JEE MAIN 2026 Previous Year Questions (PYQs) – Page 13 of 25
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Let
$S=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\cdots$ (13 terms)
If $13S=\frac{2^k}{n!}$, then $n+k$ is equal to:
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Solution
$S=\frac{1}{26!}\left({}^{26}C_1+{}^{26}C_3+\cdots+{}^{26}C_{25}\right)$
$= \frac{2^{25}}{26!}$
$13S=\frac{2^{24}}{25!}$
$\Rightarrow n=25,; k=24$
$n+k=49$
Consider an A.P.: $a,a_2,\ldots,a_n$
If $a_2-a_1=-\frac{3}{4},; a_n-a_{n-1}=\frac{1}{4}$ and
$\sum a_i=\frac{525}{2}$
then $\sum a_i$ is equal to:
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Solution
Using A.P. formulas
$a=3,; d=2$
Sum $=238$
Let \( \alpha, \beta \in \mathbb{R} \) be such that the function
\(
f(x)=
\begin{cases}
2\alpha(x^2-2)+2\beta x, & x<1 \\
(\alpha+3)x+(\alpha-\beta), & x\ge 1
\end{cases}
\)
be differentiable at all \( x\in\mathbb{R} \). Then \( 34(\alpha+\beta) \) is equal to:
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Solution
Continuity at $x=1$
$2\alpha(1-2)+2\beta = (\alpha+3)+(\alpha-\beta)$
$-2\alpha+2\beta = 2\alpha-\beta+3$
$\Rightarrow 4\alpha-3\beta+3=0$ …(1)
Differentiability
$f'(x<1)=4\alpha x+2\beta \Rightarrow f'(1)=4\alpha+2\beta$
$f'(x\ge1)=\alpha+3$
$4\alpha+2\beta=\alpha+3$
$\Rightarrow 3\alpha+2\beta-3=0$ …(2)
Solving (1) & (2)
$\alpha=\frac{3}{17},; \beta=\frac{21}{17}$
$34(\alpha+\beta)=34\cdot\frac{24}{17}=48$
From a lot containing $10$ defective and $90$ non-defective bulbs, $8$ bulbs are selected one by one with replacement.
Then the probability of getting at least $7$ defective bulbs is:
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Solution
$P(\text{defective})=\frac{10}{100}=\frac{1}{10}$
Required probability
$=P(7\text{ defective})+P(8\text{ defective})$
$=\binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right)+\left(\frac{1}{10}\right)^8$
$=\frac{8\cdot9}{10^8}+\frac{1}{10^8}$
$=\frac{72+1}{10^8}=\frac{73}{10^8}$
The mean and variance of a data of $10$ observations are $10$ and $2$, respectively.
If an observation $\alpha$ in the data is replaced by $\beta$, then the mean and variance become $10.1$ and $1.99$, respectively. Then $\alpha+\beta$ equals:
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Solution
Let observations be $x_1,x_2,\ldots,x_9,\alpha$
Mean $=10 \Rightarrow \sum x_i =100$
$\sum_{i=1}^{9}x_i =100-\alpha$
Variance
$\frac{\sum x_i^2}{10}-10^2=2$
$\Rightarrow \sum x_i^2=1020$
$\sum_{i=1}^{9}x_i^2=1020-\alpha^2$
New mean
$\frac{100-\alpha+\beta}{10}=10.1$
$\Rightarrow \beta-\alpha=1$ …(1)
New variance
$\frac{1020-\alpha^2+\beta^2}{10}-(10.1)^2=1.99$
$\Rightarrow \beta^2-\alpha^2=20$
$(\beta-\alpha)(\beta+\alpha)=20$
Using (1):
$1(\alpha+\beta)=20$
$\Rightarrow \alpha+\beta=20$
Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2,; x+y=8$ and y-axis lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2,; y^2=x,; x=2$ and y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to:
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Solution
$A_1=\int_0^2[(8-x)-(x^2+2)]dx$
$=\int_0^2(6-x-x^2),dx$
$=\left[6x-\frac{x^2}{2}-\frac{x^3}{3}\right]_0^2$
$=12-2-\frac{8}{3}=\frac{22}{3}$
$A_2=\int_0^2(x^2+2),dx-\frac{2}{3}(2\sqrt{2})$
$=\left[\frac{x^3}{3}+2x\right]_0^2-\frac{4\sqrt{2}}{3}$
$=\frac{8}{3}+4-\frac{4\sqrt{2}}{3}=\frac{20}{3}-\frac{4\sqrt{2}}{3}$
$A_1-A_2=\frac{22}{3}-\left(\frac{20}{3}-\frac{4\sqrt{2}}{3}\right)$
$=\frac{2}{3}+\frac{4\sqrt{2}}{3}=\frac{2}{3}(2\sqrt{2}+1)$
The number of the real solutions of the equation:
$x|x+3|+|x-1|-2=0$
is:
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Solution
Break into intervals
Case I: $x\ge1$
$x(x+3)+(x-1)-2=0$
$x^2+4x-3=0$
$x=-2\pm\sqrt{7}$ (reject)
Case II: $-3\le x<1$
$x(x+3)+(1-x)-2=0$
$x^2+2x-1=0$
$x=-1\pm\sqrt{2}$
Case III: $x<-3$
$x(-(x+3))+(1-x)-2=0$
$x^2+4x+1=0$
$x=-2\pm\sqrt{3}$ (reject)
Only valid solutions = $3$
Let a differentiable function $f$ satisfy the equation
$\int_0^{36} f\left(\frac{tx}{36}\right)dt=4\alpha f(x)$
If $y=f(x)$ is a standard parabola passing through the points $(2,1)$ and $(-4,\beta)$, then $\beta^\alpha$ is equal to ______.
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Solution
Put $\frac{tx}{36}=y$
$\int_0^x f(y)\frac{36}{x}dy=4\alpha f(x)$
$\Rightarrow \int_0^x f(y)dy=\frac{\alpha f(x)x}{9}$
Differentiate
$f(x)=\frac{\alpha}{9}(f(x)+xf'(x))$
$\Rightarrow (1-\frac{\alpha}{9})f(x)=\frac{\alpha x}{9}f'(x)$
$\frac{f'(x)}{f(x)}=\left(\frac{9}{\alpha}-1\right)\frac{1}{x}$
$f(x)=cx^{\frac{9}{\alpha}-1}$
For standard parabola ⇒ power = 2
$\frac{9}{\alpha}-1=2 \Rightarrow \alpha=3$
$f(2)=1 \Rightarrow c=\frac{1}{4}$
$f(x)=\frac{x^2}{4}$
$f(-4)=\beta=4$
$\beta^\alpha=4^3=64$
Let a line $L$ passing through the point $P(1,1,1)$ be perpendicular to the lines
$\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$
and
$\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$.
Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1,0,-1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to ______.
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Solution
Direction vectors:
\( \vec d_1 = \langle 4,1,1 \rangle,\quad \vec d_2 = \langle 1,1,0 \rangle \)
\(
\vec d =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
4 & 1 & 1 \\
1 & 1 & 0
\end{vmatrix}
= \langle -1,1,3 \rangle
\)
---
Equation of line \(L\) through \(P(1,1,1)\):
\( x=1-t,\; y=1+t,\; z=1+3t \)
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For \(Q\) (on yz-plane ⇒ \(x=0\)):
\( t=1 \Rightarrow Q(0,2,4) \)
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Line through \(S(1,0,-1)\) parallel to \(L\):
\( x=1-\mu,\; y=\mu,\; z=-1+3\mu \)
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For \(R\) (on yz-plane ⇒ \(x=0\)):
\( \mu=1 \Rightarrow R(0,1,2) \)
---
Vectors:
\( \vec{PQ}=\langle -1,1,3 \rangle,\quad \vec{PS}=\langle 0,-1,-2 \rangle \)
---
Area of parallelogram:
\(
\vec{PQ}\times \vec{PS} =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
-1 & 1 & 3 \\
0 & -1 & -2
\end{vmatrix}
= \langle 1,-2,1 \rangle
\)
---
Area:
\( |\vec{PQ}\times \vec{PS}| = \sqrt{1^2+(-2)^2+1^2} = \sqrt{6} \)
---
Required square:
\( = 6 \)
The number of numbers greater than $5000$, less than $9000$ and divisible by $3$, that can be formed using digits $0,1,2,5,9$, if repetition is allowed, is ______.
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Solution
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