Mathematics

Advanced Algebra and Calculus

135 Questions

Advanced algebra and calculus topics cover matrices, complex numbers, infinite geometric series, and differential equations. These mathematical concepts frequently appear in officer-level aptitude tests. Solving these questions builds a strong foundation for advanced problem solving.

Complex numbersMatrix operationsInfinite geometric seriesDifferential calculusAlgebraic identities

Advanced Algebra and Calculus Questions

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $\displaystyle A=\left[ \begin{matrix} \cos { \theta  }  & \sin { \theta  }  \ -\sin { \theta  }  & \cos { \theta  }  \end{matrix} \right] $, then $\displaystyle \underset { n\rightarrow \infty  }{ \lim } \frac { 1 }{ n } { A }^{ n }$ is?

  1. A null matrix

  2. An identity matrix
  3. $\displaystyle \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] $

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix}\lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n } { A }^{ n } } $
${ A }^{ n }={ \begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix} }^{ n }$
${ A }^{ n }={ \begin{bmatrix} \cos { n\theta  }  & \sin { n\theta  }  \\ -\sin { n\theta  }  & \cos { n\theta  }  \end{bmatrix} }$
Now,
$\lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n } { A }^{ n } } =\lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n }  } { \begin{bmatrix} \cos { n\theta  }  & \sin { n\theta  }  \\ -\sin { n\theta  }  & \cos { n\theta  }  \end{bmatrix} }$
$=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}=$Null matrix
Proof for ${ A }^{ n }={ \begin{bmatrix} \cos { n\theta  }  & \sin { n\theta  }  \\ -\sin { n\theta  }  & \cos { n\theta  }  \end{bmatrix} }=P\left( n \right) $
$P\left( n \right) $is true for $n=1$
For $n=k,k\ge 1$
${ A }^{ k }={ \begin{bmatrix} \cos { k\theta  }  & \sin { k\theta  }  \\ -\sin { k\theta  }  & \cos { k\theta  }  \end{bmatrix} }$
${ A }^{ k+1 }={ A }^{ k }A$
$={ \begin{bmatrix} \cos { k\theta  }  & \sin { k\theta  }  \\ -\sin { k\theta  }  & \cos { k\theta  }  \end{bmatrix} }{ \begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix} }$
$\Rightarrow { \begin{bmatrix} \cos { k\theta  }  & \sin { k\theta  }  \\ -\sin { k\theta  }  & \cos { k\theta  }  \end{bmatrix} }$
$\therefore P\left( n \right) $ is true for $n=k+1\left( k\ge 1 \right) $
Multiple choice maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If A is invertible, then which of the following is not true?

  1. $\displaystyle { A }^{ -1 }={ \left| A \right| }^{ -1 }$
  2. $\displaystyle { \left( { A }^{ 2 } \right) }^{ -1 }={ \left( { A }^{ -1 } \right) }^{ 2 }$
  3. $\displaystyle { \left( { A }^{ ' } \right) }^{ -1 }={ \left( { A }^{ -1 } \right) }^{ ' }$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A is invertible
$\Rightarrow { A }^{ -1 }$ exists
Option A: ${ A }^{ -1 }={ \left| A \right|  }^{ -1 }$
But we cannot write that a matrix and its determinant are both equal
$\therefore $ option A is not true
Option B: ${ \left( { A }^{ 2 } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ 2 }$
This option is true from the property
${ \left( { A }^{ n } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ 2 }$
Option C: ${ \left( { A }^{ -1 } \right)  }^{ 1 }={ \left( { A }^{ 1 } \right)  }^{ -1 }$
Consider $\left( { A }^{ T } \right) { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 }A \right)  }^{ T }={ I }^{ T }=I$
Similarly
${ \left( { A }^{ -1 } \right)  }^{ T }{ \left( { A }^{ T } \right)  }={ \left( A{ A }^{ -1 } \right)  }^{ 1 }{ I }^{ 1 }=1$
From $1$ and $2$
${ A }^{ T }{ \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 } \right)  }^{ T }{ \left( { A }^{ T } \right)  }=I$
$\Rightarrow { A }^{ 1 }$ is multiplicative inverse of ${ \left( { A }^{ -1 } \right)  }^{ 1 }$
$\Rightarrow { \left( { A }^{ T } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ T }$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

For first $n$ natural numbers we have the following results with usual notations $ \displaystyle \sum _{r=1}^{n}r =\frac{n(n+1)}{2}, \sum _{r=1}^{n}r^{2} =\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^{n}r^{3}=\left ( \sum _{r=1}^{n}r \right )^{2}$ If $\displaystyle a _{1}a _{2}....a _{n} \in A.P $ then sum to $n$ terms of the sequence $\displaystyle \frac{1}{a _{1}a _{2}},\frac{1}{a _{2}a _{3}},...\frac{1}{a _{n-1}a _{n}}$ is equal to $\displaystyle \frac{n-1}{a _{1}a _{n}}$
 and the sum to $ n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by  $\displaystyle S _{n}= \frac{lr-a}{r-1}$ for $ r \neq 1 $ for $ r =1 $ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
 $\displaystyle \frac{lr-a}{r-1} $ when $\displaystyle n \rightarrow \infty ,\left |  r \right | < l $ where $l$ is the last term of $G.P.$  On the basis of above data answer the following questionsThe sum of the series $\displaystyle 2+6+18+...+486 $ equals?

  1. 2184

  2. 1358

  3. 1456

  4. 728
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let ${ S } _{ n }=2+6+18+...+486$

$\Rightarrow { S } _{ n }=2\left( 1+{ 3+3 }^{ 2 }+...+{ 3 }^{ 5 } \right) $
$\Rightarrow { S } _{ n }=2\left( \dfrac { { 3 }^{ 6 }-1 }{ 3-1 }  \right) =729-1$     ...[ sum of G.P series ]

$\Rightarrow { S } _{ n }=728$

Ans: D

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

$\displaystyle x(x+y)+x^{2}(x^{2}+y^{2})+x^{3}(x^{3}+y^{3})+$.......to n terms.

  1. $\displaystyle x^{2}\frac{(1-x^{2n})}{1-x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$
  2. $\displaystyle x^{2}\frac{(1+x^{2n})}{1-x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$
  3. $\displaystyle x^{2}\frac{(1+x^{2n})}{1+x^{2}}+xy\frac{(1+x^{n}y^{n})}{1+xy}$
  4. $\displaystyle x^{2}\frac{(1+x^{2n})}{1+x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle x(x+y)+x^{2}(x^{2}+y^{2})+x^{3}(x^{3}+y^{3})+$.......to n terms.

$=\displaystyle \left(x^2+x^4+x^6+...n terms\right)+\left(xy+x^2y^2+... n terms\right)$

$=\displaystyle x^{2}\frac{(1-x^{2n})}{1-x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$
Hence, option A

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of $\displaystyle\sum _{r=1}^{n}log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$ is

  1. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n}} \right )$
  2. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n+1}} \right )$
  3. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n+1}} \right )$
  4. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n-1}} \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Now,
$\displaystyle\sum _{r=1}^{n}\log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$
$=\displaystyle\sum _{r=1}^{n}\left(\log a^{r}-\log b^{r-1}\right)$
$=\displaystyle\sum _{r=1}^{n}\left(r\log a-(r-1)\log b\right)$
$=(\log a)\times \dfrac{n(n+1)}{2}-\log b\times\dfrac{(n-1)n}{2}$
$=\dfrac{n}{2}\left(\log a^{n+1}-\log b^{n-1}\right)$
$=\dfrac{n}{2}\log\left(\dfrac{a^{n+1}}{b^{n-1}}\right)$
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If (x + 6) is the HCF of $\displaystyle p\left ( x \right )=x^{2}-a$ and $\displaystyle q\left ( x \right )=x^{2}-bx+6$ then $\displaystyle \frac{p\left ( x \right )}{q\left ( x \right )}$ in its lowest terms is______

  1. $\displaystyle \frac{x-6}{x-2}$
  2. $\displaystyle \frac{x+6}{x+1}$
  3. $\displaystyle \frac{x-6}{x-1}$
  4. $\displaystyle \frac{x-6}{x+1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since (x + 6) is the HCF, (x + 6) must be a factor of both polynomials. For p(x) = x^2 - a, x = -6 makes p(-6) = 36 - a = 0, so a = 36. Thus p(x) = (x - 6)(x + 6). For q(x) = x^2 - bx + 6, x = -6 makes 36 + 6b + 6 = 0, so 6b = -42, b = -7. Thus q(x) = x^2 + 7x + 6 = (x + 6)(x + 1). The ratio p(x)/q(x) simplifies to (x - 6)/(x + 1).

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

$\displaystyle 9x^{2}+2hxy+4y^{2}+6x+2fy-3=0$ represents two parallel lines if

  1. $\displaystyle h=6, f=2 $
  2. $\displaystyle h=-6, f=-2 $
  3. $\displaystyle h=-6, f=2 $
  4. $\displaystyle h=6, f=-2 $
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Since the given equation represents a pair of parallel lines, we have
$\displaystyle h^{2}=9 \times 4\Rightarrow h= \pm 6$
and $\displaystyle \begin{vmatrix}
9 & h & 3\ 
 h& 4 & f\ 
 3& f & -3
\end{vmatrix}=0$
$\displaystyle \Rightarrow 9\left ( -12-f^{2} \right )-h\left ( -3h-3f \right )+3\left ( hf-12 \right )=0$
$\displaystyle \Rightarrow 3h^{2}+6hf-9f^{2}-144=0$
$\displaystyle \Rightarrow 108 \pm 36f-9f^{2}-144=0 \ \ \ \left ( \because h= \pm 6 \right )$
$\displaystyle \Rightarrow 9f^{2} \mp 36f+36=0 \ \ \ \ \  (if \ \  h= \pm 6)$
$\displaystyle \Rightarrow f=2 \ \ \ \ if \ \ \ \ \ ( h=6)$
and $\displaystyle \Rightarrow f=-2 \ \ \ \ \ if \ \ \ \ \ \ (h=-6)$

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following is logically equivalent to $\displaystyle \sim \left (\sim p\rightarrow q\right )$?

  1. $\displaystyle p\wedge q$
  2. $\displaystyle p\wedge \sim q$
  3. $\displaystyle \sim p\wedge q$
  4. $\displaystyle \sim p\wedge \sim q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
p  q  $\sim p$  $\sim q$  $\sim p \rightarrow q$  $\sim (\sim p \rightarrow q)$  $p \wedge q$  $p \wedge \sim q$   $\sim p \wedge q$   $\sim p \wedge \sim q$  
T  T  F  F  T  F  T  F  F  F 
T F  F  T  T  F  F  T  F  F 
F T  T  F  T  F  F  F  T  F 
F F  T  T  F  T  F  F  F  T 

The values in column 6 and column 10 are same.

Hence, option D is correct.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Which of the following formula is wrong?

  1. $\displaystyle{C _{v} = \dfrac{R}{\gamma - 1}}$
  2. $\displaystyle{C _{p} = \dfrac{\gamma R}{\gamma - 1}}$
  3. $\displaystyle \dfrac{C _{p}}{ C _{v}} = \gamma$
  4. $C _{p} - C _{v} = 2R$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The different formula for specific heats is given by:

  • $\dfrac{C _{p}}{C _{v}} = \gamma$
  • $C _{p} - C _{v} = R$
Upon further simplification, we get:
  • $C _{p} = \dfrac{\gamma R}{\gamma -1}$
  • $C _{v} = \dfrac{R}{\gamma -1}$
The incorrect formula is
$C _{p} - C _{v} = 2R$
Hence option D is the answer.

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Simplify: 
$\displaystyle x-\left[ y-{ x-\left( y-1 \right) -2x}  \right] $

  1. $2y+1$
  2. $-2y+1$
  3. $2x+y-1$
  4. $2x-y-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

On simplified, we have

$\displaystyle x-\left[ y-{ x-\left( y-1 \right) -2x}  \right] $
$=x-\left[ y-{ x-y+1-2x}  \right] $
=$\displaystyle x-\left[ y-{ -x-y+1}  \right] =x-\left[ y+x+y-1 \right] $
=$\displaystyle x-\left[ 2y+x-1 \right] =x-2y-x+1=-2y+1$
Hence, simplified form of the given expression is $-2y+1$.

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $\displaystyle M=a\left ( m+n \right )$ and $\displaystyle N=b(m-n)$ then the value of  $\displaystyle \left ( \frac{M}{a}+\frac{N}{b} \right )\div \left ( \frac{M}{a}-\frac{N}{b} \right )$ is :

  1. $\displaystyle \frac{m}{n}$
  2. $\frac{n}{m}$

  3. 1
  4. $\frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \frac{M}{a}=m+n;\frac{N}{b}=m-n$
$\displaystyle \therefore \left ( \frac{M}{a}+\frac{N}{b} \right )\div \left ( \frac{M}{a}-\frac{N}{b} \right )=2m\div 2n=\frac{m}{n}$

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $A\,\colon\,B=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3},\,B\,\colon\,C=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}$, then $A\,\colon\,B\,\colon\,C$ is equal to:

  1. $\;2\,\colon\,3\,\colon\,3$
  2. $\;1\,\colon\,2\,\colon\,6$
  3. $\;3\,\colon\,2\,\colon\,6$
  4. $\;9\,\colon\,6\,\colon\,4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\;A\,\colon\,B=\displaystyle\frac{1}{2} \colon\displaystyle\frac{1}{3}=\displaystyle\frac{1}{2}\times6\ \colon\displaystyle\frac{1}{3}\times6=3\,\colon\,2$


$\;\;\;\;\;\;\;\;B\,\colon\,C=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}=3\,\colon\,2$

By taking the L.C.M. of $2$ and $3$, i.e., $6$, we can make the value of $B$ equal in both the ratio.

$\;\;\;\;\;\;\;\;\displaystyle\frac{A}{B}=\displaystyle\frac{3}{2}=\displaystyle\frac{9}{6}$ and $\displaystyle\frac{B}{C}=\displaystyle\frac{3}{2}=\displaystyle\frac{6}{4}$

$\;\;\;\;\;\;\;\therefore\,A\,\colon\,B\,\colon\,C=9\,\colon\,6\,\colon\,4$.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Determine the order relation between the following pairs of ratios.

$\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}}, \frac{2\sqrt{2}}{3\sqrt{3}}$

  1. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} > \frac{2\sqrt{2}}{3\sqrt{3}}$
  2. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} < \frac{2\sqrt{2}}{3\sqrt{3}}$

  3. Cannot be determined

  4. None of These
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle A=\pi \left ( R^{2}-r^{2} \right )$, then $R$ is equal to

  1. $\displaystyle \sqrt{\frac{A-\pi r^{2}}{\pi }}$
  2. $\displaystyle \sqrt{\frac{A+\pi r^{2}}{\pi }}$
  3. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{\pi }}$
  4. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{r}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, $A=\pi(R^2-r^2)$
Therefore, $A =$ $\displaystyle \pi R^{2}-\pi r^{2}$
$\Rightarrow  A+\pi r^{2}=\pi R^{2}$
$\displaystyle \Rightarrow R^{2}=\frac{A+\pi r^{2}}{\pi }$
$\displaystyle \Rightarrow $ $\displaystyle R=\sqrt{\frac{A+\pi r^{2}}{\pi }}$