JEE MAIN Mathematics Previous Year Questions (PYQs) – Page 12 of 275
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If the area of the bounded region $R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$ is , $\alpha {({\log _e}2)^{ - 1}} + \beta ({\log _e}2) + \gamma $, then the value of ${(\alpha + \beta - 2\lambda )^2}$ is equal to :
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$\int_{0}^{20\pi} (|\sin x| + |\cos x|)^{2} \, dx$ is equal to:
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For the system of linear equations $\alpha x+y+z=1,\quad x+\alpha y+z=1,\quad x+y+\alpha z=\beta$, which one of the following statements is **NOT** correct?
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The area (in sq. units) of the region described by $\{(x,y):y^{2}\le2x,\;y\ge4x-1\}$ is:
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Let the point $A$ divide the line segment joining the points $P(-1,-1,2)$ and $Q(5,5,10)$ internally in the ratio $r:1\ (r>0)$. If $O$ is the origin and $(\overrightarrow{OQ}\cdot\overrightarrow{OA})-\dfrac{1}{5}\lvert\overrightarrow{OP}\times\overrightarrow{OA}\rvert^{2}=10$, then the value of $r$ is:
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If the sum of the first $20$ terms of the series $\dfrac{4\cdot1}{4+3\cdot1^{2}+1^{4}}+\dfrac{4\cdot2}{4+3\cdot2^{2}+2^{4}}+\dfrac{4\cdot3}{4+3\cdot3^{2}+3^{4}}+\dfrac{4\cdot4}{4+3\cdot4^{2}+4^{4}}+\cdots$ is $\dfrac{m}{n}$, where $m$ and $n$ are coprime, then $m+n$ is equal to:
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Two vertices of a triangle are $(0,2)$ and $(4,3)$. If its orthocenter is at the origin, then its third vertex lies in which quadrant:
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A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^\circ$ with the line $x + y = 0$.
Then an equation of the line $L$ is:
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The locus of the foot of perpendicular drawn from the centre of the ellipse
$x^{2}+3y^{2}=6$ on any tangent to it is :
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Let $[ \cdot ]$ denote the greatest integer function and
$f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right]$. Then $12\sum_{j=1}^{\infty} f(j)$ is equal to:
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$\sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] \le \sum_{k=1}^{n} \frac{k^2}{3^x}$
$= \frac{n(n+1)(2n+1)}{6\cdot 3^x}$
$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] = \frac{1}{3^{x+1}}$
$\Rightarrow f(x) = \frac{1}{3^{x+1}}$
$12\sum_{j=1}^{\infty} f(j) = 12\sum_{j=1}^{\infty} \frac{1}{3^{j+1}}$
$= 12\left(\frac{1}{9} + \frac{1}{27} + \cdots\right)$
$= 12\left(\frac{1/9}{1 - 1/3}\right) = 2$
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