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 lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

If the expression $ \displaystyle (x+y)^{-1}. (x^{-1}+y^{-1})(xy^{-1}+x^{-1}y)^{-1} $ is simplefied it takes the form of which one of the following ?

  1. x+y
  2. $ \displaystyle( x^{2}+y^{2})^{-1}$

  3. xy
  4. $ \displaystyle x^{2}+y^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\left ( x+y \right )^{-1}\left ( \frac{1}{x}+\frac{1}{y} \right )^{-1}\left ( xy^{-1}+x^{-1}y \right )^{-1}$
$\frac{1}{x+y}\left ( \frac{1}{x}+\frac{1}{y} \right )\left ( \frac{x}{y}+\frac{y}{x}^{-1} \right )$
=$\frac{1}{x+y}\left ( \frac{x+y}{xy} \right )\left ( \frac{x^{2}+y^{2}}{xy} \right )^{-1}$
=$\frac{1}{x+y}\times \frac{x+y}{xy}\times \frac{xy}{x^{2}+y^{2}}$
=$\left ( \frac{1}{x^{2}+y^{2}} \right )=(x^{2}+y^{2})^{-1}$
Multiple choice mathematics and statistics set language de morgan's law for set theory complement of sets different sets de morgan's law

If $U = {3, 4, 5, 6, 7, 8, 9}, X = {3, 4}, Y = {5, 6}$ and $Z = {7, 8, 9}$, then $\displaystyle Y'\cap \left ( X\cap Z \right )'$ is equal to 

  1. $\displaystyle X\cup Y$
  2. $\displaystyle Y\cup Z$
  3. $\displaystyle X'\cap Y'$
  4. $X\cup Z$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ { Y ' } = U - Y = { 3, 4, 7, 8, 9} $

$ X\cap Z = {0} $ as there is no element common in sets X and Z

$\Rightarrow (X\cap Z) ' = U - (X\cap Z) =  {3, 4, 5, 6, 7, 8, 9} $

Now, $Y' \cap (X\cap Z)'  = { 3, 4, 7, 8, 9} \cap {3, 4, 5, 6, 7, 8, 9} ={ 3, 4, 7, 8, 9} $

This is equal to $ X \cup Z = {3,4,7,8,9} $

Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

$y _1 = A cos (2f _1t)$     and $y _2 = A cos (2 f _2t),$, then $y _{total} $ is

  1. $y _{total} = y _1 + y _2 = A {cos (2 f _1t) + cos (2 f _2t)}$
  2. $y _{total} = y _1 - y _2 = A {cos (2 f _1t) - cos (2 f _2t)}$
  3. $y _{total} =\dfrac{ y _1}{ y _2} = A\dfrac{{cos (2 f _1t)}}{{cos (2 f _2t)}}$
  4. $y _{total} = y _1 \times y _2 = A {cos (2 f _1t) \times cos (2 f _2t)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Resultant  $(y _{total})$ of the two waves is equal to the superposition of the waves and is given by,
$y _{total}  = y _1+y _2$
$\therefore$  $y _{total} = A \ cos(2f _1t)+ A \ cos(2f _2t)$

Multiple choice combining transformations transformations vectors and transformations maths

Let $\displaystyle A=(1,0)$ and $\displaystyle B=(2,1).$ The line $AB$ turns about $A$ through an angle $ \dfrac{\pi}6$ in the clockwise sense, and the new position of $B$ is $B'$. Then $B'$ has the coordinates

  1. $\displaystyle \left ( \frac{3+\sqrt{3}}{2},\frac{\sqrt{3}-1}{2} \right )$
  2. $\displaystyle \left ( \frac{3\sqrt{3}}{2},\frac{\sqrt{3}+1}{2} \right )$
  3. $\displaystyle \left ( \frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2} \right )$

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

Given, points are $A=(1,0)$ and $B=(2,1)$

Slope of $AB=\cfrac { 1-0 }{ 2-1 } =1$
Then angle of $AB$ with $x$-axis is 
$\angle BAX={ 45 }^{ 0 }$
Hence, $\angle B'AX={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$
Therefore for $B'\left( h,k \right) $
$h=1+\sqrt{2}\cos{ 15 }^{ 0 },k= \sqrt{2}\sin{ 15 }^{ 0 }\$
We have, $\sin \left(15^{0} \right) = \dfrac{\sqrt 6 -\sqrt 2 }{4} $
$\Rightarrow \cos \left (15^{0} \right) = \dfrac{\sqrt6 + \sqrt 2 }{4}$
$ \Rightarrow h=\cfrac { 3+\sqrt { 3 }  }{ 2 } ,k=\cfrac { \sqrt { 3 } -1 }{ 2 } $

Multiple choice maths drawing of different geometrical figures constructing perpendicular lines perpendicular to a line from an external point constructing an perpendicular line constructing a perpendicular bisector construction of a perpendicular bisector construction of penpendicual bisector set squares

$\displaystyle \overleftrightarrow {PQ}$ is perpendicular to $\displaystyle \overleftrightarrow {RS}$ is symbolically written as:

  1. $\displaystyle \overleftrightarrow {PQ}\perp \overleftrightarrow {RS}$
  2. $\displaystyle \overleftrightarrow {PQ}\parallel \overleftrightarrow{RS}$
  3. $\displaystyle \overleftrightarrow {PQ}\neq \overleftrightarrow{RS}$
  4. $\displaystyle \overleftrightarrow{PQ}= \overleftrightarrow {RS}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \overleftrightarrow { PQ } $ is perpendicular to $ \overleftrightarrow { RS } $ is symbolically written as $\overleftrightarrow { PQ } \bot  \overleftrightarrow { RS } $

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If the lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$ are concurrent then the value of m is

  1. $\displaystyle \frac{19}{5}$
  2. $\displaystyle 1$
  3. $\displaystyle \frac{5}{19}$

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

Given lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$
For concurrency,
$\begin{vmatrix} -1 & 1 & 5 \ 3 & 4 & 1 \ -m & 1 & 3 \end{vmatrix}=0$
$\Rightarrow -5+19m=0$
$\Rightarrow \displaystyle m =\frac{5}{19}$

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

if $ \displaystyle a,b,c $ as well as $ \displaystyle d,e,f $ are in G.P. with same common ratio then set of points $ \displaystyle \left ( a,d \right ),\left ( b,e \right ),\left ( c,f \right ) $ are

  1. collinear

  2. concurrent

  3. lies on a circle

  4. lie on an ellipse
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Are of triangle formed by the given points is,
$\Delta  = \cfrac{1}{2}\left|\begin{vmatrix}a&d&1\b&e&1\c&f&1\end{vmatrix}\right|$
Let common ratio is $r$
$\Rightarrow \Delta = \cfrac{1}{2}\left|\begin{vmatrix}a&d&1\ar&dr&1\ar^2&dr^2&1\end{vmatrix}\right|$
taking $a$ and $d$ common from first and second column respectively,
$\Delta =  \cfrac{ad}{2}\left|\begin{vmatrix}1&1&1\r&r&1\r^2&r^2&1\end{vmatrix}\right| = 0$, Since first and second column are same.
Hence given points are collinear.

Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

Let $\displaystyle A\left ( x _{1},y _{1} \right ),B\left ( x _{2},y _{2} \right ), C\left ( x _{3},y _{3} \right )$ be three points. Area of triangle with vertices $A, B,C$ is given by
$\displaystyle \frac{1}{2}\left | \Delta  \right |$ where,  

$\displaystyle \Delta =\begin{vmatrix}x _{1} &y _{1}  &1 \ x _{2} & y _{2}  & 1\ x _{3} &y _{3}  &1 \end{vmatrix}$.

If $\displaystyle a=BC,b=CA,c=AB$ and $\displaystyle 2s=a+b+c$, then $\displaystyle \Delta ^{2}$ equals

  1. $\displaystyle abc $
  2. $\displaystyle s(s-a)(s-b)(s-c)$
  3. $\cfrac {abc}{4} $
  4. $\displaystyle 4s(s-a)(s-b)(s-c)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given area$=\cfrac { 1 }{ 2 } \left| \triangle  \right| $

Heron's formula

Area$=\sqrt { S\left( S-a \right) \left( S-b \right) \left( S-c \right)  } $

$\cfrac { 1 }{ 2 } \left| \triangle  \right| =\sqrt { S\left( S-a \right) \left( S-b \right) \left( S-c \right)  } $

Squaring on both sides

$=\cfrac { { \left| \triangle  \right|  }^{ 2 } }{ 4 } =S\left( S-a \right) \left( S-b \right) \left( S-c \right) $

${ \left| \triangle  \right|  }^{ 2 }=4S\left( S-a \right) \left( S-b \right) \left( S-c \right) $

Option D
Multiple choice mathematics and statistics inverse of a matrix and linear equations application of determinants area of triangle and collinearity of three points applications of determinants

Let $\displaystyle A\left ( x _{1},y _{1} \right ),B\left ( x _{2},y _{2} \right ), C\left ( x _{3},y _{3} \right )$ be three points. Area of triangle with vertices $A, B,C$ is given by $\displaystyle \frac{1}{2}\left | \Delta  \right |$ where,  $\displaystyle \Delta =\begin{vmatrix}x _{1} &y _{1}  &1 \\
x _{2} & y _{2}  & 1\\
x _{3} &y _{3}  &1
\end{vmatrix}$.If $\displaystyle \triangle ABC$ is an equilateral triangle and $\displaystyle a = BC$ is a rational number, then $\displaystyle \triangle$ must be

  1. an integer

  2. a rational number

  3. an irrational number

  4. an imaginary number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $a$ is rational then $a^{ 2 }$ is also rational 
Now as $\Delta =\dfrac { \sqrt { 3 }  }{ 4 } a^{ 2 }$
Then $\Delta $ is irrational

Multiple choice maths numbers and place value face value of digit large numbers general form of number

For $Z _1=\displaystyle \sqrt[6]{\frac{1-i}{1+i\sqrt{3}}}; Z _2=\sqrt[6]{\frac{1-i}{\sqrt{3}+i}}; Z _3=\sqrt[6]{\frac{1+i}{\sqrt{3}-i}}$ which of the following holds good?

  1. $\displaystyle\sum|Z _1|^2=\frac{3}{2}$
  2. $\displaystyle|Z _1|^4+|Z _2|^4=|Z _3|^{-8}$
  3. $\displaystyle\sum|Z _1|^3+|Z _2|^3=|Z _3|^{-6}$
  4. $|Z _1|^4+|Z _2|^4=|Z _3|^8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$z _1 =\sqrt {\dfrac {1-i}{1+i\sqrt 3}}, z _2=\sqrt {\dfrac {1-i}{\sqrt 3 +i}}, z _3=\sqrt {\dfrac {1+i}{\sqrt 3-1}}$
$z _1 =\sqrt [6]{\dfrac {1-i}{1+\sqrt 3}}=\sqrt [6]{\dfrac {(1-i)(1-i\sqrt 3)}{1+3}}=\sqrt [6]{\dfrac {1(1-\sqrt 3)-i(1+\sqrt 3)}{4}}$
$|z _1|^2 =z _1 \bar {z} _1 =\sqrt [6]{\dfrac {(1-\sqrt 3)}{4}}\times \sqrt [6]{\dfrac {(1-\sqrt 3)+i(1+\sqrt 3)}{4}}$
$|z _1|^2 =\sqrt [6]{\dfrac {(1-\sqrt 3)^2 +(1+\sqrt 3)^2}{16}}$
$|z _1|^2 =\sqrt [6]{\dfrac {8}{16}}=\dfrac {1}{(2) 1/6}$
$z _2 =\sqrt [6]{\dfrac {1-i}{\sqrt 3+i}}=\sqrt [6]{\dfrac {(1-i) (\sqrt 3 -i)}{(3+1)}}=\sqrt [6]{\dfrac {(\sqrt 3-1)-i (1+\sqrt 3)}{4}}$
$|z _2|^2 =z _2 \bar {z} _2=\sqrt [6]{\dfrac {(\sqrt 3-1)-i (1+\sqrt 3)+i(1+\sqrt 3)}{4}}$
$|z _2|^2 =\sqrt [6]{\dfrac {(\sqrt 3-1)^2 +(1+\sqrt 3)^2}{16}}=\sqrt {\dfrac {8}{16}}=\dfrac {1}{(2) 1/6}$
$z _3 =z _3 \bar {z} _3 =\sqrt [6]{\dfrac {(\sqrt 3-1)+(1+\sqrt 3)}{4}\times \dfrac {(\sqrt 3-1)-i (1+\sqrt 3)}{4}}$
$=\sqrt [6]{\dfrac {(\sqrt 3-1)^2 +(1+\sqrt 3)^2}{16}}=\sqrt {\dfrac {8}{16}}=\dfrac {1}{(2)1/6}$
$|z _1|^4 =\dfrac {1}{2^{2/6}}\quad |z _2|^4 =\dfrac {1}{2^{2/6}}$
$|z _3|^8 =\dfrac {1}{2^{4/6}}\ \Rightarrow \ |z _3|^{-8}=2^{4/6}$
$\Rightarrow \ |z _1|^4 +|z _2|^4 =\dfrac {1}{2^{2/6}}+\dfrac {1}{2^{2/6}}=\dfrac {2}{2^{2/6}}=2^{4/6}$
$=|z _3|^{-8}$
so, $\boxed {|z _1|^4 +|z _2|^4 =|z _3|^{-8}}$ as
so, option $(B)$ is right.
Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

$\displaystyle \frac{\pi ^{c}}{5}$ in sexagesimal measure is _____

  1. $\displaystyle 18^{\circ}$
  2. $\displaystyle 36^{\circ}$
  3. $\displaystyle 54^{\circ}$
  4. $\displaystyle 72^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\text{Sexagesimal System}$, an angle is measured in degrees, minutes and seconds.
$ \pi = {180}^{0} $

So, $ \dfrac {\pi}{5} = \dfrac {{180}^{0}}{5} = {36}^{0}  $

Multiple choice mathematics and statistics binary operations properties of binary operations discrete mathematics sets and relations

If $\displaystyle M\cup N=N\cup R$ and $\displaystyle M\cap  N=N\cap R$  then which of the following is necessarily true?

  1. M=N

  2. N=R

  3. M=R

  4. M=N=R
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that Union and Intersection of sets is commutative.
We see that if $ R $ is replaced by $ M $, then the relations show the commutative property being satisfied.

So, $ M = R $

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

If $A $and $B$ are not disjoint, then $\displaystyle n\left( A \cup  B \right) $ is equal to

  1. $\displaystyle n\left( A \right) +n\left( B \right) $
  2. $\displaystyle n\left( A \right) +n\left( B \right) -n\left( A \cap B \right) $
  3. $\displaystyle n\left( A \right) +n\left( B \right) +n\left( A \cap B \right) $
  4. $\displaystyle n\left( A \right) .n\left( B \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle n\left( A\quad \cup \quad B \right) =n\left( A \right) +n\left( B \right) -n\left( A\quad \cap \quad B \right) $

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

Let $\displaystyle n\left ( u \right )=700,n\left ( A \right )=200, n\left ( B \right )=300, n\left (A\cap B \right )=100$, then $n\left ( A'\cap B' \right )=$

  1. $400$
  2. $600$
  3. $300$

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

$\displaystyle n \left ( A' \cap B'\right )=n\left ( A\cup  B\right

)'$ $\displaystyle =n\left ( u \right )-n\left ( A\cup B \right

)$ $\displaystyle =n\left ( u \right )-\left {n \left ( A \right

)+n\left ( B \right )-n\left ( A\cap B \right ) \right

}$ $\displaystyle =700-\left { 200+300-100 \right }=300$