Mathematics

Advanced Algebra and Calculus

138 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 maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of the pair of lines represented by $\displaystyle  3x^{2}+4xy+5y^{2}=0 $ in the line mirror x = 0 is

  1. $\displaystyle 3x^{2}-4xy+5y^{2}=0 $
  2. $\displaystyle 3x^{2}-4xy-5y^{2}=0 $
  3. $\displaystyle 5y^{2}-4xy-3x^{2}=0 $

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given pair
$3x^2+4xy+5y^2=0$
$x=0$ is the $Y-axis $ hence the $x$-coordinates will become $-x$ and $y$-coordinates remains same 
Hence 
$3(-x)^2+4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

If $\displaystyle \left ( -2, 6 \right )$ is the image of the point $\displaystyle \left ( 4,2 \right )$ with respect to the line $\displaystyle L=0$, then $\displaystyle L=$

  1. $\displaystyle 6x-4y-7=0$
  2. $\displaystyle 2x-3y-5=0$
  3. $\displaystyle 3x-2y+5=0$
  4. $\displaystyle 3x-2y+10=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Slope of line joining the image $Q(-2,6)$ and the point $P(4,2)$ is $\displaystyle -\frac{2}{3}$

So, the slope of mirror $L$ is $\displaystyle \frac{3}{2}$

Mid-point of $PQ$ is $(1,4)$

Since, the image and point are equidistant from mirror. So, this point $(1,4)$ lies on the mirror.

So, the equation of mirror is
$y-4=\displaystyle \frac{3}{2} (x-1)$

$\Rightarrow 3x-2y+5=0$

Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

If $\displaystyle { a }^{ 2 }$ ends in 5, then $\displaystyle { a }^{ 3 }$ ends in 25.

  1. True

  2. False

  3. Ambiguous

  4. Insufficient information
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

False. 15 x 15=225 and 15 x 15 x 15=3375. so the statement $\displaystyle { a }^{ 2 }$ ends in 5, then $\displaystyle { a }^{ 3 }$ ends in 25 is not true in all cases.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

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 to infinite terms of the series $\displaystyle \frac{1}{2}+\frac{1}{6}+\frac{1}{18}+.. $ is equal to ?

  1. $\displaystyle \frac{4}{3}$
  2. $\displaystyle \frac{3}{4}$
  3. $\displaystyle \frac{8}{3}$

  4. Does not exit
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let, ${ S } _{ \infty  }=\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 6 } +d\frac { 1 }{ 18 } +..\infty $

$\Rightarrow { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( 1+\dfrac { 1 }{ 3 } +\dfrac { 1 }{ { 3 }^{ 2 } } +....\infty  \right) $

As we know that, sum of infinite G.P series $=\dfrac { a }{ 1-r } $

Therefore, $ { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ 1-\left( 1/3 \right)  }  \right) =\dfrac { 3 }{ 4 } $

Ans: B

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $\displaystyle x=\sum _{a=0}^{\infty }a^{n},y=\sum _{a=0}^{\infty }b^{n},z=\sum _{a=0}^{\infty }c^{n}$ Where $a,b,c $ are in A.P and $\displaystyle \left | a \right |<1,\left | b \right |<1,\left | c \right |<1$ then $x,y,z$ are in

  1. H.P

  2. Arithmetic-Geometric progression

  3. A.P

  4. G.P
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $\displaystyle \left | a \right |< 1,\left | b \right |< 1,\left | c \right |< 1,\ \ a,b,c \in A.P$


and $\displaystyle \sum _{n=0}^{\infty }a^{n}=\frac{1}{1-a},\sum _{n=0}^{\infty }b^{n}=\frac{1}{1-b},\sum _{r=0}^{\infty }c^{n}=\frac{1}{1-c}$

$\displaystyle \therefore x=\frac{1}{1-a},y=\frac{1}{1-b},c=\frac{1}{1-c}$

$\displaystyle \Rightarrow a=\frac{x-1}{x},b=\frac{y-1}{y},c=\frac{z-1}{z}$

$\displaystyle \because 2b=a+c \ as \ a,b,c \in A.P$

$\displaystyle 2\left ( \frac{y-1}{y} \right )=\frac{x-1}{x}+\frac{z-1}{z}\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$\displaystyle \Rightarrow x,y,z \in H.P$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $R \subset\left ( 0,\pi  \right )$ denote the set of values of which satisfies the equation $ \displaystyle 2^{\left ( 1+\left | \cos x \right |+\left | cos^{2}x \right |+\left | cos^{3}x \right | \right )+\left | cos^{4}x  \right |...............\infty}=4$ then $R$ equals

  1. $\displaystyle\left \{ -\frac{\pi }{3} \right \}$
  2. $\displaystyle\left \{ \frac{\pi }{3},\frac{2\pi }{3} \right \}$
  3. $\displaystyle\left \{ \frac{-\pi }{3},\frac{2\pi }{3} \right \}$
  4. $\displaystyle\left \{ \frac{\pi }{3},\frac{-2\pi }{3} \right \}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ 2 }^{ \left( 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty  \right)  }={ 2 }^{ 2 }\ \Rightarrow 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty =2\ \Rightarrow \dfrac { 1 }{ 1-\left| \cos { x }  \right|  } =2\ \Rightarrow 1-\left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow \left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow x=\dfrac { \pi  }{ 3 } ,\dfrac { 2\pi  }{ 3 } $
  in the range $\left( 0,\pi  \right) $

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Calculate the sum of the infinite geometric series $2+\left(-\displaystyle\frac{1}{2}\right)+\left(\displaystyle\frac{1}{8}\right)+\left(-\displaystyle\frac{1}{32}\right)+...$

  1. $1\displaystyle\frac{3}{8}$
  2. $1\displaystyle\frac{2}{5}$
  3. $1\displaystyle\frac{1}{2}$
  4. $1\displaystyle\frac{3}{5}$
  5. $1\displaystyle\frac{5}{8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given the geometric series is $2,\left ( -\dfrac{1}{2} \right ),\left ( \dfrac{1}{8} \right ),\left ( -\dfrac{1}{32} \right ).......................$

Then common ratio $=-\dfrac{1}{4}$
And first term is $2$.
Then sum of the infinite geometric series $=$ $S=\dfrac{a _{1}}{1-r}=\dfrac{2}{1-(-\frac{1}{4})}=\dfrac{2\times 4}{4+1}=\dfrac{8}{5}=1\dfrac{3}{8}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $e^{\displaystyle \left [ \left ( \sin^{2}x + \sin^{4}x + \sin^{6}x + .... + \infty \right ) \log _{e}2\right ]}$ satisfies the equation $\displaystyle x^{2} -9x + 8 = 0$,then the value of $\displaystyle g \left ( x \right ) = \frac{\cos x}{\cos x + \sin x}$ is

  1. $\displaystyle \frac{\sqrt{3} + 1}{2}$
  2. $\displaystyle \frac{\sqrt{3} - 1}{2}$
  3. $\displaystyle 8$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider, $y=exp\left[ \left( \sin ^{ 2 } x+\sin ^{ 4 } x+\sin ^{ 6 } x+....+\infty  \right) \log _{ e } 2 \right] $
$\displaystyle\Rightarrow y=exp\left[ \left( \frac { \sin ^{ 2 }{ x }  }{ 1-\sin ^{ 2 }{ x }  }  \right) \log _{ e } 2 \right] =exp\left[ \tan ^{ 2 }{ x } \log _{ e } 2 \right] ={ 2 }^{ \tan ^{ 2 }{ x }  }$
Since, $y$ satisfies $x^{ 2 }-9x+8=0$, then
        $y=1,8$
$\Rightarrow { 2 }^{ \tan ^{ 2 }{ x }  }={ 2 }^{ 0 },{ 2 }^{ 3 }$
$\Rightarrow \tan ^{ 2 }{ x } =0,3$
$\Rightarrow \tan { x } =0,\pm \sqrt { 3 } $

Now, $\displaystyle g\left( x \right) =\frac { \cos  x }{ \cos  x+\sin  x } =\frac { 1 }{ 1+\tan { x }  } =1,\frac { 1 }{ 1\pm \sqrt { 3 }  } =1,\frac { -\sqrt { 3 } -1 }{ 2 } ,\frac { \sqrt { 3 } -1 }{ 2 } $

Ans: B

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

Let $\displaystyle A=\begin{pmatrix}1 &2 \3  &4
\end{pmatrix}$ and $\displaystyle B=\begin{pmatrix}a &0 \0  &b \end{pmatrix} a,b \epsilon N.$Then

  1. there cannot exist any B such that $\displaystyle AB = BA $
  2. there exist more than one but finite number of B's such that $\displaystyle AB = BA$
  3. there exists exactly One B such that $\displaystyle AB = BA$
  4. there exist infinitely many B's such that $\displaystyle AB = BA.$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A=\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ and $B=\begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}$

$AB = \begin{bmatrix} a & 2b \ 3a & 4b \end{bmatrix}$

$BA = \begin{bmatrix} a & 2a \ 3b & 4b \end{bmatrix}$

$AB\quad =\quad BA \Rightarrow a=b$

$\therefore$ there exist infinitely many  $B's$  such that $AB=BA$.

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

If AB=KI where $\displaystyle K\in R$ then $\displaystyle A^{-1}$= _____

  1. B

  2. KB
  3. $\displaystyle \frac{1}{K}B$
  4. $\displaystyle \frac{1}{K^{2}}B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given $AB=KI\quad K\epsilon R$
i.e., K is constant
Now ${ A }^{ -1 }=\cfrac { I }{ A } $
I is identity matrix
$AB=KI$
$\Rightarrow \cfrac { 1 }{ K } B=\cfrac { I }{ A } \Rightarrow { A }^{ -1 }=\cfrac { 1 }{ K } B$
OPTION C
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