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 logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\displaystyle y= a^{\left(\frac{1}{1-\log _{a}x}\right)}$ and $\displaystyle z= a^{\left(\frac{1}{1-\log _{a}y}\right)}$, then relation between $x$ and $z$ is

  1. $\displaystyle x= a^{\left(\frac{1}{1-\log _{a}z}\right)}$
  2. $\displaystyle x= a^{\left(\frac{1}{1+\log _{a}z}\right)}$
  3. $\displaystyle x= a\left(\frac{1}{1-\log _{a}z}\right)$
  4. $\displaystyle x= a\left(\frac{1}{1+\log _{a}z}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle y={ a }^{ \frac { 1 }{ 1-\log _{ a }{ x }  }  }\Rightarrow \log _{ a }{ y } =\frac { 1 }{ 1-\log _{ a }{ x }  } $, taking log both sides on base 'a'

$\displaystyle z={ a }^{ \frac { 1 }{ 1-\log _{ a }{ y }  }  }\Rightarrow \log _{ a }{ z } =\frac { 1 }{ 1-\log _{ a }{ y }  } =\frac { 1 }{ 1-\frac { 1 }{ 1-\log _{ a }{ x }  }  } $

$\displaystyle \Rightarrow \log _{ a }{ z } =\frac { 1-\log _{ a }{ x }  }{ 1-\log _{ a }{ x } -1 } \Rightarrow -\log _{ a }{ x } \log _{ a }{ z } =1-\log _{ a }{ x } $

$\displaystyle \Rightarrow \log _{ a }{ x } =\frac { 1 }{ 1-\log _{ a }{ z }  } \Rightarrow x={ a }^{ \frac { 1 }{ 1-\log _{ a }{ z }  }  }$

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

If $\displaystyle \frac{^{n}P _{r-1}}{a}=\frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$,then which of the following holds good 

  1. $c^{2}=a(b+c)$
  2. $a^{2}=c(a+b)$
  3. $b^{2}=a(b+c)$
  4. $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \frac{^{n}p _{r-1}}{a}=\frac{^{n}P _{r}}{b}$

$\displaystyle \Rightarrow n-r=\frac{b}{a}-1$

and $\displaystyle \frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$

$\displaystyle \Rightarrow n-r=\frac{c}{b}$
On dividing (i) and (ii) we get
$b^{2}=a(b+c)$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $\displaystyle x^{3}-mx^{2}-3x+2=0$ has two roots equal in magnitude but opposite in sign, then $m$ is:

  1. $\displaystyle \frac{3}{2}$
  2. $\displaystyle \frac{2}{3}$
  3. $\displaystyle -\frac{2}{3}$

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

Let $\alpha ,-\alpha ,\beta $ be the roots of $x^{ 3 }-mx^{ 2 }-3x+2=0$
Then
${ s } _{ 1 }=\alpha -\alpha +\beta =m\ \Rightarrow \beta =m$
Substituting $x=m$ in equation, we get
$m^{ 3 }-m.m^{ 2 }-3.m+2=0\ \Rightarrow m=\cfrac { 2 }{ 3 } $
Hence, option 'B' is correct.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

The value of $\displaystyle\ \alpha^{4n-1}+\alpha^{4n-3}, n\epsilon\mathbb{N}$ and $\displaystyle\ \alpha$ is a nonreal fourth root of unity is 

  1. $0$
  2. $-1$
  3. $3$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{4}=1$
$x^{2}=\pm1$
$x=\pm i$ and $x=\pm 1$
Hence
$\alpha^{4n-1}+\alpha^{4n-3}$
$=\alpha^{4n}[\alpha^{-1}+\alpha^{-3}]$
$=[\alpha^{-1}+\alpha^{-3}]$
$=\alpha^{-1}[1+\alpha^{-2}]$
$=\alpha^{-3}[\alpha^{2}+1]$
$=\alpha^{-3}[(\pm i)^{2}+1]$
$=0$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $\displaystyle \alpha$ is a non-real root of $\displaystyle x^{5}+1=0$ then $\displaystyle \alpha ^{10n+2}+\alpha ^{5n+2}+\alpha ^{5n}$, where n is an odd positive integer,has the value

  1. $1$
  2. $0$
  3. $-1$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^{5}=-1$
Hence
$\alpha^{5}=-1$
Therefore
$\alpha^{10n+2}+\alpha^{5n+2}+\alpha^{5n}$
$=(\alpha^{5n})^{2}\alpha^{2}+(\alpha^{5n}).\alpha^{2}+\alpha^{5n}$
$=(-1)^{2}\alpha^{2}+(-1)\alpha^{2}+\alpha^{5n}$ .... Since n is odd
$=-\alpha^{2}+\alpha^{2}-1$
$=-1$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $\displaystyle \alpha $ is non-real and $\displaystyle \alpha=\sqrt[5]{1} ,$ then the value of $\displaystyle 2^{\left | 1+\alpha +\alpha ^{2}+\alpha ^{3}-\alpha ^{-1} -\alpha^{-2}\right |} $ is equal to

  1. 4

  2. 2

  3. 1

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\alpha =\sqrt [ 5 ]{ 1 } ,$   ${ 2 }^{ \left| 1+\alpha +{ \alpha  }^{ 2 }+{ \alpha  }^{  3}-{ \alpha  }^{ -1 } -\alpha^{-2}\right|  }$

$\Rightarrow \alpha =1,$

$\Rightarrow { 2 }^{ \left| 1+1+1+1-1-1 \right|  }=2^2=4.$

${ 2 }^{ \left| 1+\alpha +{ \alpha  }^{ 2 }+{ \alpha  }^{  3}-{ \alpha  }^{ -1 } -\alpha^{-2}\right|  }=4$

Hence, the answer is $4.$
Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $\displaystyle w\neq 1 $ is $n^{th}$ root of unity, then value of $\displaystyle \sum _{k=0}^{n-1}\left | z _{1}+w^{k}z _{2} \right |^{2} $ is

  1. $\displaystyle n\left ( \left | z _{1} \right |^{2}+\left | z _{2} \right |^{2} \right )$
  2. $\displaystyle \left | z _{1} \right |^{2}+\left | z _{2} \right |^{2}$
  3. $\displaystyle \left ( \left | z _{1} \right |+\left | z _{2} \right | \right )^{2}$
  4. $\displaystyle n\left ( \left | z _{1} \right |+\left | z _{2} \right | \right )^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Expanding the square |z1 + w^k z2|^2 gives |z1|^2 + |z2|^2 + z1*conj(w^k)conj(z2) + conj(z1)w^k z2. Summing over k=0 to n-1, the cross terms involve the sum of powers of w, which is 0 for n-th roots of unity (w not equal to 1). Thus, the sum becomes n(|z1|^2 + |z2|^2).

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

Solve the equation $\displaystyle z^{n-1}=\bar{z},n\epsilon N.$

  1. $\displaystyle z =\sin \frac{2m\pi }{n}+i\cos \frac{2m\pi }{n}$
  2. $\displaystyle z =\sin \frac{2m\pi }{n}-i\cos \frac{2m\pi }{n}$
  3. $\displaystyle z =\cos \frac{2m\pi }{n}-i\sin \frac{2m\pi }{n}$
  4. $\displaystyle z =\cos \frac{2m\pi }{n}+i\sin \frac{2m\pi }{n}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$z^{n-1}=\overline{z}$
Or 
$z^{n}=1$
This describes $nth$ roots of unity.
Hence let $z=e^{i\theta}$
$e^{in\theta}=e^{i2k\pi}$
Hence
$\theta=\dfrac{2k\pi}{n}$ Where $k\epsilon N$ and $0\leq K\leq n$
Hence
$z=cos\theta+isin\theta$
$=cos(\dfrac{2k\pi}{n})+isin(\dfrac{2k\pi}{n})$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

lf 1, $a _{1},\ a _{2},...,\ a _{n-1}$ are $n^{th}$ roots of unity then $\displaystyle \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+\ldots+\frac{1}{1-a _{n-1}}$ equals?

  1. $\displaystyle \frac{2^{n}-1}{n}$
  2. $\displaystyle \frac{n-1}{2}$
  3. $\displaystyle \frac{n}{n-1}$
  4. $\displaystyle \frac{n}{n+1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We know that $w,w^{2},1$ are the cube roots of unity
Hence,
$\cfrac{1}{1-w}+\cfrac{1}{1-w^{2}}$
$=\cfrac{1}{1-w}+\cfrac{1}{(1-w)(1+w)}$
Now,
$=\cfrac{1}{1-w}(1+\cfrac{1}{1+w})$
$=\cfrac{1}{1-w}(\cfrac{2+w}{1+w})$
$=\cfrac{2+w}{1-w^{2}}$
We know 
$1+w+w^{2}=0$
$2+w+w^{2}=1$ ...(by adding 1 on both sides).
$2+w=1-w^{2}$ substituting we get
$=\cfrac{2+w}{2+w}$
$=1$
$=\cfrac{3-1}{2}$
Hence, the correct alternative is $\cfrac{n-1}{2}$
Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $\displaystyle 1,a _{1},a _{2}...,a _{n-1} $ are $\displaystyle n^{th}$ roots of unity, then $\displaystyle \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+...+\frac{1}{1-a _{n-1}}$ equals

  1. $\displaystyle \frac{2^{n}-1 }{n}$
  2. $\displaystyle \frac{n-1 }{2}$
  3. $\displaystyle \frac{n}{n-1}$

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

Given $\displaystyle z^{n}= 1,z= 1,a _{1}, a _{2},..., a _{n-1}$
Let $\displaystyle a= \frac{1}{1-z}\Rightarrow z= 1-\frac{1}{a}$
 $\displaystyle \therefore \left ( 1-\frac{1}{a} \right )^{n}= 1$
$\displaystyle \Rightarrow \left ( a-1 \right )^{n}-a^{n}= 0$
$\displaystyle \Rightarrow -C _{1}a^{n-1}+C _{2}a^{n-2}+...+\left ( -1 \right )^{n}= 0$ where  $\displaystyle a= \frac{1}{1-a _{1}}, \frac{1}{1-a _{2}}.....\frac{1}{1-a _{n-1}}$

$\displaystyle \Rightarrow \frac{1}{1-a _{1}}+\frac{1}{1-a _{2}}+.....+\frac{1}{1-a _{n-1}}= \frac{^{n}c _{2}}{n}= \frac{n-1}{2}$

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Write the following transformation in matrix form
$\quad x _1 = \displaystyle\frac{\sqrt 3}{2}y _1 + \displaystyle\frac{1}{2}y _2; \quad x _2 = -\displaystyle\frac{1}{2}y _1 + \displaystyle\frac{\sqrt 3}{2}y _2$.
Hence find the transformation in matrix form which expresses $y _1, y _2$ in terms of $x _1, x _2$.

  1. $y _1 = \displaystyle\frac{\sqrt 3}{2}x _1 + \displaystyle\frac{1}{2}x _2; \quad y _2 = \displaystyle\frac{1}{2}x _1 + \displaystyle\frac{\sqrt 3}{2}x _2$
  2. $y _1 = \displaystyle\frac{\sqrt 3}{2}x _1 - \displaystyle\frac{1}{2}x _2; \quad y _2 = \displaystyle\frac{1}{2}x _1 + \displaystyle\frac{\sqrt 3}{2}x _2$
  3. $y _1 = \displaystyle\frac{\sqrt 3}{2}x _1 - \displaystyle\frac{1}{2}x _2; \quad y _2 = \displaystyle\frac{1}{2}x _1 - \displaystyle\frac{\sqrt 3}{2}x _2$

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

$ \displaystyle  { x } _{ 1 }=\frac { \sqrt { 3 }  }{ 2 } { y } _{ 1 }+\frac { 1 }{ 2 } { y } _{ 2 }  $ and $\displaystyle { x } _{ 2 }=\frac { -1 }{ 2 } { y } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { y } _{ 2 } $ 
We observe $ \displaystyle \frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 }=\frac { 3 }{ 4 } { y } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } .\frac { 1 }{ 2 } { y } _{ 2 }+\frac { 1 }{ 4 } { y } _{ 1 }-\frac { \sqrt { 3 }  }{ 2 } \frac { 1 }{ 2 } { y } _{ 2 } $
$ \displaystyle \Rightarrow \frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 }={ y } _{ 1 } $
Similarly $ \displaystyle \frac { 1 }{ 2 } { x } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { x } _{ 2 }=\frac { 1 }{ 4 } { y } _{ 2 }+\frac { 3 }{ 4 } { y } _{ 2 }={ y } _{ 2 } $
$ \displaystyle \therefore { y } _{ 1 }=\frac { \sqrt { 3 }  }{ 2 } { x } _{ 1 }-\frac { 1 }{ 2 } { x } _{ 2 };{ y } _{ 2 }=\frac { 1 }{ 2 } { x } _{ 1 }+\frac { \sqrt { 3 }  }{ 2 } { x } _{ 2 }  $ 

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let $P(x, y)$ be any given point and $\displaystyle P(x _{1},y _{1})$ be the image of $P(x,y)$ after reflection.
The matrix of reflection of point $P$ through the line $\displaystyle y = x $ is given by

  1. $\displaystyle \begin{bmatrix}1 &0 \\0 &1\end{bmatrix}$
  2. $\displaystyle \begin{bmatrix}-1 &0 \\0 &-1\end{bmatrix}$
  3. $\displaystyle \begin{bmatrix}-1 &0 \\0 &1\end{bmatrix}$
  4. $\displaystyle \begin{bmatrix}0 &1 \\1&0\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Reflection of matrix $P(x,y)$ through the line $y=mx$ making an angle $\theta$ with $x-$axis is

 
$\begin{bmatrix} \cos  2\theta  & \sin { 2\theta  }  \ \sin{ 2\theta  } & -\cos { 2\theta  }  \end{bmatrix}$

Given line is $y=x$ which makes an angle of $45^{\circ}$ with $x-$axis

Hence, the transformation matrix is $\begin{bmatrix} 0&1\1&0\end{bmatrix}$

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

Let $\displaystyle A\equiv \left( 2,0 \right) $ and $\displaystyle B\equiv \left( 3,1 \right) $. The line $\displaystyle AB$ turns about $\displaystyle A$ through an angle $\displaystyle \frac { \pi  }{ 12 } $ in the clockwise sense, and the new position of $\displaystyle B$ is $\displaystyle B'$. Then $\displaystyle B'$ has the co-ordinates :-

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

Slope of the line $\displaystyle AB=\frac { 0-1 }{ 2-3 } =1$
$\therefore \angle BAX={ 45 }^{ o }$
Given $\angle B'AB={ 15 }^{ o }\Rightarrow \angle B'AX={ 30 }^{ o }$
Therefore slope of the line $\displaystyle AB'=\tan { { 30 }^{ o } } =\frac { 1 }{ \sqrt { 3 }  } $
Now line $AB'$ makes an angle of ${ 30 }^{ o }$ with positive direction of $x$-axis and 
$AB'=AB=\sqrt { { \left( 3-2 \right)  }^{ 2 }+{ \left( 1-0 \right)  }^{ 2 } } =\sqrt { 2 } $
Therefore coordinates are $\displaystyle \left( 2+\sqrt { 2 } \cos { { 30 }^{ o } } ,0+\sqrt { 2 } \sin { { 30 }^{ o } }  \right) =\left( \frac { 2\sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2 }  } ,\frac { 1 }{ \sqrt { 2 }  }  \right) $

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

Here the mirror image of line $3x^2+4xy+5y^2=0$ in mirror $x=0$

i.e about $Y-axis $
So the x-coordinates About $Y-axis $ becomes negative and the y-coordiantes remains same Hence putting $x=-x$ in given eq 
$3(-x)^2-4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$