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 angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Given the line $\displaystyle L:\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-3 }{ -1 } $ and the plane $\pi :x-2y=0$. Of the following assertion, the only one that is always true is

  1. $L$ is $\bot$ to $\pi$
  2. $L$ lies in $\pi$
  3. $L$ is parallel to $\pi$

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

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0,$
$\therefore$ given line is $\bot $ to the normal to the plane i.e., given line is parallel to the given plane.
Also $\left( 1,-1,3 \right) $ lies on the plane $x-2y-z=0$, if $1-2\left( -1 \right) -3=0$ i.e., $1+2-3=0$
which is true 

$\therefore L$ lies in plane $\pi .$

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If $\displaystyle \alpha ,\beta $ are the roots of $\displaystyle x^{2}+x+1=0$ and $\displaystyle \gamma ,\delta $ are the roots $\displaystyle x^{2}+3x+1=0,$ then $\displaystyle (\alpha -\gamma)(\beta +\delta )(\alpha +\delta )(\beta -\gamma )=$

  1. $2$`
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Using $\alpha \beta =1$     and       $\gamma \delta =1$

$(\alpha \beta + \alpha \delta - \beta \gamma - \gamma \delta) (\alpha \beta - \alpha \gamma +\beta \delta - \gamma \delta) $

$\Rightarrow (1+\alpha \delta - \beta \gamma - 1)(1-\alpha \gamma +\beta \delta - 1)$

$\Rightarrow (\alpha \delta - \beta \gamma) (\beta \delta - \alpha \gamma) $

$\Rightarrow \delta ^{2}-\alpha ^{2}-\beta ^{2}+\gamma ^{2}$

$\Rightarrow - (\alpha ^{2}+\beta ^{2})+(\gamma ^{2}+\delta ^{2})$

$\Rightarrow - (-1)+7=8$

Final answer is 8
Multiple choice the nth roots of unity complex numbers maths

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

  1. $\displaystyle\ 4$
  2. $\displaystyle\ 2$
  3. $\displaystyle\ 1$

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

We have $1 + \alpha +\alpha^2 +\alpha^3 +\alpha^4 = 0$


$ 1 + \alpha +\alpha^2 + \alpha^3 +\alpha^{-1} $
$= -\alpha^4 + \dfrac{1}{\alpha} $
$ = \dfrac{1 -\alpha^5}{\alpha} $
$ =0 $
Hence, 
$2^{| 1+ \alpha + \alpha^2 +\alpha^3 + \alpha^{-1} | } =1 $

Multiple choice the nth roots of unity complex numbers maths

if $\displaystyle\ z _{\gamma }=\cos \frac{2\gamma \pi}{5}+i\sin \frac{2\gamma \pi}{5}=0$, $\displaystyle\ \gamma = 0,1,2,3,4.....$ then $\displaystyle\ z _{1}z _{2}z _{3}z _{4}z _{5}$ is equal to

  1. $\displaystyle\ -1$
  2. $\displaystyle\ 0$
  3. $\displaystyle\ 1$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $z _\gamma = \cos{\frac{2 \pi \gamma}{5}} + i \sin{\frac{2 \pi \gamma}{5}}$ where $\gamma = 1, 2, 3, 4, 5.$
Thus, $z _\gamma = e^(i \frac{2 \pi \gamma}{5})$
Thus, $z _1z _2z _3z _4z _5 = e^(\frac{2 \pi}{5} + \frac{4 \pi}{5} + \frac{6 \pi}{5} + \frac{8 \pi}{5} + \frac{10 \pi}{5}) $
$= e^{6 \pi}$
$= \cos(6 \pi) + i \sin(6 \pi)$
$= 1$

Multiple choice the nth roots of unity complex numbers maths

If $\displaystyle z=\cos \frac{8\pi }{11}+i\sin\frac{8\pi }{11},$ then Real $\displaystyle \left ( z+z^{2}+z^{3}+z^{4}+z^{5} \right )$ is

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

  2. 0
  3. $\displaystyle \frac{1}{2}$

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

Real (z) $\displaystyle =\frac{z+\bar{z}}{2}=\frac{1}{2}\left ( z+\frac{1}{z} \right )$


$\displaystyle \because z\bar{z}=\cos ^{2}\frac{8\pi }{11}+\sin ^{2}\frac{8\pi }{11}=1\ \ \therefore \bar{z}=\frac{1}{z}$

$\displaystyle \because$ E=Real part of $\displaystyle \left ( z+z^{2}+z^{3}+z^{4}+z^{5}+\frac{1}{z}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+\frac{1}{z^{4}}+\frac{1}{z^{5}} \right )$

Now $\displaystyle z^{11}= \cos 8\pi +i\sin 8\pi = 1$

$\displaystyle \therefore \frac{1}{z^{4}}= \frac{z^{7}}{z^{11}}= z^{7}$ etc.
$\displaystyle \therefore E= \frac{1}{2}\left [z+z^{2}+z^{3}+z^{4}+z^{5}+z^{10}+z^{9}+z^{8}+z^{7}+z^{6} \right ]$

Add and subtract $\displaystyle z^{11}.$
$\displaystyle \therefore E= \frac{1}{2}$ [sum of G.P. of 11 terms-$\displaystyle z^{11}$]

$\displaystyle = \frac{1}{2}\left [ \frac{z\left ( 1-z^{11} \right )}{1-z}-z^{11} \right ]= \frac{1}{2}\left ( 0-1 \right )= -\frac{1}{2}$ by (I)

Multiple choice the nth roots of unity complex numbers maths

lf $\alpha$ be the $n^{th}$ root of unity then the sum of the series $1+2\alpha+3\alpha^{2}+\ldots.+n\alpha^{n-1}$ equals?

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

$s=1+2\alpha+3\alpha^{2}- -n\alpha^{n-1}$
$s\alpha=\alpha+2\alpha^{2}+3\alpha^{3}- - n\alpha
\So, s-s\alpha=1+\alpha+\alpha^{2}- -\alpha^{n-1}-n\alpha^{n}$
(Sum of roots of unity)
$\Rightarrow s (1-\alpha)=-n\alpha^{n}$
$\Rightarrow s=\dfrac{-n\alpha^{n}}{1-\alpha}=\dfrac{-n}{1-\alpha}$

as $\alpha^{n}=1$
$\alpha$ being of unity root

Multiple choice the nth roots of unity complex numbers maths

If $\displaystyle \alpha = \cos\frac{8\pi}{11}+i\sin\frac{8\pi }{11}$ then $\displaystyle Re(\alpha +\alpha^{2}+\alpha^{3}+\alpha^{4}+\alpha^{5})$ equals

  1. 0
  2. $\displaystyle -\frac{1}{2}$
  3. $\displaystyle \frac{1}{2}$

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

Given : $\displaystyle \alpha = \cos\dfrac{8\pi}{11}+i\sin\dfrac{8\pi }{11}$
$\displaystyle \therefore \alpha ^{11}= \cos 8\pi +i\sin 8\pi = 1$
$\displaystyle \sum _{n= 0}^{10}\alpha ^{n}= \dfrac{1-\alpha ^{11}}{1-\alpha }= 0$ (sum of 11th roots of unity)
Now $\displaystyle Re(z)= \dfrac{z+\bar{z}}{2}=\dfrac{sum \ of\  11 \ roots\  of\  unity -1}{2}= -1/2 $

Multiple choice the nth roots of unity complex numbers maths

If $\displaystyle \alpha  $ be the $\displaystyle n^{th}  $ root of unity then the sum of the series
$\displaystyle 1+2\alpha+3\alpha^{2}+...n\alpha ^{n-1}$ equals.

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

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

$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
Here $S$ forms an A.G.P
Therefore
$S=1+2\alpha+3\alpha^{2}+...n\alpha^{n-1}$
$S(\alpha)=\alpha+2\alpha^{2}+3\alpha^{3}+...(n-1)\alpha^{n-1}+n\alpha^{n}$
Hence
$S(1-\alpha)=1+\alpha+\alpha^{2}+....\alpha^{n-1}-n\alpha^{n}$
$S(1-\alpha)=\dfrac{1-\alpha^{n}}{1-\alpha}-n\alpha^{n}$
$S(1-\alpha)=0-n$
$S=\dfrac{-n}{1-\alpha}$
Hence, option 'A' is correct.

Multiple choice the nth roots of unity complex numbers maths

If $\displaystyle \alpha _{1}, \alpha _{2}, \cdots \alpha _{100}$ are all the 100th roots of unity, then $\displaystyle \sum \sum \left ( \alpha _{i}\alpha _{j} \right )^{5}$ is $\displaystyle 1\leq i< j\leq 100$

  1. $20$
  2. $\displaystyle \left ( 20 \right )^{1/20}$
  3. $0$

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

$\displaystyle 2\sum ab= \left ( \sum a \right )^{2}-\sum a^{2}$
$\displaystyle \therefore 2\sum \sum \left ( \alpha _{i}\alpha _{j} \right )^{5}= \left ( \alpha _{1}^{5}+\alpha _{2}^{5}+\cdots  \right )^{2}-\left ( \alpha _{1}^{10}+\alpha _{2}^{10}+\cdots  \right )$
$\displaystyle = 0-0$ ($\displaystyle \because \sum \alpha _{i}^{r}= 100$ if $\displaystyle r= 100k$
and $ \sum \alpha _{i}^{r} = 0$ if $\displaystyle r\neq 100k$)
Here both 5 and 10 are not multiples of 100.

Multiple choice the nth roots of unity complex numbers maths

The value of $\displaystyle \sum _{k= 1}^{6}\left ( \sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7} \right )$ is

  1. -1

  2. 0

  3. -i

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$\displaystyle = -i^{2}\sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7}$
$\displaystyle = -i\left ( \cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7} \right )= -i e^{i2\pi k/7}= -iz^{k}$
where, $\displaystyle z= \cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}$ or $\displaystyle z^{7}= 1$
or $\displaystyle z= \left ( 1 \right )^{1/7} \therefore 1+z+2^{2}+\cdots +z^{6}= 0$ ...(1)
Above being the sum of seven, seventh roots of unity
$\displaystyle \sum = -i \sum _{i= 1}^{k} z^{k}= -1\left ( z+z^{2}+\cdots +z^{6} \right )= -i\left ( -1 \right )= i$ by (I)
Alternative Method. $\displaystyle \sin \theta -i\cos \theta $
$\displaystyle = -i^{2}\sin \theta -i\cos \theta = -i\left ( \cos \theta +i\sin \theta  \right )$
$\displaystyle = -ie^{i\theta }$ where $\displaystyle \theta = \frac{2\pi }{7}$ or $\displaystyle 7\theta = 2\pi $
Hence the given sigma is
$\displaystyle \sum _{k= 1}^{6}-ie^{ik\theta }= -i\left [ e^{i\theta +}e^{2i\theta }+\cdots +e^{6i\theta } \right ]$
$\displaystyle = -ie^{i\theta }\left [ \frac{1-\left ( e^{i\theta } \right )^{6}}{1-e^{i\theta }} \right ]= -i\left [ \frac{e^{i\theta }-e^{7i\theta }}{1-e^{i\theta }} \right ]$ (G.P.)
$\displaystyle = -i\left [ \frac{e^{i\theta }-1}{1-e^{i\theta }} \right ]= -i\left ( -1 \right )= i$
$\displaystyle \because 7i\theta = 2\pi i \therefore e^{7i\theta }= \left ( \cos 2\pi +i\sin 2\pi  \right )= 1.$

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $\displaystyle 2^{2^{3}}=j, 2^{3^{2}}=k, 3^{2^{2}}=\varphi ,$ then 

  1. $k = 2j$
  2. $j < k$
  3. $\displaystyle \varphi < k$

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

  $j=2^{2{^3}}$, $k=2^{3^{2}}$, $\varphi=3^{2^{2}}$

$\Rightarrow$  $j=256,\,\,k=512,\,\,\varphi=81$
$\Rightarrow$  Option A says $k=2j$, which is true because $j=256$, so $k=2\times 256=512$
$\Rightarrow$  Option B says $j<k$, which is true because value of $j$ is $255$ and value of $k$ is $512$
$\Rightarrow$   Option C says $\varphi <k$, which is true because value of $\varphi$ is $81$ and value of $k$ is $512$

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $\displaystyle a^{m}=b^{m}$ and $(m > 0)$, then which of the following options could be true:

  1. $a = -b$
  2. $a + b= 0$
  3. $2a - b = 0$
  4. $\displaystyle \dfrac{a^{2}}{b^{2}}=1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $a^m=b^m$


Dividing by $b^m$

$\dfrac{a^m}{b^m}=1$
${\left(\dfrac{a}{b}\right)}^m=1$

Only Option D satisfies this answer.

Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If $\displaystyle :A= \left [ a _{ij} \right ]$ is a scalar matrix, then trace of A is

  1. $\displaystyle \:\sum _{i} \sum _{i} a _{ij}$
  2. $\displaystyle \:\sum _{i} a _{ij}$
  3. $\: \sum _{ i } a _{ ij }\times { a } _{ ji }$

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

By definition of trace of a matrix of order n, 
$tr(A)=a _{11}+ a _{22}+a _{33}+.....+a _{nn}$
$\displaystyle =: \sum _{ i=j }  { a } _{ ij } =: \sum _{ i } a _{ ij }$
Hence, option 'B' is correct.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Let $a, b$ be non-zero real numbers. The equation $\displaystyle \left ( ax^{2}+by^{2}+c \right )\left ( x^{2}-5xy+6y^{2} \right )$ represents

  1. four straight lines, when $\displaystyle c=0$ and $a, b$ are of the same sign
  2. two straight lines and a circle, when $\displaystyle a=b$ and $c$ is of sign opposite to that of $a$
  3. two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$
  4. a circle and an ellipse, when $a$ and $b$ are of the same sign
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From given expression, $\displaystyle { x }^{ 2 }-5xy+6{ y }^{ 2 }=0$ are pair of straight lines $\displaystyle y=\frac { x }{ 2 } $ and $ y=\dfrac { x }{ 3 } $
Now, $\displaystyle a{ x }^{ 2 }+b{ y }^{ 2 }+c=0$ will be cirlce of radius $\displaystyle \sqrt { -\frac { c }{ a }  } $.

If $\displaystyle a=b$ and $\displaystyle c$ is of opposite sign of $\displaystyle a  $ and $  b$, then $\displaystyle { x }^{ 2 }+{ y }^{ 2 }=-\frac { c }{ a } $

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

If the line $\displaystyle \frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z + 1}{\lambda}$ lies in the plane $\displaystyle 3x - 2y + 5z = 0$ then $\displaystyle \lambda$ is

  1. $\displaystyle 1$
  2. $\displaystyle -\frac{7}{5}$
  3. $\displaystyle \frac{5}{7}$

  4. no possible value
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have equation of plane,
$3x-2y+5z=0.......(1)$

We have line,

$\dfrac{x-1}{1}=\dfrac{y+1}{-2}=\dfrac{z+1}{\lambda}=\mu......(2)$
General point on line is,
$P=(\mu+1,-2\mu-1,\lambda\mu-1)$
Since line (2) lies on line plane (1),so point P satisfy equation (1)
Therefore,
$3(\mu+1)-2(2\mu-1)+5(\lambda\mu-1)=0$
$3\mu+3+4\mu+2+5\mu\lambda-5=0$
$7\mu+5\mu\lambda=0$
$\lambda=\dfrac{-7}{5}$ 
Therefore option (B) is Correct.