If α+β=π2 and β+γ=α, then tanα equals [2001]

Q 7 : The maximum value of (cosα1).(cosα2)⋯(cosαn), under the restrictions 0≤α1,α2,…,αn≤π2 and (cotα1).(cotα2)⋯(cotαn)=1 is [2001]

Q 8 : Let f(θ)=sinθ(sinθ+sin3θ). Then f(θ) is [2000]

Q 9 : Let α and β be real numbers such that -π4<β<0<α<π4. If sin(α+β)=13 and cos(α-β)=23, then the greatest integer less than or equal to (sinαcosβ+cosβsinα+cosαsinβ+sinβcosα)2 is _______. [2022]

Q 10 : The maximum value of the expression 1sin2θ+3sinθcosθ+5cos2θ is _______. [2010]

Q 11 : Let α=1sin60°sin61°+1sin62°sin63°+⋯+1sin118°sin119°. Then the value of (cosec1°α)2 is ________. [2025]

Q 12 : Let f:[0,2]→ℝ be the function defined by f(x)=(3-sin(2πx))sin(πx-π4)-sin(3πx+π4). If α,β∈[0,2] are such that {x∈[0,2]:f(x)≥0}=[α,β], then the value of β-α is _______. [2020]

Q 13 : Let f(x)=xsin(πx), x>0. Then for all natural numbers n, f'(x) vanishes at [2013]

Q 14 : Let θ,φ∈[0,2π] be such that 2cosθ(1-sinφ)=sin2θ(tanθ2+cotθ2)cosφ-1,tan(2π-θ)>0 and -1<sinθ<-32, then φ cannot satisfy [2012]

Q 15 : If sin4x2+cos4x3=15, then [2009]

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Q 1 :

Let $f : ℝ \to ℝ$ be a function defined by

$f (x)$ $=$ ${\begin{matrix} x^{2} \sin (\frac{π}{x^{2}}), & if x \neq 0 \\ 0, & if x = 0 \end{matrix}$

Then which of the following statements is TRUE? [2024]

Q 2 :

Let $\frac{π}{2} < x < π$ be such that $\cot x = \frac{- 5}{\sqrt{11}}$ . Then

$(\sin \frac{11 x}{2}) (\sin 6 x - \cos 6 x) + (\cos \frac{11 x}{2}) (\sin 6 x + \cos 6 x)$ is equal to [2024]

Q 3 :

The value of $\sum_{k = 1}^{13} \frac{1}{\sin (\frac{π}{4} + \frac{(k - 1) π}{6}) \sin (\frac{π}{4} + \frac{k π}{6})}$ is equal to [2016]

Q 4 :

Let $θ \in (0, \frac{π}{4})$ and $t_{1} = {(\tan θ)}^{\tan θ}, t_{2} = {(\tan θ)}^{\cot θ}, t_{3} = {(\cot θ)}^{\tan θ}, t_{4} = {(\cot θ)}^{\cot θ},$ then [2006]

Q 5 :

The values of $θ \in (0, 2 π)$ for which $2 \sin^{2} θ - 5 \sin θ + 2 > 0$ , are [2006]

Q 6 :

If $α + β = \frac{π}{2}$ and $β + γ = α$ , then $\tan α$ equals [2001]

Q 7 :

The maximum value of $(\cos α_{1}) . (\cos α_{2}) \dots (\cos α_{n}),$ under the restrictions

$0 \leq α_{1}, α_{2}, \dots, α_{n} \leq \frac{π}{2} and (\cot α_{1}) . (\cot α_{2}) \dots (\cot α_{n}) = 1$ is [2001]

Q 8 :

Let $f (θ) = \sin θ (\sin θ + \sin 3 θ)$ . Then $f (θ)$ is [2000]

Q 10 :

The maximum value of the expression $\frac{1}{\sin^{2} θ + 3 \sin θ \cos θ + 5 \cos^{2} θ}$ is _______. [2010]

Q 11 :

Let $α = \frac{1}{\sin 60 ° \sin 61 °} + \frac{1}{\sin 62 ° \sin 63 °} + \dots + \frac{1}{\sin 118 ° \sin 119 °} .$ Then the value of ${(\frac{c o s e c 1 °}{α})}^{2}$ is ________. [2025]

Q 12 :

Let $f : [0, 2] \to ℝ$ be the function defined by $f (x) = (3 - \sin (2 π x)) \sin (π x - \frac{π}{4}) - \sin (3 π x + \frac{π}{4}) .$

If $α, β \in [0, 2]$ are such that ${x \in [0, 2] : f (x) \geq 0} = [α, β],$ then the value of $β - α$ is _______. [2020]

Q 13 :

Let $f (x) = x \sin (π x), x > 0$ . Then for all natural numbers $n, f' (x)$ vanishes at [2013]

Q 14 :

Let $θ, φ \in [0, 2 π]$ be such that $2 \cos θ (1 - \sin φ) = \sin^{2} θ (\tan \frac{θ}{2} + \cot \frac{θ}{2}) \cos φ - 1,$ $tan (2 π - θ) > 0$

and $- 1 < \sin θ < - \frac{\sqrt{3}}{2}$ , then $φ$ cannot satisfy [2012]

Q 15 :

If $\frac{\sin^{4} x}{2} + \frac{\cos^{4} x}{3} = \frac{1}{5}$ , then [2009]