JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 3 of 275
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Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2},; \ldots, \frac{a_{10}}{2^9}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \cdots + a_{10} = 62$, then $a_1$ is equal to:
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Solution
$\frac{a_2}{2a_1} = \frac{a_3}{2a_2} = \cdots = \frac{a_{10}}{2a_9} = \frac{1}{\sqrt{2}}$
$\Rightarrow a_1, a_2, a_3, \ldots, a_{10}$ are in G.P. with common ratio $\sqrt{2}$
$\sum_{i=1}^{10} a_i = a_1 \cdot \frac{(\sqrt{2})^{10} - 1}{\sqrt{2} - 1} = 62$
$\Rightarrow a_1 = 2(\sqrt{2}-1)$
$\lim_{x \to 1} \left( \dfrac{\int_{0}^{(x-1)^{2}} t \cos(t^{2}) \, dt}{(x-1)\sin(x-1)} \right)$
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Solution
If $P = \begin{bmatrix} 1 & 0 \\ \tfrac{1}{2} & 1 \end{bmatrix}$, then $P^{50}$ is :
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Solution
$\tan\!\left(2\tan^{-1}\!\tfrac{1}{5}+\sec^{-1}\!\tfrac{\sqrt{5}}{2}+2\tan^{-1}\!\tfrac{1}{8}\right)$ is equal to:
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Solution
Let $f(x)=\begin{vmatrix}
1+\sin^{2}x & \cos^{2}x & \sin 2x\\
\sin^{2}x & 1+\cos^{2}x & \sin 2x\\
\sin^{2}x & \cos^{2}x & 1+\sin 2x
\end{vmatrix},\ x\in\left[\dfrac{\pi}{6},\dfrac{\pi}{3}\right].$ If $\alpha$ and $\beta$ respectively are the maximum and the minimum values of $f$, then
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Solution
Let $f:\mathbb{R}\to\mathbb{R}$ be a function given by
$
f(x)=
\begin{cases}
\dfrac{1-\cos 2x}{x^2}, & x<0,\\[6pt]
\alpha, & x=0,\\[6pt]
\dfrac{\beta\sqrt{\,1-\cos x\,}}{x}, & x>0,
\end{cases}
$
where $\alpha,\beta\in\mathbb{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to:
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Solution
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is:
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Solution
First term $a=3$, common difference $d$
Sum of first 4 terms:
$S_4=\frac{4}{2}[2a+3d]=2(6+3d)=12+6d$
Sum of first 8 terms:
$S_8=\frac{8}{2}[2a+7d]=4(6+7d)=24+28d$
Sum of next 4 terms:
$=S_8-S_4=(24+28d)-(12+6d)=12+22d$
Given:
$12+6d=\frac{1}{5}(12+22d)$
$60+30d=12+22d$
$8d=-48 \Rightarrow d=-6$
Now sum of first 20 terms:
$S_{20}=\frac{20}{2}[2a+19d]=10(6+19(-6))$
$=10(6-114)=10(-108)=-1080$
$\boxed{-1080}$
Sum of first 4 terms:
$S_4=\frac{4}{2}[2a+3d]=2(6+3d)=12+6d$
Sum of first 8 terms:
$S_8=\frac{8}{2}[2a+7d]=4(6+7d)=24+28d$
Sum of next 4 terms:
$=S_8-S_4=(24+28d)-(12+6d)=12+22d$
Given:
$12+6d=\frac{1}{5}(12+22d)$
$60+30d=12+22d$
$8d=-48 \Rightarrow d=-6$
Now sum of first 20 terms:
$S_{20}=\frac{20}{2}[2a+19d]=10(6+19(-6))$
$=10(6-114)=10(-108)=-1080$
$\boxed{-1080}$
Considering the principal values of the inverse trigonometric functions, $\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2}< x<\frac{1}{\sqrt{2}}$, is equal to
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Solution
Let $x=\sin\theta,\quad -\frac{\pi}{6}<\theta<\frac{\pi}{4}$
Then $\sqrt{1-x^2}=\cos\theta$
Expression becomes:
$\sin^{-1}\left(\frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta\right)$
$=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{6}\right)\right)$
Since $\theta\in\left(-\frac{\pi}{6},\frac{\pi}{4}\right)$
$\Rightarrow \theta+\frac{\pi}{6}\in\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
So principal value:
$\sin^{-1}(\sin(\theta+\frac{\pi}{6}))=\theta+\frac{\pi}{6}$
$=\sin^{-1}x+\frac{\pi}{6}$
$\boxed{\sin^{-1}x+\frac{\pi}{6}}$
Then $\sqrt{1-x^2}=\cos\theta$
Expression becomes:
$\sin^{-1}\left(\frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta\right)$
$=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{6}\right)\right)$
Since $\theta\in\left(-\frac{\pi}{6},\frac{\pi}{4}\right)$
$\Rightarrow \theta+\frac{\pi}{6}\in\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
So principal value:
$\sin^{-1}(\sin(\theta+\frac{\pi}{6}))=\theta+\frac{\pi}{6}$
$=\sin^{-1}x+\frac{\pi}{6}$
$\boxed{\sin^{-1}x+\frac{\pi}{6}}$
The curve amongst the family of curves represented by the differential equation
$(x^2 - y^2)dx + 2xy\,dy = 0$
which passes through $(1, 1)$ is :
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Solution
If $m$ is the minimum value of $k$ for which the function $f(x)=x\sqrt{kx-x^{2}}$ is increasing in the interval $[0,3]$ and $M$ is the maximum value of $f$ in $[0,3]$ when $k=m$, then the ordered pair $(m,M)$ is equal to:
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Solution
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