JEE MAIN 2025 Previous Year Questions (PYQs) – Page 1 of 38

Let $t=\frac{x-11}{x+15}$

Then, $dx=\frac{26}{(1-t)^2}dt$

Also, $(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}=t^{\frac{11}{13}}(x+15)^2$

and $x+15=\frac{26}{1-t}$

So, $I=\int \frac{\frac{26}{(1-t)^2}dt}{t^{\frac{11}{13}}\cdot \frac{26^2}{(1-t)^2}}$

$I=\frac{1}{26}\int t^{-\frac{11}{13}}dt$

$I=\frac{1}{26}\cdot\frac{13}{2}t^{\frac{2}{13}}$

$I=\frac14\left(\frac{x-11}{x+15}\right)^{\frac{2}{13}}$

Now, $I(37)-I(24)=\frac14\left[\left(\frac{26}{52}\right)^{\frac{2}{13}}-\left(\frac{13}{39}\right)^{\frac{2}{13}}\right]$

$=\frac14\left[\left(\frac12\right)^{\frac{2}{13}}-\left(\frac13\right)^{\frac{2}{13}}\right]$

$=\frac14\left[\frac{1}{4^{\frac{1}{13}}}-\frac{1}{9^{\frac{1}{13}}}\right]$

So, $b=4,\ c=9$

$3(b+c)=3(4+9)=39$

$\boxed{39}$

Triangle vertices:
$A\left(-\frac{5}{3},\frac{20}{3}\right),\ B(1,2),\ C\left(\frac{25}{11},\frac{30}{11}\right)$

Slope $AC=-1$ ⇒ altitude from $B$ has slope $1$ ⇒ $y=x+1$

Solving with another altitude gives orthocentre $H_1=(1,2)$

Second triangle $(0,0),(1,0),(0,1)$ ⇒ right angled ⇒ $H_2=(0,0)$

Distance $=\sqrt{(1)^2+(2)^2}=\sqrt{5}$

$\boxed{\sqrt{5}}$

Given line: $3x-2y+12=0 \Rightarrow y=\frac{3x+12}{2}$
Parabola: $4y=3x^2 \Rightarrow y=\frac{3x^2}{4}$

Equate:
$\frac{3x^2}{4}=\frac{3x+12}{2}$
$\Rightarrow 3x^2=6x+24$
$\Rightarrow x^2-2x-8=0$
$\Rightarrow (x-4)(x+2)=0$

Points: $A(4,12),\ B(-2,3)$
Vertex $V(0,0)$

Slope $VA=\frac{12}{4}=3,\quad VB=\frac{3}{-2}=-\frac{3}{2}$

$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{3+\frac{3}{2}}{1- \frac{9}{2}}\right| =\left|\frac{\frac{9}{2}}{-\frac{7}{2}}\right| =\frac{9}{7}$

$\theta=\tan^{-1}\left(\frac{9}{7}\right)$

$\boxed{\tan^{-1}\left(\frac{9}{7}\right)}$

$T_r=\binom{n}{r-1}a^{,n-r+1}b^{,r-1}$ for $(a+b)^n$. 

First term $a=3$, common difference $d$

Sum of first 4 terms:
$S_4=\frac{4}{2}[2a+3d]=2(6+3d)=12+6d$

Sum of first 8 terms:
$S_8=\frac{8}{2}[2a+7d]=4(6+7d)=24+28d$

Sum of next 4 terms:
$=S_8-S_4=(24+28d)-(12+6d)=12+22d$

Given:
$12+6d=\frac{1}{5}(12+22d)$

$60+30d=12+22d$
$8d=-48 \Rightarrow d=-6$

Now sum of first 20 terms:
$S_{20}=\frac{20}{2}[2a+19d]=10(6+19(-6))$
$=10(6-114)=10(-108)=-1080$

$\boxed{-1080}$

Let $x=\sin\theta,\quad -\frac{\pi}{6}<\theta<\frac{\pi}{4}$

Then $\sqrt{1-x^2}=\cos\theta$

Expression becomes:
$\sin^{-1}\left(\frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta\right)$

$=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{6}\right)\right)$

Since $\theta\in\left(-\frac{\pi}{6},\frac{\pi}{4}\right)$
$\Rightarrow \theta+\frac{\pi}{6}\in\left(0,\frac{5\pi}{12}\right)\subset\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

So principal value:
$\sin^{-1}(\sin(\theta+\frac{\pi}{6}))=\theta+\frac{\pi}{6}$

$=\sin^{-1}x+\frac{\pi}{6}$

$\boxed{\sin^{-1}x+\frac{\pi}{6}}$

Let radius $=1$ and take $O$ as origin

Arc $AC$ subtends $\frac{\pi}{2}$ ⇒ take $A(1,0),\ C(0,1)$

Given $\frac{AB}{BC}= \frac{1}{5}$ ⇒ $\angle AOB=\frac{\pi}{12}$
So $B=\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$

$\overrightarrow{OC}=\alpha \overrightarrow{OA}+\beta \overrightarrow{OB}$

$(0,1)=\alpha(1,0)+\beta\left(\cos\frac{\pi}{12},\sin\frac{\pi}{12}\right)$

Comparing components:
$\alpha+\beta\cos\frac{\pi}{12}=0$
$\beta\sin\frac{\pi}{12}=1$

$\beta=\frac{1}{\sin\frac{\pi}{12}}=\frac{1}{\frac{\sqrt6-\sqrt2}{4}}=\sqrt6+\sqrt2$

$\alpha=-\beta\cos\frac{\pi}{12}=-(\sqrt6+\sqrt2)\cdot\frac{\sqrt6+\sqrt2}{4}=-2-\sqrt3$

Now,
$\alpha+\sqrt2(\sqrt3-1)\beta$
$=(-2-\sqrt3)+\sqrt2(\sqrt3-1)(\sqrt6+\sqrt2)$

$\sqrt2(\sqrt6+\sqrt2)=2\sqrt3+2$

So expression:
$=(-2-\sqrt3)+(2\sqrt3+2)(\sqrt3-1)$
$=(-2-\sqrt3)+(6-2\sqrt3+2\sqrt3-2)$
$=(-2-\sqrt3)+4=2-\sqrt3$

$\boxed{2-\sqrt3}$

Given $A=\{-3,-2,-1,0,1,2,3\}$ and $xRy \iff 2x-y\in\{0,1\}$

So $y=2x$ or $y=2x-1$

All valid pairs in $A$:
$(-1,-2),\ (-1,-3),\ (0,0),\ (0,-1),\ (1,2),\ (1,1),\ (2,3)$

So $l=7$

Reflexive pairs needed: $(x,x)$ for all 7 elements
Present: $(0,0),(1,1)$ ⇒ missing $5$ ⇒ $m=5$

For symmetry, add reverse of non-symmetric pairs:
Missing: $(-2,-1),\ (-3,-1),\ (-1,0),\ (2,1),\ (3,2)$ ⇒ $n=5$

$l+m+n=7+5+5=17$

$\boxed{17}$

Given $P(5,4),\ Q(-2,4)$ and orthocentre $O\left(2,\frac{14}{5}\right)$

$PQ$ is horizontal ⇒ altitude from $R$ is vertical ⇒ $x=2$
So $a=2$

Area $=\frac{1}{2}\times |PQ|\times \text{height}$
$=\frac{1}{2}\times 7 \times |b-4|=35$
$\Rightarrow |b-4|=10 \Rightarrow b=14 \text{ or } -6$

Using orthocentre condition (slopes), valid point is $R(2,-6)$

Centroid:
$c=\frac{5+(-2)+2}{3}=\frac{5}{3},\quad d=\frac{4+4-6}{3}=\frac{2}{3}$

$c+2d=\frac{5}{3}+\frac{4}{3}=3$

$\boxed{3}$

For $f(x)=\log_{4}\big(\log_{3}\big(\log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)\big)\big)$

Condition:
$\log_{3}(\cdot)>0 \Rightarrow \log_{7}\big(8-\log_{2}(x^{2}+4x+5)\big)>1$

$\Rightarrow 8-\log_{2}(x^{2}+4x+5)>7$
$\Rightarrow \log_{2}(x^{2}+4x+5)<1$
$\Rightarrow x^{2}+4x+5<2$
$\Rightarrow (x+2)^2<1$
$\Rightarrow -3$
So $(\alpha,\beta)=(-3,-1)$

For $g(x)=\sin^{-1}\left(\frac{7x+10}{x-2}\right)$

Condition: $-1\le \frac{7x+10}{x-2}\le 1$

Solve:
$\frac{7x+10}{x-2}\le 1 \Rightarrow \frac{6x+12}{x-2}\le 0 \Rightarrow \frac{x+2}{x-2}\le 0$
$\Rightarrow -2\le x<2$

$\frac{7x+10}{x-2}\ge -1 \Rightarrow \frac{8x+8}{x-2}\ge 0 \Rightarrow \frac{x+1}{x-2}\ge 0$
$\Rightarrow x\le -1 \text{ or } x>2$

Intersection: $[-2,-1]$
So $(\gamma,\delta)=(-2,-1)$

Now:
$\alpha^2+\beta^2+\gamma^2+\delta^2=9+1+4+1=15$

$\boxed{15}$