WBJEE 2026 Paper 1 (Mathematics) Question Paper with Solutions

${𝑃}^{\prime }\left(𝑥\right)=4{𝑥}^3+3𝑎{𝑥}^2+2𝑏𝑥+𝑐$

$\begin{array}{cc}& {𝑃}^{\prime }\left(𝑎\right)=0\Rightarrow 𝐶=0\\ & {𝑃}^{\prime }\left(𝑥\right)=𝑥(4{𝑥}^2+3𝑎𝑥+2𝑏)\end{array}$

Roots of $4{𝑥}^2+3𝑎𝑥+2𝑏=0$ are imaginary

$\begin{array}{cc}& \Rightarrow 9{𝑎}^2-32𝑏<0\dots \dots \left(1\right)\Rightarrow 𝑏>0\\ & P\left(x\right)=x^4+𝑎{𝑥}^3+𝑏{𝑥}^2+𝑑\\ & P(-1)<P\left(1\right)\Rightarrow a>0\\ & P^{\prime }\left(x\right)>\forall x\in (0,1),P^{\prime }\left(x\right)<0\forall x\in [-1,0]\\ & x=0\text{ is minima }\end{array}$

$𝛼+𝛽={𝑝}_1,𝛼𝛽=𝑞$

$(\sqrt{\surd 𝛼}+\sqrt{\surd 𝛽}{)}^2=𝛼+𝛽+2\sqrt{\surd 𝛼𝛽}=𝑝+2\sqrt{\surd 𝑞}=1\sqrt{\surd 𝛼}+\sqrt{\surd 𝛽}=\sqrt{\surd 𝑝+2\sqrt{\surd 𝑞}}$

Let $𝑥={𝛼}^{1/4}+{𝛽}^{1/4}$

${𝑥}^2=\sqrt{\surd 𝛼}+\sqrt{\surd 𝛽}+2(𝛼𝛽{)}^{1/4}<text>=</text><text>=</text>\sqrt{\surd 𝑃+2\sqrt{\surd 𝑞}}+2{𝑞}^{1/4}$

Simplifying we get $𝑘=\frac{1}{4}$

$\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2r^2}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{2}{4r^2}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)$

$\begin{array}{cc}& \sum _{𝑟=1}^{\infty }({\tan }^{-1}(2𝑟+1)-{\tan }^{-1}(2𝑟-1\left)\right)=\frac{𝜋}{4}\\ & 𝑎=\frac{𝜋}{4}\Rightarrow \tan 𝑎=1\end{array}$

$𝑓\left(𝑥\right)=𝑔\left(𝑥\right)\cdot (𝑥-𝑎{)}^2$

$\underset{𝑥\to 𝑎}{lim}\frac{(2𝜆-1)𝑔\left(𝑥\right)+𝜆(𝑥-𝑎){𝑔}^{\prime }\left(𝑥\right)}{𝑔\left(𝑥\right)\cdot (𝑥-𝑎)}=𝑚$

for $𝑚\ne 0\Rightarrow 2𝜆-1=0,𝜆=\frac{1}{2}$

$\begin{array}{cc}& \text{ Hint: }|\bar{P}{|}^2=f^2(t)+g^2(t)+1=h(t)\\ & \begin{array}{cc}& {ℎ}^{\prime }\left(𝑡\right)=2𝑓\left(𝑡\right)\cdot {𝑓}^{\prime }\left(𝑡\right)+2𝑔\left(𝑡\right)\cdot {𝑔}^{\prime }\left(𝑡\right)\\ & \overrightarrow{𝑃}=𝜆\overline{𝑄}\Rightarrow 𝑓\left(𝑡\right)=-𝜆{𝑓}^{\prime \prime }\left(𝑡\right)...\left(𝑖\right)\end{array}\\ & 𝑔\left(𝑡\right)=𝜆{𝑓}^{\prime }\left(𝑡\right)...\left(𝑖𝑖\right)\\ & \begin{array}{cc}& 1=𝜆\\ & \Rightarrow h^{\prime }\left(t\right)=0\\ & \Rightarrow h\left(t\right)\text{ is constant }\end{array}\end{array}$

$\sum _{10}^{𝑟=0}(\cos \frac{2𝜋𝑟}{11}+𝑖\sin \frac{2𝜋}{11})=0$

$\begin{array}{cc}& 1+\sum _{𝑟=1}^{10}(\cos \frac{2𝜋𝑟}{11}+𝑖\sin \frac{2𝜋𝑟}{11})=0\\ & =\sum _{𝑘=1}^{32}(3𝑘+2)(-𝑖{)}^{𝑘}\cdot (-1{)}^{𝑘}=\sum _{𝑘=1}^{32}{𝑖}^{𝑘}\cdot (3𝑘+2)=48(1-𝑖)\end{array}$

Let $𝑚={𝑥}^2+{𝑦}^2$

$\begin{array}{cc}𝑚{\cos }^3𝜃& =𝑐{\cos }^3𝜃+𝑦{\sin }^3𝜃\\ 𝑚{\cos }^3𝜃& =𝑦\cos 3𝜃-𝑥\sin 3𝜃\end{array}$

Solving $𝑥=𝑚({\cos }^3𝜃\cos 3𝜃-{\sin }^3𝜃\sin 3𝜃)$

$\begin{array}{cc}& 𝑦=𝑚({\cos }^2𝜃\sin 3𝜃+{\sin }^3𝜃\cos 3𝜃)\\ & \Rightarrow \cos 4𝜃=\frac{4𝑥-𝑚}{3𝑚}...\left(𝑖\right)\\ & 𝑦=𝑚[\left(\frac{\cos 3𝜃+3\cos 𝜃}{4}\right)\sin 3𝜃+\left(\frac{3\sin 𝜃-\sin 3𝜃}{4}\right)\cos 3𝜃]=\frac{3𝑚\sin 4𝜃}{4}\\ & \Rightarrow \sin 4𝜃=\frac{4𝑦}{3𝑚}\Rightarrow ({𝑥}^2+{𝑦}^2+2𝑥)({𝑥}^2+{𝑦}^2-𝑥)=2{𝑦}^2\end{array}$

Sum of the coefficients of the given quadratic equation is equal to zero. Hence $𝑥=1$ is a root of the equation.

$\begin{array}{cc}& {𝑡}_{𝑝𝑞}={𝑡}_{𝑝}+(𝑝𝑞-𝑟)𝑑\\ & =\frac{1}{𝑞}+𝑝(𝑞-1)\left(\frac{\frac{1}{𝑝}-\frac{1}{𝑞}}{𝑞-𝑝}\right)=\frac{1}{𝑞}+\frac{𝑞-1}{𝑞}=1\end{array}$

Hence option A is correct

$\{1,2,3,\dots \dots ..13\}$

$\begin{array}{cc}& 1\le 𝑑\le 6\\ & 𝐴𝑃=1+3+5+7+9+11=36\\ & \text{ Required probability }=\frac{36}{{}^{13}{𝐶}_3}=\frac{18}{143}\end{array}$

$\begin{array}{cc}& \text{ Hint }:𝑓\left(𝑥\right)=\frac{1+𝑥}{1-𝑥}=(1+𝑥)(1-𝑥{)}^{-1}=(1+𝑥)(1+𝑥+{𝑥}^2+{𝑥}^3+\dots \dots )\\ & 𝑓\left(𝑥\right)=(1+𝑥+{𝑥}^2+{𝑥}^3+\dots \dots )+(𝑥+{𝑥}^2+{𝑥}^3+\dots .)\\ & 𝑓\left(𝑥\right)=1+2𝑥+2{𝑥}^2+2{𝑥}^3+\dots \dots ..\\ & 𝑓\left(𝐴\right)=𝐼+2𝐴+2{𝐴}^2,(𝑎𝑠{𝐴}^3=0)\end{array}$

$𝑓\left(𝑥\right)>0,𝑔\left(𝑥\right)>0,{𝑓}^{\prime }\left(𝑥\right)<0,{𝑔}^{\prime }\left(𝑥\right)>0$

Let $ℎ\left(𝑥\right)=\frac{𝑓\left(𝑥\right)}{𝑔\left(𝑥\right)},{ℎ}^{\prime }\left(𝑥\right)=\frac{{𝑓}^{\prime }\left(𝑥\right)𝑔\left(𝑥\right)-{𝑔}^{\prime }\left(𝑥\right)𝑓\left(𝑥\right)}{{𝑔}^2\left(𝑥\right)}<0$

$\Rightarrow ℎ\left(𝑥\right)$ is decreasing

Let $\sin 𝛼=a,\sin 𝛽=b,\sin 𝛾=c$

$\frac{1}{𝑥-𝑎}+\frac{1}{𝑥-𝑏}+\frac{1}{𝑥-𝑐}=\frac{(𝑥-𝑏)(𝑥-𝑐)+(𝑥-𝑎)(𝑥-𝑐)+(𝑥-𝑎)(𝑥-𝑏)}{(𝑥-𝑎)(𝑥-𝑏)(𝑥-𝑐)}$

Let $𝑓\left(𝑥\right)=(𝑥-𝑏)(𝑥-𝑐)+(𝑥-𝑎)(𝑥-𝑐)+(𝑥-𝑎)(𝑥-𝑏)$

$f\left(a\right)>0$

$f\left(b\right)<0$

$f\left(b\right)>0$

$\Rightarrow 2$ real and unequal roots

(A) $|𝑥-𝑦|<2$ is symmetric

(B) $\left|𝑥\right|\ge 𝑦$ is not transitive

$[\because |b|\ge a$ and $\left|c\right|\ge b$ together do not confirm $\left|c\right|\ge a]$

(C) Correct

(D) Not symmetric

[ $a-b<2$ does not confirm $b-a<2$ ]

$\int \frac{{\mathrm{cosec}}^{2}x-2010}{{\cos }^{2010}x}\mathrm{dx}$

$\begin{array}{cc}& =\int \underset{2}{\underset{}{{\mathrm{cosec}}^2𝑥}}\cdot \underset{1}{\underset{}{{\sec }^{2010}𝑥}}𝑑𝑥-2010\int {\sec }^{2010}𝑥𝑑𝑥\\ & =-{\sec }^{2010}𝑥\cot 𝑥+\int 2010\cdot {\sec }^{2009}𝑥\cdot \sec 𝑥\tan 𝑥\cot 𝑥𝑑𝑥-2010\int {\sec }^{2010}𝑥𝑑𝑥\\ & =-{\sec }^{2010}𝑥\cot 𝑥+𝑐\\ & =-\frac{\cot 𝑥}{{\cos }^{2010}𝑥}+𝑐,𝑓\left(𝑥\right)=\cot 𝑥\\ & 𝑔\left(𝑥\right)=\cos 𝑥\end{array}$

$\frac{𝑓\left(𝑥\right)}{𝑔\left(𝑥\right)}=\left\{𝑥\right\}\Rightarrow \mathrm{cosec}𝑥=\left\{𝑥\right\}$ No solution.

Let the midpoint of the normal chord $𝐴𝐵$ be $(ℎ,𝑘)$.

For the parabola ${𝑦}^2=4𝑥$, the chord with midpoint $(ℎ,𝑘)$ is given by $𝑇={𝑆}_1$.

So, the equation of chord $𝐴𝐵$ is $𝑘𝑦-2(𝑥+ℎ)={𝑘}^2-4ℎ$

$\Rightarrow 2𝑥-𝑘𝑦-2ℎ+{𝑘}^2=0...\left(𝑖\right)$

Since this chord is a normal chord, its equation must also be the equation of a normal to the parabola.

The equation of normal to ${𝑦}^2=4𝑥$ in parametric form is $𝑦=-𝑡𝑥+2𝑡+{𝑡}^3$.

$\Rightarrow 𝑡𝑥+𝑦-2𝑡-{𝑡}^3=0...\left(𝑖𝑖\right)$

Now both equations represent the same line, so their coefficients are proportional.

Equating (1), (2) $\frac{𝑡}{2}=\frac{1}{-𝑘}=\frac{-2𝑡-{𝑡}^3}{-2ℎ+{𝑘}^2}\Rightarrow 𝑡=\frac{-2}{𝑘},\frac{1}{2}=\frac{2+{𝑡}^2}{2ℎ-{𝑘}^2}$

$\begin{array}{cc}& \Rightarrow 2ℎ-{𝑘}^2=4+2{𝑡}^2\\ & \Rightarrow 2ℎ-{𝑘}^2=4+2\cdot \frac{4}{{𝑘}^2}\\ & \Rightarrow 2ℎ-{𝑘}^2-4-\frac{8}{{𝑘}^2}=0\\ & \Rightarrow 2ℎ-4=\frac{8}{{𝑘}^2}+{𝑘}^2\end{array}$

Now divide by $2$:

$ℎ-2=\frac{4}{{𝑘}^2}+\frac{{𝑘}^2}{2}$

Replacing $(ℎ,𝑘)$ by $(𝑥,𝑦)$, the locus becomes

so locus is $x-2=\frac{4}{y^2}+\frac{y^2}{2}𝜇=4,𝜈=2,𝜆=2𝜆+𝜇+𝜈=8$

$1+0\le 1+{\sin }^2𝜃\le 1+1$

$\begin{array}{cc}& \frac{1}{2}\le \frac{1}{1+{\sin }^2𝜃}\le 1\\ & {\sin }^{-1}\frac{1}{2}\le {\sin }^{-1}\left(\frac{1}{1+{\sin }^2𝜃}\right)\le {\sin }^{-1}1\\ & \because {\sin }^{-1}x\text{ is increasing function }\\ & \Rightarrow \frac{𝜋}{6}\le \frac{k𝜋}{6}\le \frac{𝜋}{2}\\ & \Rightarrow 1\le k\le 3\end{array}$

Number of function $=n^n$

Number of injective function $=𝑛!$

$\begin{array}{cc}& \therefore \frac{𝑛!}{{𝑛}^{𝑛}}=\frac{3}{32}\\ & \text{ for }𝑛=4\frac{𝑛!}{{𝑛}^{𝑛}}=\frac{24}{4^4}=\frac{24}{256}=\frac{3}{32}\end{array}$

$𝑓\left(𝑥\right)=2{𝑥}^3-3{𝑥}^2+6$

${𝑓}^{\prime }\left(𝑥\right)=6{𝑥}^2-6𝑥=6𝑥(𝑥-1)$

$\therefore f\left(x\right)$ is monotonically increasing in $x\in [1,\infty )$

$\therefore f\left(x\right)$ is invertible in $x\in [1,\infty )$

$\therefore 𝛼=1𝑓\left(1\right)=2-3+6=5$

∴ for $𝑥\in [1,\infty )𝑓\left(𝑥\right)\in [5,\infty )$

∴ Range $=$ Codomain of $𝑓\left(𝑥\right)$ is $[5,\infty )$

$\begin{array}{cc}& 𝐴={\mathrm{Lt}}_{𝑛\to 𝛼}{\cos }^{2𝑛}𝑥,𝑥=𝑛𝜋\\ & 𝑥=𝑛𝜋\therefore \cos 𝑥=\pm 1\therefore {\cos }^{2𝑛}𝑥=1\therefore 𝑎=1\\ & \text{ if }𝑥\ne 𝑛𝜋\cos 𝑥\in [-1,1)\therefore {\cos }^{2𝑛}𝑥\in [0,1)\\ & 𝑛\to 𝛼{\cos }^{2𝑛}𝑥=0\\ & \therefore 𝑎=1,𝑏=0\end{array}$

Area of the triangle formed by $(1,0),(-2,1),(2,1)$

$=\frac{1}{2}\times 4\times 1=2$

$\begin{array}{cc}& \begin{array}{cc}& \therefore |\overline{𝑎}\times \overline{𝑏}|=\mid 𝑎𝑏\sin \frac{𝜋}{2}\mid =\left|𝑎𝑏\right|=\left|\overline{𝑎}\right|\cdot \left|\overline{𝑏}\right|\\ & \text{ given }16𝑎𝑏=3(𝑎+𝑏{)}^2\\ & 3{𝑎}^2-10𝑎𝑏+3{𝑏}^2=0\\ & (𝑎-3𝑏)(3𝑎-𝑏)=0\\ & \therefore 𝑎=3𝜆𝑖,𝑏=𝜆𝑗\end{array}\\ & \text{ We can take }𝑎=3𝑖,𝑏=𝑗\text{ angle will be same. }\\ & \begin{array}{cc}& \therefore \overline{\mathrm{OC}}=3𝑖+𝑗\overline{\mathrm{AB}}=3𝑖-𝑗\\ & \therefore \cos 𝜃=\frac{(3𝑖-𝑗)\cdot (3𝑖+𝑗)}{\sqrt{\surd 10}\cdot \sqrt{\surd 10}}=\frac{4}{5}\end{array}\end{array}$

$\frac{1-{\tan }^2\frac{𝜃}{2}}{1+{\tan }^2\frac{𝜃}{2}}=\frac{4}{5}\Rightarrow {\tan }^2\frac{𝜃}{2}=\frac{1}{9}\tan \frac{𝜃}{2}=\frac{1}{3}$

${𝐿}_1:\overline{𝑟}\times \overline{𝑎}=\overline{𝑏}\times \overline{𝑎}$

$\begin{array}{cc}& (\overline{𝑟}-\overline{𝑏})\times \overline{𝑎}=0\\ & \overline{𝑟}=\overline{𝑏}+𝜆\overline{𝑎}=(2𝑖-𝑘)+𝜆(𝑖+𝑗)\\ & {𝐿}_2:\overline{𝑟}\times \overline{𝑏}=\overline{𝑎}\times \overline{𝑏}\\ & (\overline{𝑟}-\overline{𝑎})\times \overline{𝑏}=0\\ & \overline{𝑟}=\overline{𝑎}+𝜇\overline{𝑏}\\ & =(𝑖+𝑗)+𝜇(2𝑖-𝑘)\\ & {𝐿}_1:\frac{𝑥-1}{2}=\frac{𝑦-0}{1}=\frac{𝑧-1}{0}\\ & {𝐿}_2:\frac{𝑥-2}{1}=\frac{𝑦-1}{0}=\frac{𝑧}{-1}\end{array}$

∴ Point of intersection is $(𝛼,1,-1)$

from ${𝐿}_1:\frac{𝑥-2}{1}=\frac{1-0}{1}$

$𝑥=3\Rightarrow 𝛼=3$

∴ Point of intersection is $(3,1,-1)$ or $3i+j-k$

${𝑎}_1+{𝑎}_1{𝑟}^{𝑛-1}=66$

$\begin{array}{cc}& {𝑎}_2\cdot {𝑎}_{𝑛-1}={𝑎}_1𝑟\cdot {𝑎}_1{𝑟}^{𝑛-2}=128\\ & {𝑎}_2^1{𝑟}^{𝑛-1}=128\\ & {𝑎}_1{𝑟}^{𝑛-1}=\frac{128}{{𝑎}_1}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& {𝑎}_1+\frac{128}{{𝑎}_1}=66\\ & {𝑎}_2^1-66{𝑎}_1+128=0\\ & \Rightarrow {𝑎}_1=2,64\end{array}\\ & \text{ As it is increasing G.P. }\\ & \begin{array}{cc}& \therefore {𝑎}_1=2\therefore 2+2{𝑟}^{𝑛-1}=66\therefore {𝑟}^{𝑛-1}=32\\ & {𝑎}_1+{𝑎}_2+\dots +{𝑎}_{𝑛}\\ & ={𝑎}_1\cdot \frac{{𝑟}^{𝑛}-1}{𝑟-1}=126\\ & \Rightarrow {𝑟}^{𝑛}-1=63(𝑟-1)\\ & 32𝑟-1=63𝑟-63\\ & \Rightarrow 31𝑟=62\Rightarrow 𝑟=2\\ & \therefore {𝑎}_1=2,𝑟=2\text{ and }{𝑟}^{𝑛-1}=322^{𝑛-1}=32\Rightarrow 𝑛=6\end{array}\end{array}$

The tangent to the ellipse $\frac{{𝑥}^2}{4}+\frac{{𝑦}^2}{9}=1$ in parametric form is

$\begin{array}{cc}& \frac{𝑥\cos 𝜃}{2}+\frac{𝑦\sin 𝜃}{3}=1\\ & 3𝑥\cos 𝜃+2𝑦\sin 𝜃=6\end{array}$

This tangent cuts the circle ${𝑥}^2+{𝑦}^2=25$ and forms a chord.

To get the shortest intercept on the circle, we need the shortest chord made by this tangent.

For a circle, the chord is shortest when its distance from the centre is greatest.

The distance of the line $3𝑥\cos 𝜃+2𝑦\sin 𝜃=6$ from the origin is

$𝑑=\frac{6}{\sqrt{\surd 9{\cos }^2𝜃+4{\sin }^2𝜃}}=\frac{6}{\sqrt{\surd 4+5{\cos }^2𝜃}}$

For $𝑑$ to be maximum, the denominator must be minimum.

The smallest value of $4+5{\cos }^2𝜃$ is $4$, when $\cos 𝜃=0$.

So,

${𝑑}_{max}=\frac{6}{\sqrt{\surd 4}}=3$

Now the radius of the circle is $5$.

If the distance of the chord from the centre is $𝑑=3$, then half of the chord length is

$𝐴𝑀=\sqrt{\surd 25-3^2}=\sqrt{\surd 16}=4$

So the full chord length is

$𝐴𝐵=2𝐴𝑀=2\times 4=8$

The minimum length of the intercept is $8$.

$\begin{array}{cc}& \overrightarrow{𝑟}=𝑥\widehat{𝑖}+𝑦\widehat{𝑗}+𝑧\widehat{𝑘}\\ & \overrightarrow{𝑛}=𝑎\widehat{𝑖}+𝑏\widehat{𝑗}+𝑐\widehat{𝑘}\end{array}$

∴ Equation of the plane

$\begin{array}{cc}& 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑\\ & \frac{𝑥}{𝑑/𝑎}+\frac{𝑦}{𝑑/𝑏}+\frac{𝑧}{𝑑/𝑐}=1\end{array}$

Now $\widehat{𝑖}\cdot \overrightarrow{𝑛}=𝑎,\widehat{𝑗}\cdot \overrightarrow{𝑛}=𝑏,\widehat{𝑘}\cdot \overrightarrow{𝑛}=𝑐$

$\begin{array}{cc}\therefore 𝑥\text{ intercept }& =\frac{𝑑}{\widehat{𝑖}\cdot \overrightarrow{𝑛}}\\ 𝑦\text{ intercept }& =\frac{𝑑}{\widehat{𝑗}\cdot \overrightarrow{𝑛}}\\ 𝑧\text{ intercept }& =\frac{𝑑}{\widehat{𝑘}\cdot \overrightarrow{𝑛}}\end{array}$