JEE MAIN 2023 Previous Year Questions (PYQs) – Page 18 of 34
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Let $f(x)=\displaystyle \int \frac{2x}{(x^{2}+1)(x^{2}+3)}\,dx$.
If $f(3)=\dfrac{1}{2}(\log_{e}5-\log_{e}6)$, then $f(4)$ is equal to:
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Solution
Let the lines
\[
\ell_1:\ \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}
\quad\text{and}\quad
\ell_2:\ 3x+2y+z-2=0\;=\;x-3y+2z-13
\]
be coplanar. If the point $P(a,b,c)$ on $\ell_1$ is nearest to the point $Q(-4,-3,2)$,
then $|a|+|b|+|c|$ is equal to:
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Solution
The distance of the point $P(4,6,-2)$ from the line passing through the point $(-3,2,3)$ and parallel to a line with direction ratios $3,3,-1$ is equal to:
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Solution
The number of five digit numbers, greater than 40000 and divisible by 5 , which can be formed using the digits 1, 3, 7 and 9 without repetition, is equal to :
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Solution
Let $y(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})(1+x^{16})$.
Then $y' - y''$ at $x=-1$ is equal to:
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Solution
Let $C$ be the circle in the complex plane with centre $z_0=\tfrac{1}{2}(1+3i)$ and radius $r=1$.
Let $z_1=1+i$ and the complex number $z_2$ be outside the circle $C$ such that
$\lvert z_1-z_0\rvert\,\lvert z_2-z_0\rvert=1$.
If $z_0,z_1$ and $z_2$ are collinear, then the smaller value of $\lvert z_2\rvert^2$ is equal to:
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Solution
Consider the lines $L_{1}$ and $L_{2}$ given by
$L_{1}:\ \dfrac{x-1}{2}=\dfrac{y-3}{1}=\dfrac{z-2}{2}$
$L_{2}:\ \dfrac{x-2}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$
A line $L_{3}$ having direction ratios $1,-1,-2$ intersects $L_{1}$ and $L_{2}$ at the points $P$ and $Q$ respectively.
Then the length of line segment $PQ$ is:
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Solution
If the point $(\alpha, \dfrac{7\sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x\cos\theta + y\sin\theta = 7, \theta \in (0, \dfrac{\pi}{2})$ between the co-ordinates axes, then $\alpha$ is equal to:
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Solution
The points of intersection of the line $ax+by=0,\ (a\ne b)$ and the circle $x^{2}+y^{2}-2x=0$ are $A(\alpha,0)$ and $B(1,\beta)$.
The image of the circle with $AB$ as a diameter in the line $x+y+2=0$ is:
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Solution
Let $\alpha, \beta$ be the roots of the quadratic equation $x^{2}+\sqrt{6}x+3=0$. Then $\dfrac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ is equal to:
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Solution
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