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 brackets order operations and algebra using brackets in algebraic expressions order of operations

If $a$ and $ b $ are any two real numbers with opposite signs, which of the following is the greatest?

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

$(a-b)^2=a^2+b^2-2ab$

as a and b are of oppsite sign ab<0 and -2ab>0,it means $(a-b)^2>a^2+b^2-2|a||b|=(|a|-|b|)^2>(|a2+b^2|)>(|a^2-b^2|)$

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 maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

If $\displaystyle \triangle ABC\sim \triangle DEF$ BC=4 cm, EF=5 cm and ar $\displaystyle \left ( \triangle ABC \right )=80cm2$,the ar$\displaystyle \left ( \triangle DEF \right )$ is

  1. $\displaystyle 120cm^{2}$
  2. $\displaystyle 125cm^{2}$
  3. $\displaystyle 150cm^{2}$
  4. $\displaystyle 200cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

$\therefore \dfrac{arc(\triangle ABC)}{arc(\triangle DEF)}=\dfrac{BC^2}{EF^2}$
$\Rightarrow \dfrac{80}{arc(\triangle DEF)}=\dfrac{16}{25}$
$\Rightarrow arc(\triangle DEF)=\dfrac{80\times 25}{16}=125 cm^2$


Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The H.C.F. of two expressions is x and their L.C.M is $ \displaystyle x^{3}-9x  $  IF one of the expression is $ \displaystyle x^{2}+3x  $  then,the other expression is 

  1. $ \displaystyle x^{2}-3x $
  2. $ \displaystyle x^{3}-3x $
  3. $ \displaystyle x^{2}+9x $
  4. $ \displaystyle x^{2}-9x $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let two expressions $p(x)$ and $q(x)$ then

$p(x)\times q(x)=L.C.M.\ \times\ H.C.F.$

Since $p(x)=x^2+3x$
$(x^2+3x)\times q(x)=(x^3-9x) \times\ x$
$(x^2+3x)\times q(x)=(x^2-3x) \times\ (x^2+3x)$

$q(x)=(x^2-3x)$
Hence, this is the required solution.

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.)
$\displaystyle = -i\left [ \frac{e^{i\theta }-1}{1-e^{i\theta }} \right ]= -i\left ( -1 \right )= i$
$\displaystyle \because 7i\theta = 2\pi i \therefore e^{7i\theta }= \left ( \cos 2\pi +i\sin 2\pi  \right )= 1.$