UP PGT 2026 Mathematics Previous Year Question Paper (May 10, 2026) Questions

For the standard basis vectors, $T(1,0)=(2,1)$ and $T(0,1)=(-3,1)$.

These vectors form the columns of the matrix. Therefore, $[T]=\begin{bmatrix}2&-3\\1&1\end{bmatrix}$.

Statement a is true. If $G=\langle g\rangle$ is cyclic, then every pair of elements can be written as $g^m$ and $g^n$. Hence, $g^mg^n=g^{m+n}=g^{n+m}=g^ng^m$. Therefore, every cyclic group is abelian.

Statement b is true. In the Klein $4$-group, every non-identity element has order $2$. Therefore, no element generates all $4$ elements, so the group is not cyclic.

Statement c is false. A cyclic group of order $n$ has $\varphi(n)$ generators, where $\varphi$ is Euler’s totient function. This number is not always $2$.

Statement d is true. Every subgroup of a cyclic group is itself cyclic.

Thus, statements a, b and d are correct.

here are $7$ invited people and $1$ host.

Therefore, the total number of people is $7+1=8$.

The number of ways to arrange $n$ distinct people around a circular table is $(n-1)!$.

Hence, the required number of arrangements is $(8-1)!=7!=5040$.

Given $\vec a+\vec b+\vec c=\vec0$, take the dot product of the vector with itself:

$(\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)=0$

Expanding:

$|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0$

Since $m=\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a$:

$|\vec a|^2+|\vec b|^2+|\vec c|^2+2m=0$

Thus, $m=-\dfrac12\left(|\vec a|^2+|\vec b|^2+|\vec c|^2\right)$.

Because all three vectors are non-zero, the sum of their squared magnitudes is positive. Therefore, m is negative.

The expression inside the modulus becomes zero at $x=-1$.

Thus, $|x+1|=-(x+1)$ for $-2\leq x

Since the forces act in the same direction, their resultant is their sum:

$\dfrac{P}{Q}+\dfrac{Q}{P}=2$

Multiplying by $PQ$:

$P^2+Q^2=2PQ$

Therefore, $P^2-2PQ+Q^2=0$, so $(P-Q)^2=0$.

Hence, $P=Q$.

The $r$th term of the series is

$a_r=10^r+(2r-1)$.

Therefore,

$S_n=\sum_{r=1}^n10^r+\sum_{r=1}^n(2r-1)$.

Now, $\sum_{r=1}^n10^r=\dfrac{10(10^n-1)}{9}$, while the sum of the first $n$ odd numbers is $\sum_{r=1}^n(2r-1)=n^2$.

Hence, $S_n=\dfrac{10}{9}(10^n-1)+n^2$.

Correct option: (C) $\dfrac{10}{9}(10^n-1)+n^2$

On rotating the upper half of the cardioid about the initial line,

$V=2\pi\iint_R y,dA$.

In polar coordinates, $y=r\sin\theta$ and $dA=r,dr,d\theta$. Therefore,

$V=2\pi\int_0^\pi\int_0^{a(1+\cos\theta)}r^2\sin\theta,dr,d\theta$

$=\dfrac{2\pi a^3}{3}\int_0^\pi(1+\cos\theta)^3\sin\theta,d\theta$.

Since $\int_0^\pi(1+\cos\theta)^3\sin\theta,d\theta=4$,

$V=\dfrac{8\pi a^3}{3}$.

The general term is


$T_{r+1}=\binom{10}{r}\left(\dfrac{x}{2}\right)^{10-r}\left(-\dfrac{3}{x^2}\right)^r$.


The power of $x$ in this term is $10-r-2r=10-3r$.


For the term containing $x^4$, $10-3r=4$, giving $r=2$.


Therefore, the required coefficient is


$\binom{10}{2}\left(\dfrac{1}{2}\right)^8(-3)^2=\dfrac{45\cdot9}{256}=\dfrac{405}{256}$.

The characteristic equation is


$\left|\begin{matrix}1-\lambda&3\\0&4-\lambda\end{matrix}\right|=0$.


Thus, $(1-\lambda)(4-\lambda)=0$.


Hence, $\lambda=1,4$.

First, equate the equations to determine their points of intersection:

$2x-x^2=-x$

$\Rightarrow 3x-x^2=0$

$\Rightarrow x(3-x)=0$

Therefore, $x=0$ or $x=3$.

For $0\leq x\leq3$, the curve $y=2x-x^2$ lies above the line $y=-x$.

Therefore, the required area is:

$A=\int_0^3\left[(2x-x^2)-(-x)\right],dx$

$A=\int_0^3(3x-x^2),dx$

$A=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3$

$A=\left(\frac{3(3)^2}{2}-\frac{(3)^3}{3}\right)-0$

$A=\frac{27}{2}-9$

$A=\frac{27-18}{2}$

$A=\frac{9}{2}$

Hence, the area bounded by the given curve and line is $\frac{9}{2}$ square units.

Let the roots be $\alpha$ and $\beta$. By Vieta’s formulas, $\alpha+\beta=a-2$ and $\alpha\beta=-a-1$.

Now, $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$.

Thus, $\alpha^2+\beta^2=(a-2)^2-2(-a-1)=a^2-2a+6=(a-1)^2+5$.

Since $(a-1)^2\geq0$, the sum is minimum when $(a-1)^2=0$. Hence, $a=1$.

The minimum value of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$.

Therefore, $\min(3\cos x+4\sin x)=-\sqrt{3^2+4^2}=-5$.

Hence, $\min(3\cos x+4\sin x+8)=-5+8=3$.

Using the standard identities:

$\sinh^{-1}x=\log\left(x+\sqrt{x^2+1}\right)$, therefore $a-\mathrm{IV}$.

$\cosh^{-1}x=\log\left(x+\sqrt{x^2-1}\right)$, therefore $b-\mathrm{III}$.

$\tanh^{-1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$, therefore $c-\mathrm{II}$.

$\coth^{-1}x=\frac{1}{2}\log\left(\frac{x+1}{x-1}\right)$, therefore $d-\mathrm{I}$.

Let $T(x,y,z)=(u,v,w)$. Then $u=3x$, $v=x-y$ and $w=2x+y+z$.

From $u=3x$, we get $x=\frac{u}{3}$.

From $v=x-y$, we get $y=\frac{u}{3}-v$.

From $w=2x+y+z$, we get $z=w-u+v$.

Therefore, $T^{-1}(u,v,w)=\left(\frac{u}{3},\frac{u}{3}-v,w-u+v\right)$.