Trigonometry · JEE Main previous year questions

2024 31 questions

1. Q132024 · 29 Jan 2024 · Shift 1

   For $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, if $y(x)=\int \frac{\text{cosec }x+\sin x}{\text{cosec }x\sec x+\tan x{\sin }^{2}x}\mathrm{d}x$, and ${\lim }_{x\to {\left(\frac{\pi }{2}\right)}^{-}}y(x)=0$ then $y\left(\frac{\pi }{4}\right)$ is equal to

   1. A.$-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
   2. B.${\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
   3. C.$\frac{1}{2}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
   4. D.$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(-\frac{1}{2}\right)$
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   Answer: D

2. Q182024 · 29 Jan 2024 · Shift 2

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

   1. A.3
   2. B.1
   3. C.0
   4. D.$-1$
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   Answer: D

3. Q192024 · 29 Jan 2024 · Shift 2

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

   1. A.$3x-2y=-2$
   2. B.$3x+2y=2$
   3. C.$5x+8y=9$
   4. D.$5x-8y=-9$
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   Answer: C

4. Q242024 · 27 Jan 2024 · Shift 1· Numerical answer

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

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   Answer: 48

5. Q12024 · 27 Jan 2024 · Shift 2

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

   1. A.more than 2
   2. B.$2$
   3. C.$0$
   4. D.$1$
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   Answer: D

6. Q132024 · 27 Jan 2024 · Shift 2

   If $2{\tan }^2\theta -5\sec \theta =1$ has exactly 7 solutions in the interval $[0,\frac{n\pi }{2}]$, for the least value of $n\in \mathbf{N}$, then ${\sum }_{k=1}^n\frac{k}{2^k}$ is equal to:

   1. A.$\frac{1}{2^{14}}(2^{15}-15)$
   2. B.$1-\frac{15}{2^{13}}$
   3. C.$\frac{1}{2^{15}}(2^{14}-14)$
   4. D.$\frac{1}{2^{13}}(2^{14}-15)$
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   Answer: D