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 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 
$3(-x)^2-4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$

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 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