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JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.
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Simulator
Previous Years AIEEE/JEE Main Questions
Q. Let A and B denote the statements
$\mathbf{A}: \cos \alpha+\cos \beta+\cos \gamma=0$
$\mathbf{B}: \sin \alpha+\sin \beta+\sin \gamma=0$
If $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2},$ then $:-$
(1) Both A and B are true
(2) Both A and B are false
(3) A is true and B is false
(4) A is false and B is true
[AIEEE 2009]
Ans. (1)
$\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$
$2 \cos (\beta-\gamma)+2 \cos (\gamma-\alpha)+2 \cos (\alpha-\beta)=-3$
$1+1+1+2(\cos \beta \cos \gamma+\sin \beta \sin \gamma)+2(\cos \gamma \cos \alpha+\sin \gamma \sin \alpha)+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)$
$=0$
$\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+\left(\sin ^{2} \gamma+\cos ^{2} \gamma\right)+2 \cos \alpha \cos \beta+2 \cos \beta \cos \gamma+2 \cos$
$\gamma \cos \alpha$
$+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma+2 \sin \gamma \sin \alpha=0$
$\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+2 \sin \alpha \sin \beta+2 \sin \beta \sin \gamma\right.$
$+2 \sin \gamma \sin \alpha)+\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right.$
$+2 \cos \alpha \cos \beta+\cos \beta \cos \gamma+\cos \gamma \cos \alpha)=0$
$(\sin \alpha+\sin \beta+\sin \gamma)^{2}+(\cos \alpha+\cos \beta+\cos \gamma)^{2}=0$
Only Possible when
$\sin \alpha+\sin \beta+\sin \gamma=0$
$\cos \alpha+\cos \beta+\cos \gamma=0$
Q. The possible values of $\theta \in(0, \pi)$ such that $\sin (\theta)+\sin (4 \theta)+\sin (7 \theta)=0$ are:
(1) $\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{4 \pi}{9}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$
(2) $\frac{\pi}{4}, \frac{5 \pi}{12}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$
(3) $\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{3 \pi}{4}, \frac{35 \pi}{36}$
(4) $\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{2 \pi}{3}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$
[AIEEE 2011]
Ans. (1)
$\sin \theta+\sin 4 \theta+\sin 7 \theta=0$
$2 \sin \left(\frac{\theta+7 \theta}{2}\right) \cos \left(\frac{7 \theta-\theta}{2}\right)+\sin 4 \theta=0$
$\Rightarrow \sin 4 \theta[2 \cos 3 \theta+1]=0$
$\Rightarrow \sin 4 \theta=0 \Rightarrow 4 \theta=0, \pi, 2 \pi, 3 \pi, 4 \pi$
$\Rightarrow \theta=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi$
but 0 and $\pi$ are not included.
and $2 \cos 3 \theta+1=0 \Rightarrow \cos 3 \theta=\frac{-1}{2}$
$\Rightarrow 3 \theta=\frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3} \quad \Rightarrow \quad \theta=\frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}, \frac{10 \pi}{9}$
but $\frac{10 \pi}{9} \notin(0, \pi)$
So, $\theta=\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}$
Q. If $0 \leq x<2 \pi,$ then the number of real values of $x,$ which satisfy the equation
$\cos x+\cos 2 x+\cos 3 x+\cos 4 x=0,$ is : -
(1) 9 (2) 3 (3) 5 (4) 7
[JEE Mains 2016]
Ans. (4)
$2 \cos 2 x \cos x+2 \cos 3 x \cos x=0$
$\Rightarrow 2 \cos x(\cos 2 x+\cos 3 x)=0$
$2 \cos x 2 \cos 5 x / 2 \cos x / 2=0$
$x=\frac{\pi}{2}, \frac{3 \pi}{2}, \pi, \frac{\pi}{5}, \frac{3 \pi}{5}, \frac{7 \pi}{5}, \frac{9 \pi}{5}$
7 Solutions
Q. If sum of all the solutions of the equation $8 \cos x \cdot\left(\cos \left(\frac{\pi}{6}+x\right) \cdot \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right)=1$ in
$[0, \pi]$ is $k \pi,$ then $k$ is equal to :
(1) $\frac{13}{9}$
(2) $\frac{8}{9}$
(3) $\frac{20}{9}$
( 4)$\frac{2}{3}$
[JEE Mains 2016]
Ans. (1)
$8 \cos x\left(\cos ^{2} \frac{\pi}{6}-\sin ^{2} x-\frac{1}{2}\right)=1$
$\Rightarrow 8 \cos x\left(\frac{1}{4}-\left(1-\cos ^{2} x\right)\right)=1$
$\Rightarrow 8 \cos x\left(\cos ^{2} x-\frac{3}{4}\right)=1$
$\Rightarrow 2 \cos 3 x=1 \Rightarrow \cos 3 x=\frac{1}{2}$
$\therefore 3 x+2 n \pi \pm \frac{\pi}{3}, n \in I$
$\Rightarrow \mathrm{x}=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}$
$\ln \mathrm{x} \in[0, \pi]: \mathrm{x}=\frac{\pi}{9}, \frac{2 \pi}{3}+\frac{\pi}{9}, \frac{2 \pi}{3}-\frac{\pi}{9}$ only
$\operatorname{sum}=\frac{13 \pi}{9}$
Comments
Namo kaul from unacademy
April 15, 2021, 2:03 p.m.
pure 15 saal me 4 questions hi aae hai kya jee main bc?
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