UP TGT 2026 Mathematics Previous Year Question Paper (Jun 03, 2026) Questions

Fisher's Ideal Index Number is the geometric mean of Laspeyres' and Paasche's index numbers.

Its formula is:

$P_F=\sqrt{P_L\times P_P}$

Where:

$P_F$ = Fisher's Ideal Price Index

$P_L$ = Laspeyres' Price Index

$P_P$ = Paasche's Price Index

Therefore:

$\boxed{P_F=\sqrt{P_LP_P}}$

Hence, option (C) is correct.

The number of arrangements of 5 different books is calculated using permutations.

The number of ways to arrange $n$ distinct objects is:

$n!$

Here:

$n=5$

Therefore:

$5!=5\times4\times3\times2\times1$

$\boxed{5!=120}$

Hence, option (D) is correct.

The given polynomial is:

$p(x)=x^2-5x+6$

By Vieta's formulas:

$\alpha+\beta=5$

$\alpha\beta=6$

Using the identity:

$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$

Substituting the values:

$\alpha^2+\beta^2=5^2-2(6)$

$=25-12$

$\boxed{\alpha^2+\beta^2=13}$

Hence, option (C) is correct.

The given expression is:

$\frac{1}{\sqrt{5}-\sqrt{3}}$

A rationalising factor is a factor that gives a rational number when multiplied by the given expression.

Multiplying by $\sqrt{5}-\sqrt{3}$:

$\frac{1}{\sqrt{5}-\sqrt{3}}\times(\sqrt{5}-\sqrt{3})=1$

Since 1 is a rational number and $\sqrt{5}-\sqrt{3}$ is positive, the required factor is:

$\boxed{\sqrt{5}-\sqrt{3}}$

Hence, option (B) is correct.

In the additive group of integers, the subgroup generated by 6 and 9 is:

$\langle6,9\rangle=\{6m+9n:m,n\in\mathbb{Z}\}$

This subgroup is generated by the greatest common divisor of 6 and 9.

$\langle6,9\rangle=\gcd(6,9)\mathbb{Z}$

Since:

$\gcd(6,9)=3$

Therefore:

$\boxed{\langle6,9\rangle=3\mathbb{Z}}$

Hence, option (C) is correct.

The given sequence is:

$a_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$

It represents the partial sums of the harmonic series.

$a_n=\sum_{k=1}^{n}\frac{1}{k}$

The harmonic series is divergent.

$\sum_{k=1}^{\infty}\frac{1}{k}=\infty$

Therefore:

$\lim_{n\to\infty}a_n=\infty$

Every Cauchy sequence in $\mathbb{R}$ is convergent.

Since the given sequence diverges, it cannot be Cauchy.

$\boxed{\text{Not Cauchy and divergent in }\mathbb{R}}$

Hence, option (B) is correct.

Let:

$f(x)=x^3-3x^2+ax-6$

According to the Factor Theorem, if $(x-2)$ is a factor, then:

$f(2)=0$

$2^3-3(2)^2+2a-6=0$

$8-12+2a-6=0$

$2a-10=0$

$\boxed{a=5}$

Hence, option (A) is correct.

Pipe A fills $\frac{1}{10}$ of the tank in one hour.

Pipe B fills $\frac{1}{15}$ of the tank in one hour.

Their combined filling rate is:

$\frac{1}{10}+\frac{1}{15}=\frac{1}{6}$

Tank filled in 3 hours:

$3\times\frac{1}{6}=\frac{1}{2}$

Remaining part of the tank:

$1-\frac{1}{2}=\frac{1}{2}$

Time taken by pipe B to fill the remaining part:

$T=\frac{\frac{1}{2}}{\frac{1}{15}}$

$\boxed{T=7.5\text{ hours}}$

Hence, option (B) is correct.

The given cubic equation is:

$x^3-6x^2+11x-6=0$

By Vieta's formula, the sum of the roots is:

$\alpha+\beta+\gamma=-\frac{b}{a}$

Here, $a=1$ and $b=-6$.

$\alpha+\beta+\gamma=-\frac{-6}{1}$

$\boxed{\alpha+\beta+\gamma=6}$

Hence, option (D) is correct.

$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}$

$\frac{1}{x^2-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$

$\int\frac{1}{x^2-1}\,dx=\frac{1}{2}\ln|x-1|-\frac{1}{2}\ln|x+1|+C$

$\boxed{\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C}$

By the forward difference definition, the derivative of $f(x)$ at $x=a$ is

$f'(a)=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$.

Here, $f(a+h)-f(a)$ represents the forward change in the function, and $h$ represents the corresponding change in $x$.

Hence, the correct answer is A.

The element $2$ is located in the first row and first column. Therefore, its cofactor is

$C_{11}=(-1)^{1+1}M_{11}$

After deleting the first row and first column, the remaining minor is

$M_{11}=5$

Therefore,

$C_{11}=(-1)^2(5)=5$

Hence, the correct answer is 5.

The absolute value of the scalar triple product of three vectors gives the volume of the parallelepiped formed by them.

$V=\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|$

Therefore, $\vec{a}\cdot(\vec{b}\times\vec{c})$ represents the signed volume, while its absolute value represents the actual volume.

Hence, the correct answer is $\boxed{\text{(C) Volume of a parallelepiped}}$.

The given vector is $\vec{a}=\hat{i}+\hat{j}+\hat{k}$.

Its magnitude is

$|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$.

The direction cosines are

$\left(\dfrac{1}{|\vec{a}|},\dfrac{1}{|\vec{a}|},\dfrac{1}{|\vec{a}|}\right)
=\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)$.

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