JEE MAIN 2023 Previous Year Questions (PYQs) – Page 3 of 34
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Let $f:\mathbb{R}-\{0,1\}\to\mathbb{R}$ be a function such that
$f(x)+f\!\left(\frac{1}{1-x}\right)=1+x.$
Then $f(2)$ is equal to:
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Solution
Let $S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$.
If $\mathrm{n(S)}$ denotes the number of elements in $\mathrm{S}$ then :
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Solution
Let $A = \{x \in \mathbb{R} : [x + 3] + [x + 4] \le 3\}$,
and $B = \left\{ x \in \mathbb{R} : 3^x \left( \sum_{r=1}^{\infty} \frac{3}{10^r} \right)^{x - 3} < 3^{-3x} \right\}$,
where $[\,]$ denotes the greatest integer function.
Then,
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Solution
If the system of equations
$x + y + a z = b$
$2x + 5y + 2z = 6$
$x + 2y + 3z = 3$
has infinitely many solutions, then $2a + 3b$ is equal to :
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Solution
The straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line
$L : 9x + 5y = 45$ between the axes.
If $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$, then the point of intersection of the line
$y = (m_1 + m_2)x$ with $L$ lies on :
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Solution
One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along the $x$, $y$ and $z$ axes are $3,\,4$ and $5$ units respectively.
Let $P$ be the vertex $(3,4,5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$–axis, not passing through $O$ or $P$, is:
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Solution
The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively.
The mean and variance of another set of $15$ numbers are $14$ and $\sigma^{2}$ respectively.
If the variance of all the $30$ numbers in the two sets is $13$, then $\sigma^{2}$ is equal to:
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Solution
Let $A = [a_{ij}]_{2\times 2}$, where $a_{ij}\ne 0$ for all $i,j$ and $A^{2}=I$.
Let $a$ be the sum of all diagonal elements of $A$ and $b=\lvert A\rvert$ (i.e., $b=\det A$).
Then $3a^{2}+4b^{2}$ is equal to:
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Solution
Let $5f(x)+4f\!\left(\dfrac{1}{x}\right)=\dfrac{1}{x}+3,\; x>0.$
Then $18\displaystyle\int_{1}^{2} f(x)\,dx$ is equal to:
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Solution
Let $I(x)=\displaystyle \int \frac{x^{2}\big(x\sec^{2}x+\tan x\big)}{(x\tan x+1)^{2}}\,dx.$
If $I(0)=0$, then $I\!\left(\frac{\pi}{4}\right)$ is equal to:
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Solution
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