JEE MAIN 2025 Previous Year Questions (PYQs) – Page 2 of 38

Use identity: $\operatorname{adj}(M^{-1})=\dfrac{\operatorname{adj}(M)}{|M|}$

So,
$\operatorname{adj}(A^{-1})=\dfrac{A}{|A|},\quad \operatorname{adj}(B^{-1})=\dfrac{B}{|B|}$

Hence,
$\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})=\dfrac{A}{|A|}+\dfrac{B}{|B|}$

Given expression:
$X=A\Big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\Big)^{-1}B$

$X^{-1}=B^{-1}\Big(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\Big)A^{-1}$

Substitute:
$=B^{-1}\left(\dfrac{A}{|A|}+\dfrac{B}{|B|}\right)A^{-1}$

$=\dfrac{1}{|A|}B^{-1}AA^{-1}+\dfrac{1}{|B|}B^{-1}BA^{-1}$

$=\dfrac{1}{|A|}B^{-1}+\dfrac{1}{|B|}A^{-1}$

$=\dfrac{1}{|A|} \dfrac{adjB}{|B|}$+$\dfrac{1}{|B|} \dfrac{adjA}{|A|}$
$=\dfrac{adjB}{|AB|}$+$\dfrac{adjA}{|AB|}$

Given data: $2,3,3,4,5,7,a,b$ (8 observations)

Mean $=4 \Rightarrow$ total sum $=8\times 4=32$
Sum of known terms $=24 \Rightarrow a+b=8$

Variance $=2 \Rightarrow \sum (x-4)^2=8\times 2=16$
Known contribution $=16 \Rightarrow (a-4)^2+(b-4)^2=0$
⇒ $a=b=4$

Data: $2,3,3,4,4,4,5,7$
Mode $=4$

Mean deviation about mode:
$=\frac{1}{8}[|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|7-4|]$
$=\frac{1}{8}[2+1+1+0+0+0+1+3]=\frac{8}{8}=1$

Given sum of focal distances: $SP_1 + SP_2 = 2ex = 8\sqrt{\frac{5}{3}}$
At $P(4,3)$ ⇒ $2e \cdot 4 = 8\sqrt{\frac{5}{3}}$
⇒ $e=\sqrt{\frac{5}{3}}$

Now, $b^2 = a^2(e^2 - 1) = a^2\left(\frac{5}{3}-1\right)=\frac{2}{3}a^2$

Using point $(4,3)$: $\frac{16}{a^2}-\frac{9}{b^2}=1$
Substitute $b^2=\frac{2}{3}a^2$:
$\frac{16}{a^2}-\frac{9}{(2/3)a^2}=1$
⇒ $\frac{16}{a^2}-\frac{27}{2a^2}=1$
⇒ $\frac{5}{2a^2}=1$
⇒ $a^2=\frac{5}{2},\ b^2=\frac{5}{3}$

Latus rectum: $l=\frac{2b^2}{a}$ ⇒ $l^2=\frac{4b^4}{a^2}$
⇒ $l^2=\frac{4(25/9)}{5/2}=\frac{40}{9}$
⇒ $9l^2=40$

Product of focal distances: $m=(ex+a)(ex-a)=e^2x^2-a^2$
⇒ $m=\frac{5}{3}\cdot16-\frac{5}{2}=\frac{145}{6}$
⇒ $6m=145$

$9l^2+6m=40+145=185$

$\boxed{185}$

$I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx$

$I+\int_{0}^{\pi/2}\frac{\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\frac{\pi}{2}$
⇒ $2I=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$

Required integral:
$J=\int_{0}^{2I}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx=\int_{0}^{\pi/2}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$

Using symmetry:
$2J=\frac{\pi}{2}\int_{0}^{\pi/2}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx$

Now,
$\int_{0}^{\pi/2}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx=\frac{\pi}{4}$

So,
$2J=\frac{\pi}{2}\cdot\frac{\pi}{4}=\frac{\pi^2}{8}$
⇒ $J=\frac{\pi^2}{16}$

$\boxed{\frac{\pi^2}{16}}$

Given $A^n=A^{n-2}+A^2-I$ for $n\ge 3$

Let $S_n$ = sum of all elements of $A^n$

Taking sum on both sides:
$S_n=S_{n-2}+S_2-3$ (since sum of $I=3$)

Compute base values:
$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\end{bmatrix}$
⇒ $S_1=3$

$A^2=\begin{bmatrix}1&0&0\\1&1&0\\1&0&1\end{bmatrix}$
⇒ $S_2=5$

So recurrence:
$S_n=S_{n-2}+2$

Even terms:
$S_2=5,\ S_4=7,\ S_6=9,\dots$
⇒ $S_{2k}=5+(k-1)\cdot2=2k+3$

For $n=50$, $k=25$:
$S_{50}=2(25)+3=53$

Given $|z|=1 \Rightarrow z=e^{i\theta}$

$\frac{z}{\overline{z}}=e^{2i\theta},\quad \frac{\overline{z}}{z}=e^{-2i\theta}$

So,
$\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right| =|e^{2i\theta}+e^{-2i\theta}|=|2\cos 2\theta|$

Given $|2\cos 2\theta|=1 \Rightarrow |\cos 2\theta|=\frac{1}{2}$

Now in $[0,2\pi)$:
$2\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$

⇒ $\theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}$

But $z=e^{i\theta}$ is periodic with $2\pi$, so each valid $\theta$ gives **2 distinct $z$** over full cycle of $2\theta$

Hence total solutions $=4\times 2=8$

$f(x)=6x^3-45ax^2+108a^2x+1$

$f'(x)=18x^2-90ax+108a^2$
$=18(x^2-5ax+6a^2)$

So critical points:
$x_1,x_2=2a,3a$

Given $x_1x_2=54$
⇒ $6a^2=54 \Rightarrow a^2=9$
$ \Rightarrow a=3$ (since $a>0$)

$x_1+x_2=5a=15$

$a+x_1+x_2=3+15=18$

Given ellipse: $\frac{x^2}{4}+\frac{y^2}{2}=1$
Midpoint $(1,\frac{1}{2})$

Use midpoint (chord) formula: $T=S_1$
$\frac{xx_1}{4}+\frac{yy_1}{2}= \frac{x_1^2}{4}+\frac{y_1^2}{2}$

RHS:
$\frac{1}{4}+\frac{1/4}{2}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$

So chord equation:
$\frac{x}{4}+\frac{y}{4}=\frac{3}{8}$
⇒ $2x+2y=3$ ⇒ $x+y=\frac{3}{2}$

Now find intersection with ellipse:
$y=\frac{3}{2}-x$

Substitute:
$\frac{x^2}{4}+\frac{(3/2-x)^2}{2}=1$

Multiply by 4:
$x^2+2\left(\frac{9}{4}-3x+x^2\right)=4$

$x^2+\frac{9}{2}-6x+2x^2=4$

$3x^2-6x+\frac{9}{2}-4=0$

$3x^2-6x+\frac{1}{2}=0$

$6x^2-12x+1=0$

Roots difference:
$|x_1-x_2|=\frac{\sqrt{144-24}}{6}$

Given: $(7x^{4}\cot y - e^{x}\csc y)\frac{dx}{dy}=x^{5}$

⇒ $\frac{dy}{dx}=\frac{7x^{4}\cot y - e^{x}\csc y}{x^{5}}$

Multiply by $\sin y$:
$\sin y\frac{dy}{dx}=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$

Now note:
$\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}$

So,
$-\frac{d}{dx}(\cos y)=\frac{7}{x}\cos y-\frac{e^{x}}{x^{5}}$

⇒ $\frac{d}{dx}(\cos y)+\frac{7}{x}\cos y=\frac{e^{x}}{x^{5}}$

This is linear in $\cos y$

IF $=e^{\int \frac{7}{x}dx}=x^{7}$

$\frac{d}{dx}(x^{7}\cos y)=x^{2}e^{x}$

Integrate:
$x^{7}\cos y=\int x^{2}e^{x}dx$

$\int x^{2}e^{x}dx=e^{x}(x^{2}-2x+2)$

So,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)+C$

Use $(1,\frac{\pi}{2})$ ⇒ $\cos y=0$
$0=e(1-2+2)+C=e+C$ ⇒ $C=-e$

Thus,
$x^{7}\cos y=e^{x}(x^{2}-2x+2)-e$

At $x=2$:
$2^{7}\cos y=e^{2}(4-4+2)-e=2e^{2}-e$

$\cos y=\frac{2e^{2}-e}{128}=\frac{e(2e-1)}{128}$

$\boxed{\frac{e(2e-1)}{128}}$