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 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.

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

The Foot of the $\displaystyle \perp$ from origin to the plane $\displaystyle 3x + 4y - 6z + 1 = 0$ is

  1. $\displaystyle - \frac {3}{61}, \frac {4}{61}, \frac {6}{61}$
  2. $\displaystyle \frac {-3}{61}, \frac {-4}{61}, \frac {-6}{61}$
  3. $\displaystyle \frac {4}{61}, \frac {-3}{61}, \frac {5}{61}$

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

Clearly direction ratios of perpendicular drawn from origin to the given plane are $3,4,-6$
Hence equation of perpendicular line to the given plane and  passing through origin is given by,
$\cfrac{x}{3}=\cfrac{y}{4}=\cfrac{z}{-6}=k$ (say)
Now let foot of perpendicular be $P(3k, 4k, -6k)$
Also this point lie in the given plane $\Rightarrow 3(3k)+4(4k)-6(-6k)+1=0\Rightarrow k = -\cfrac{1}{61}$
Hence $P \equiv \left(-\cfrac{3}{61}, -\cfrac{4}{61}, \cfrac{6}{61}\right)$

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

Let the line $\displaystyle \frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta = 0$. Then $\left ( \alpha ,\beta  \right )$ equals :

  1. $\left ( -6,7 \right )$
  2. $\left ( 5,-15 \right )$
  3. $\left ( -5,5 \right )$
  4. $\left ( 6,-17 \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line is $\displaystyle \frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}2{}$


The direction ratios of the line are $(3,-5,2)$

As the line is in the plane $x+3y-az+ \beta =0$,

We have $\left ( 3 \right )\left ( 1 \right )+\left ( -5 \right )\left ( 3 \right )+2\left ( -\alpha  \right )=0$

$\Rightarrow-12-2 \alpha =0$

$ \therefore \alpha = -6$

Again $(2,1,-2)$ lies on the plane

$\Rightarrow 2+3+2 \alpha + \beta =0$

$\Rightarrow \beta = -2 \alpha -5=12-5=7$

Hence, $\left ( \alpha ,\beta  \right )$ is $\left ( -6,7 \right )$

Multiple choice maths numbers and sequences series introduction to series introduction to sequences and series

If $\left| x \right| <1$ and $\left| y \right| <1$, the sum to infinity of the series $x+y,({ x }^{ 2 }+xy+{ y }^{ 2 }),({ x }^{ 3 }+{ x }^{ 2 }y+x{ y }^{ 2 }+{ y }^{ 3 }),.........$ is

  1. $\frac { x+y-xy }{ 1-x-y+xy } $
  2. $\frac { x+y+xy }{ 1-x-y+xy } $
  3. $\frac { x }{ 1-x } +\frac { y }{ 1-y } $
  4. $\frac { (x-y)(x+y-xy) }{ 1-x-y+xy } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is a sum of geometric series. The n-th term is (x^(n+1) - y^(n+1)) / (x-y). Summing this from n=1 to infinity gives the sum of two infinite geometric series: x/(1-x) + y/(1-y) is incorrect; the correct sum is (x+y-xy)/((1-x)(1-y)).

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

If for $n\in I, n > 10; 1+(1+x)+(1+x)^2+.....+(1+x)^n=\displaystyle\sum^n _{k=0}a _k\cdot x^k, x\neq 0$ then?

  1. $\displaystyle\sum^n _{k=0}a _k=2^{n+1}$
  2. $a _{n-2}=\dfrac{n(n+1)}{2}$
  3. $a _p > a _{p-1}$ for $p < \dfrac{n}{2}, p \in N$
  4. $(a _9)^2-(a _8)^2={^{n+2}C _{10}}({^{n+1}C _{10}}-{^{n+1}C _9})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum is a geometric series: ( (1+x)^(n+1) - 1 ) / ( (1+x) - 1 ) = ( (1+x)^(n+1) - 1 ) / x. The sum of coefficients a_k is the value of the polynomial at x=1, which is ((1+1)^(n+1) - 1) / 1 = 2^(n+1) - 1. Option A is the standard result for this series.

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

If $\displaystyle A\cap B=A$ and $\displaystyle B\cap C=B$ then $\displaystyle A\cap C$ is equal to :

  1. $B$
  2. $C$
  3. $\displaystyle B\cup C$
  4. $A$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given:-

$A\cap B=A$ and $B\cap C$
So,$A$ is subset of $B$.
B is a subset of C.Since $B\cap C =B$
$A$ is a subset of $B$ and $B$ is subset of $C$.
So, $A$ and $B$ is subset of $C$.
So, $A\cap C=A$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

$\displaystyle \left ( \frac{1 + i}{1 - i} \right )^2 + \left(\frac{1 - i}{1 + i} \right )^2$ is equal to

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

$\left(\dfrac{1+i}{1-i}\right)^2+\left(\dfrac{1-i}{1+i}\right)^2=\left[\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\right]^2+\left[\dfrac{(1-i)(1-i)}{(1+i)(1-i)}\right]^2$


$=\left[\dfrac{1+2i-1}{2}\right]^2+\left[\dfrac{1-2i-1}{2}\right]^2$

$=\dfrac{4i^2}{4}+\dfrac{4i^2}{4}=[-1]+[-1]=-2$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Let $\displaystyle \Delta =\left | \begin{matrix}a _{11} & a _{12} & a _{13}\a _{21}  &a _{22}  &a _{23} \a _{31}  &a _{32}  &a _{33} \end{matrix} \right |$ and $\displaystyle a _{pq}= i^{p+q}$ where $\displaystyle i= \sqrt{-1}.$ The value of $\displaystyle \Delta $ is 

  1. real and positive

  2. real and negative
  3. $0$

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

$\triangle =\left| \begin{matrix} { a } _{ 11 }\quad \quad  & { a } _{ 12 }\quad \quad  & { a } _{ 13 } \ { a } _{ 21 }\quad \quad  & { a } _{ 22 }\quad \quad  & { a } _{ 23 } \ { a } _{ 31 }\quad \quad  & { a } _{ 32 }\quad \quad  & { a } _{ 33 } \end{matrix} \right| \quad &amp; \quad { a } _{ pq }={ i }^{ p+q }$

$\Rightarrow \quad \triangle =\left| \begin{matrix} { i }^{ 2 }\quad \quad  & { i }^{ 3 }\quad \quad  & { i }^{ 4 } \ { i }^{ 3 }\quad \quad  & { i }^{ 4 }\quad \quad  & { i }^{ 5 } \ { i }^{ 4 }\quad \quad  & { i }^{ 5 }\quad \quad  & { i }^{ 6 } \end{matrix} \right| ={ i }^{ 2+3+4 }\left| \begin{matrix} { 1 }\quad \quad  & 1\quad \quad  & 1 \ { i }\quad \quad  & { i }\quad \quad  & { i } \ { i }^{ 2 }\quad \quad  & { i }^{ 2 }\quad \quad  & { i }^{ 2 } \end{matrix} \right| $


$=i\left| \begin{matrix} 1 & 1 & 1 \ i & i & i \ -1 & -1 & -1 \end{matrix} \right| =-i\left| \begin{matrix} 1 & 1 & 1 \ i & i & i \ 1 & 1 & 1 \end{matrix} \right| $

$\therefore \quad \triangle =0$
Hence, option 'C' is correct.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

 $1+\displaystyle \frac{x}{a _{1}}+\frac{x(x+a _{1})}{a _{1}a _{2}}+\ldots +\displaystyle \frac{x(x+a _{1})(x+a _{2}.).\cdot.\cdots\cdots\cdot(x+a _{n})}{a _{1}a _{2}...a _{n}}=$

  1. $ \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}a _{n}} \left\{x^{2}+ a _{n}x+a _{n}\right\}$
  2. $ \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}a _{n}} \left\{x^{2}+ a _{n}x\right\}$
  3. $ \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}} \left\{x^{2}+ a _{n}x-a _{n}\right\}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ 1+ \dfrac{x}{a _{1}}+\dfrac{x(x+a _{1})}{a _{1}a _{2}} + ...+ \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ = \dfrac{a _{1}+x}{a _{1}}+ \dfrac{x(x+a _{1})}{a _{1}a _{2}} + ...+ \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ = \dfrac{(x+a _{2})(x+a _{1})}{a _{1}a _{2}} + ...+ \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ =\dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}} + \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ = \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}} \left\{1 +\dfrac{x(x+a _{n})}{a _{n}}\right\}$
$ = \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}a _{n}} \left\{x^{2}+ a _{n}x+a _{n}\right\}$
Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $\displaystyle A=\begin{bmatrix} \frac{1}{2}\left ( e^{ix}+ e^{-ix}\right )&\frac{1}{2}\left ( e^{ix}- e^{-ix}\right ) \\frac{1}{2}\left ( e^{ix}- e^{-ix}\right ) &\frac{1}{2}\left ( e^{ix}+ e^{-ix}\right ) \end{bmatrix}$ then $A^{-1}$ exists

  1. for all real $x$
  2. for positive real $x$ only
  3. for negative real $x$ only

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

$\displaystyle A=\begin{bmatrix} \frac{1}{2}\left ( e^{ix}+ e^{-ix}\right )&\frac{1}{2}\left ( e^{ix}- e^{-ix}\right ) \\frac{1}{2}\left ( e^{ix}- e^{-ix}\right ) &\frac{1}{2}\left ( e^{ix}+ e^{-ix}\right ) \end{bmatrix}$

$\Rightarrow A=\begin{bmatrix} coshx&sinhx\sinhx & coshx\end{bmatrix}$

$|A|=cosh^2x-sinh^2x=1$

$\therefore A^{-1}$ exists or all $x$

Hence, option A.