Maths 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, $[T]=\begin{bmatrix}2&-3\\1&1\end{bmatrix}$.

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.

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

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

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

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

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, $\lambda=1,4$.