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 power and exponent power of powers laws of exponents and powers law of indices

On simplifying  $\displaystyle 3^{3}\times a^{3}\times b^{3}$, we get

  1. $\displaystyle \left ( 3ab \right )^{3} $
  2. $\displaystyle 3\left ( ab \right )^{3} $
  3. $\displaystyle \left ( 27ab \right )^{3} $

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

$\displaystyle 3^{2}\times a^{3}\times b^{3}=\left ( 3ab \right )^{3}$.

This is the power of product law of exponents.
So, option $A$ is correct.

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Find the expression which equals $\displaystyle a^{x}\times b^{x}$.

  1. $\left [\displaystyle a^{x}+ b^{x} \right ]$
  2. $\displaystyle \left ( ab\right )^x $
  3. $\displaystyle \left (a+b \right )^{x} $
  4. $\displaystyle a\left ( b \right )^{x} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle a^{x}\times b^{x}=\left ( ab \right )^{n}$

So, option $B$ is correct.

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Which of the law does not stand true ?

  1. $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$
  2. $\displaystyle \left ( \frac{a^{m}}{a^{n}} \right )^{x}=\frac{a^{mx}}{a^{nx}}$
  3. $\displaystyle \frac{a^{m}}{b^{m}}=\left ( \frac{a}{b} \right )^{m}$
  4. $\displaystyle \frac{a^{m}}{a^{m}}=a^{m}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \frac{a^{m}}{a^{m}}=1$
$\displaystyle\therefore  \frac{a^{m}}{a^{m}}\neq a^{m}$

Multiple choice position of point wrt ellipse ellipse maths
$C: x^{2}+y^{2}=9$, $\displaystyle E: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, $L: y=2x$

Let $L$ intersect $x=1$ at point $R$. Then which of the following is correct :
  1. $R$ lies inside both $C$ and $E$
  2. $R$ lies outside both $C$ and $E$
  3. $R$ lies on both $C$ and $E$
  4. $R$ lies inside $C$ but outside $E$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$y=2x$, intersects $x=1$ at $(1,2)$
Coordinate of $R$ are $(1,2)$
$C(1,2)=1+22-9<0$ Since $C(1,2)$ is $<0, R $ lies inside $C$
$E(1,2)=\dfrac{1}9+1-1>0$ Since $E(1,2)$ is $>0, R $ lies outside $E$.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

The product of the values of $\displaystyle{\left[ {\cos {\pi  \over 3} + i\sin {\pi  \over 3}} \right]^{{3 \over 4}}}$ is

  1. $-1$
  2. $1$
  3. $i$
  4. $-i$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: $\displaystyle{\left[ {\cos {\pi  \over 3} + i\sin {\pi  \over 3}} \right]^{{3 \over 4}}}$


$=[e^{i(\pi/3)}]^{(3/4)}=e^{i\pi(1/3)(3/4)}=e^{4\pi i}=cos4\pi+isin{4\pi}=1-0i=1$ 

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $\displaystyle z=1+\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$, then

  1. $\displaystyle Re(z^{5})=\frac{\sqrt{3}}{2}$
  2. $\displaystyle Re(z^{5})=\frac{1}{2}$
  3. $\displaystyle Im(z^{5})=\frac{1}{2}$
  4. $\displaystyle Im(z^{5})=\frac{\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$z=1+\cos { \frac { 2\pi  }{ 3 }  } +i\sin { \frac { 2\pi  }{ 3 }  } =2\cos ^{ 2 }{ \frac { \pi  }{ 3 }  } +2i\sin { \frac { \pi  }{ 3 } \cos { \frac { \pi  }{ 3 }  }  } $

$\displaystyle \Rightarrow z=2\cos { \frac { \pi  }{ 3 }  } \left( \cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  }  \right) =\cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  } $

$\displaystyle \Rightarrow { z }^{ 5 }={ \left( \cos { \frac { \pi  }{ 3 }  } +i\sin { \frac { \pi  }{ 3 }  }  \right)  }^{ 5 }=\cos { \frac { 5\pi  }{ 3 }  } +i\sin { \frac { 5\pi  }{ 3 }  } $        ...{De Moivre's Theorem}
 
$\displaystyle \Rightarrow { z }^{ 5 }=\frac { 1-i\sqrt { 3 }  }{ 2 } $

$\displaystyle \therefore \quad Re\left( { z }^{ 5 } \right) =\frac { 1 }{ 2 } \quad &amp; \quad Im\left( { z }^{ 5 } \right) =\frac { -\sqrt { 3 }  }{ 2 } $
Hence, option B is correct.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

The roots of $\displaystyle \left ( -64a^{4} \right )^{\tfrac14}$ are

  1. $\displaystyle \pm 2a\left ( 1\pm i \right ).$
  2. $\displaystyle \pm a\left ( 1\pm i \right ).$
  3. $\displaystyle \pm 2a\left ( 1\pm 2i \right ).$
  4. $\displaystyle \pm a\left ( 1\pm 2i \right ).$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\displaystyle \left ( -64a^{4} \right )^{\tfrac14}= \left ( 2\sqrt{2} \right )a\left ( -1 \right )^{\tfrac14}$
We know that $\displaystyle -1= \cos \pi +i\sin \pi $
Now put $\displaystyle -1= r\cos \theta , 0= r\sin \theta $
$\displaystyle \therefore \left ( -64a^{4} \right )^{\tfrac14}= 2\sqrt{2}a.\left [ \cos \pi +i\sin \pi  \right ]^{\tfrac14}$
$\displaystyle = 2\sqrt{2a}\left [ \cos \left ( 2n\pi +\pi  \right )+i\sin \left ( 2n\pi +\pi  \right ) \right ]^{\tfrac14}$
$\displaystyle = 2\sqrt{2a}\left [ \cos \cfrac{2n\pi +\pi }{4}+i\sin \cfrac{2n\pi +\pi }{4} \right ],$
where n=0, 1, 2 and 3.Hence the required roots are
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( \cfrac{\pi}{4} \right )+i\sin \left ( \cfrac{\pi}{4} \right ) \right ],$
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( 3\cfrac{\pi}{4} \right )+i\sin \left ( 3\cfrac{\pi}{4} \right ) \right ],$
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( 5\cfrac{\pi}{4} \right )+i\sin \left ( 5\cfrac{\pi}{4} \right ) \right ],$
$\displaystyle 2\sqrt{2}a\left [ \cos \left ( 7\cfrac{\pi}{4} \right )+i\sin \left ( 7\cfrac{\pi}{4} \right ) \right ],$
Thus the roots on putting the values are
$\displaystyle 2\sqrt{2}a\left ( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right ), 2\sqrt{2}a\left (\dfrac{-1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right ),$
$\displaystyle 2\sqrt{2}a\left ( \dfrac{-1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right ), 2\sqrt{2}a\left (\dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right ).$
Hence the roots are $\displaystyle \pm 2a\left ( 1\pm i \right ).$

Ans: $A$
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

The value of $\displaystyle \left ( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right )^{8}$

  1. -1

  2. 1

  3. 0

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

$z={ \left( \sin { \frac { \pi  }{ 8 } +i } \cos { \frac { \pi  }{ 8 }  }  \right)  }^{ 8 }={ \left[ i\left( \cos { \frac { \pi  }{ 8 } -i\sin { \frac { \pi  }{ 8 }  }  }  \right)  \right]  }]^8$
     ...{$\because \quad { i }^{ 8 }=1$}
$\Rightarrow z=\cos { \pi -i\sin { \pi  }  } =-1$        ...{De Moivre's Theorem}
Hence, option 'A' is correct.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $z = \left(\displaystyle\frac{\sqrt3}{2}+\displaystyle\frac{i}{2}\right)^5 + \left(\displaystyle\frac{\sqrt3}{2}-\displaystyle\frac{i}{2}\right)^5,$ then

  1. $Re(z) = 0$
  2. $Im(z) = 0$
  3. $Re(z) > 0, \space Im(z) > 0$
  4. $Re(z) > 0, \space Im(z) < 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As we know that,
$\dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  } $
and $\dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 } =\cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  } $

$z=\left( \dfrac { \sqrt { 3 }  }{ 2 } +\dfrac { i }{ 2 }  \right) ^{ 5 }+\left( \dfrac { \sqrt { 3 }  }{ 2 } -\dfrac { i }{ 2 }  \right) ^{ 5 }$

$\Rightarrow z=\left( \cos { \dfrac { \pi  }{ 6 }  } +i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 5 }+\left( \cos { \dfrac { \pi  }{ 6 }  } -i\sin { \dfrac { \pi  }{ 6 }  }  \right) ^{ 5 }$         ......{ De Moivre's Theorem}

$\Rightarrow z=\cos { \dfrac { 5\pi  }{ 6 }  } +i\sin { \dfrac { 5\pi  }{ 6 }  } +\cos { \dfrac { 5\pi  }{ 6 }  } -i\sin { \dfrac { 5\pi  }{ 6 }  } $

$\Rightarrow z=-\dfrac { \sqrt { 3 }  }{ 2 } $
Therefore, $Im(z)=0$

Ans: B

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

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

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

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

$\displaystyle\alpha =\cos { \left( \frac { 8\pi  }{ 11 }  \right)  } +i\sin { \left( \frac { 8\pi  }{ 11 }  \right)  } $


We know, $z=\cos\theta+i\sin\theta=e^{i\theta}$
$\therefore$  $\alpha=e^{i\frac{8\pi}{11}}$
$\Rightarrow$  $\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5=\dfrac{\alpha(\alpha^5-1)}{\alpha-1}$             ........................... as forming G.P.

                                                  $=\dfrac{\alpha^6-\alpha}{\alpha-1}$

                                                  $=\dfrac{\left(e^{i\frac{8\pi}{11}}\right)^6-e^{i\frac{8\pi}{11}}}{e^{i\frac{8\pi}{11}}-1}$           ---- ( 1 )

$e^{-\frac{48\pi}{11}}=\cos\dfrac{48\pi}{11}+i\sin\dfrac{48\pi}{11}$

          $=\cos\left(4\pi+\dfrac{4\pi}{11}\right)+i\sin\left(4\pi+\dfrac{4\pi}{11}\right)$

          $=\cos\dfrac{4\pi}{11}+i\sin\dfrac{4\pi}{11}$

          $=e^{i\frac{4\pi}{11}}$

Substituting above value in ( 1 ) we get,
$\Rightarrow$  $\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5=\dfrac{e^{i\frac{4\pi}{11}}-e^{i\frac{8\pi}{11}}}{e^{i\frac{8\pi}{11}}-1}$

                                                  $=\dfrac{t-t^2}{t^2-1}$                       [ Let $e^{i\frac{4\pi}{11}}=t]$

                                                  $=\dfrac{-t(1-t)}{(t-1)(t+1)}$

                                                  $=\dfrac{-t}{t+1}$

                                                  $=\dfrac{-(\cos\frac{4\pi}{11}+i\sin\dfrac{4\pi}{11})}{\cos\dfrac{4\pi}{11}+i\sin\dfrac{4\pi}{11}+1}$

Let $a=\cos\dfrac{4\pi}{11}=a$ and $b=\sin\dfrac{4\pi}{11}$

                                                  $=\left(\dfrac{a+ib}{(a+1)+ib}\right)\times\dfrac{(a+1)-ib}{(a+1)-ib}$

                                                   $=-\dfrac{(1+ib)(a+1)-ib}{[(a+1)+ib][(a+1)-ib]}$

                                                   $=-\dfrac{(a(a+1)+b^2)+i(b(a+1)-ab)}{(a+1)^2+b^2}$

                                                   $=-\dfrac{a(a+1)+b^2}{(a+1)^2+b^2}$           [ Taking real part only ]

                                                   $=-\dfrac{a^2+b^2+a}{a^2+b^2+1+2a}$

                                                   $=-\dfrac{1+a}{2+2a}$

                                                   $=-\dfrac{(1+a)}{2(1+a)}$

                                                   $=\dfrac{-1}{2}$

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $\left ( 2+z \right )^{6}+\left ( 2-z \right )^{6}=0$ and $\omega =\dfrac{2+z}{2-z}$

  1. $\displaystyle \omega =e^{i}\tfrac{\left (2p+1 \right )\pi }{6},p=0,1,2,3,4,5$
  2. $\displaystyle z=\frac{2\left ( \omega -1 \right )}{\omega +1}$
  3. $\displaystyle \omega = ( -1 )^(\frac{1}{6})$

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

$\displaystyle { \left( 2+z \right)  }^{ 6 }{ +\left( 2-z \right)  }^{ 6 }=0\quad \quad &amp; \quad w=\frac { 2+z }{ 2-z } $

$\displaystyle \Rightarrow { \left( \frac { 2+z }{ 2-z }  \right)  }^{ 6 }=-1\ \Rightarrow { w }^{ 6 }=-1$

$\displaystyle { \therefore \quad w=\left( -1 \right)  }^{ \frac { 1 }{ 6 }  }$

$\displaystyle \because \quad \frac { 2+z }{ 2-z } =w\ \Rightarrow 2\left( w-1 \right) =z\left( w+1 \right) $

$\displaystyle \therefore \quad z=\frac { 2\left( w-1 \right)  }{ w+1 } $

$\displaystyle { \because \quad w=\left( -1 \right)  }^{ \frac { 1 }{ 6 }  }$

$\displaystyle w={ \left( \cos { \pi  } +i\sin { \pi  }  \right)  }^{ \frac { 1 }{ 6 }  }=\cos { \left( \frac { 2p\pi +\pi  }{ 6 }  \right) +i } \sin { \left( \frac { 2p\pi +\pi  }{ 6 }  \right)  } $       ..{De Moivre's Theorem}

Where$ p=0,1,2,3,4,5.$

$\displaystyle \Rightarrow w={ e }^{ i\frac { \left( 2p+1 \right) \pi  }{ 6 }  }$
Hence, option 'D' is correct.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

For positive integers $\displaystyle n _{1}$ and $\displaystyle n _{2}$ the value of the expression $\displaystyle (1+i)^{n _{1}}+(1+i^{3})^{n _{1}}+(1+i^{5})^{n _{2}}+(1+i^{2})^{n _{2}}$ where
$\displaystyle i= \sqrt{-1}$ is a real number iff

  1. $\displaystyle n _{1}= n _{2}$
  2. $\displaystyle n _{2}= n _{2}-1$
  3. $\displaystyle n _{1}= n _{2}+1$
  4. $\displaystyle \forall n _{1}$ and $\displaystyle n _{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(1+i)^{n _{1}}  = $ $  ^{n _{1}}C _{0} + ^{n _{1}}C _{1} i +^{n _{1}}C _{2} i^2 + ......+^{n _{1}}C _{n _{1}} i^{n _{1}}$ --------(1)


$(1+i^3)^{n _{1}}  =(1-i)^{n _{1}}=  $ $ ^{n _{1}}C _{0} - ^{n _{1}}C _{1} i +^{n _{1}}C _{2} i^2 - ......+^{n _{1}}C _{n _{1}} i^{n _{1}}$--------(2)

$(1+i^5)^{n _{2}}  =(1+i)^{n _{2}}=  $ $ ^{n _{2}}C _{0} + ^{n _{2}}C _{1} i +^{n _{2}}C _{2} i^2 + ......+^{n _{2}}C _{n _{2}} i^{n _{2}}$--------(3)

$(1+i^7)^{n _{2}}  =(1-i)^{n _{2}}=  $ $ ^{n _{2}}C _{0} - ^{n _{2}}C _{1} i +^{n _{2}}C _{2} i^2 - ......+^{n _{2}}C _{n _{2}} i^{n _{2}}$--------(4)

Adding (1),(2),(3) and (4),

$(1+i)^{n _{1}} +(1+i^3)^{n _{1}} +(1+i^5)^{n _{2}} +(1+i^7)^{n _{2}}$ 
$= 2(^{n _{1}}C _{0} +^{n _{1}}C _{2}i^2 +^{n _{1}} C _{4}i^4 +...........)+2(^{n _{2}}C _{0} +^{n _{2}}C _{2}i^2 +^{n _{2}} C _4 i^4+...........)$

$ = 2(^{n _{1}}C _{0} -^{n _{1}}C _{2} +^{n _{1}} C _{4} +...........)+2(^{n _{2}}C _{0} -^{n _{2}}C _{2} +^{n _{2}} C _4 +...........)$
$\Rightarrow$  As there are only even powers of $i$, the expression is real for all  positive integers $n _{1}$ and $n _{2}$

Multiple choice maths average arithmetic mean of ap introduction to averages means

If AM between $\displaystyle p^{th}$ and $\displaystyle q^{th}$ terms of an AP be equal to the AM between $\displaystyle r^{th}$ and $\displaystyle s^{th}$ term of the AP, then $p + q$ is equal to

  1. $r + s$
  2. $\displaystyle \frac{r-s}{r+s}$
  3. $\displaystyle \frac{r+s}{r-s}$
  4. $r + s + 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know A.P formula for nth terms with 'a' as the first term and 'd' as the common difference as shown below:


${ t } _{ n }=a+\left( n-1 \right) d$

Also AM is given between two numbers a and b. 
       $A=\dfrac { a+b }{ 2 } $

So arithmetic mean of pth and qth terms of AP is as shown below:

$=\dfrac { a+\left( p-1 \right) d+a+\left( q-1 \right) d }{ 2 } $

Similarly we can have AM of rth term and sth term of AP as shown below:

$=\dfrac { a+\left( r-1 \right) d+a+\left( s-1 \right) d }{ 2 } $

Applying the given conditions we get,

$\dfrac { a+\left( p-1 \right) d+a+\left( q-1 \right) d }{ 2 } =\dfrac { a+\left( r-1 \right) d+a+\left( s-1 \right) d }{ 2 } $

      $\dfrac { a+pd-d+a+qd-d }{ 2 } =\dfrac { a+rd-d+a+sd-d }{ 2 } $

$a+pd-d+a+qd-d=a+rd-d+a+sd-d$

         $2a+d\left( p+q \right) -2d=2a+d\left( r+s \right) -2d$

                           $d\left( p+q \right) =d\left( r+s \right) d$

                                 $p+q=r+s$ 

Hence option A is correct.

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If the $m^{th}$ term and the $n^th$ term of an AP are respectively $\displaystyle \frac { 1 }{ n } $ and $\displaystyle \frac { 1 }{ m } $, then the $mn^{th}$ term of the AP is

  1. $\displaystyle \frac { 1 }{ mn } $
  2. $\displaystyle \frac { m }{ n } $
  3. $\displaystyle 1$
  4. $\displaystyle \frac { n }{ m } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let a and d be the first term and common difference of an AP.
Since, $\displaystyle { T } _{ m }=\frac { 1 }{ n } $
$\displaystyle \therefore a+\left( m-1 \right) d=\frac { 1 }{ n } $....(i)
and $\displaystyle { T } _{ n }=\frac { 1 }{ m } $
$\displaystyle \Rightarrow a+\left( n-1 \right) d=\frac { 1 }{ m } $.....(ii)
On solving Eqs. (i) and (ii), we get
$\displaystyle a=\frac { 1 }{ mn } and\quad d=\frac { 1 }{ mn } $
$\displaystyle \therefore \quad { T } _{ mn }=a+\left( mn-1 \right) d$


$\displaystyle =\frac { 1 }{ mn } +\frac { \left( mn-1 \right)  }{ mn } $

$\displaystyle =\frac { mn }{ mn } =1$