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 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.

Multiple choice maths ratio, proportion and unitary method more on proportion terms related to proportion proportion

The third proportional to $(x^2\, -\, y^2)$ and $(x - y)$ is

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

Let third proportional is x
$\left ( x^{2}-y^{2} \right ):\left ( x-y \right )=\left ( x-y \right ):x$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left ( x^{2}-y^{2} \right )}$
$\Rightarrow x=\frac{\left (x-y  \right )\left (x-y  \right )}{\left (x+y  \right )\left (x-y  \right )}$
$\Rightarrow x=\frac{\left ( x-y \right )}{\left (x+y  \right )}$
 

Multiple choice maths ratio, proportion and unitary method more on proportion terms related to proportion proportion

Find the third proportional to $\displaystyle (x^{2}-y^{2}): and: (x+y)$.

  1. $\displaystyle \frac{x+y}{x-y}$
  2. $x-y$
  3. $\displaystyle \frac{x-y}{x+y}$

  4. 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the third proportional to $\displaystyle (x^{2}-y^{2}): and: (x+y)$ be A.

Then,
$\displaystyle \left ( x^{2}-y^{2} \right ):\left ( x+y \right )::\left ( x+y \right ):A$
$\displaystyle \Rightarrow (x^{2}-y^{2})\times A=(x+y)^{2}$

$\Rightarrow A=\cfrac{(x+y)^{2}}{x^{2}-y^{2}}$
$\Rightarrow A=\cfrac{(x+y)^{2}}{(x+y)(x-y)}=\cfrac{x+y}{x-y}$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

Let $z$ be any point in $\displaystyle A\cap B\cap C$ and let $w$ be any point satisfying $\displaystyle \left | w-2-i \right |< 3.$ Then, $\displaystyle \left | z \right |-\left | w \right |+3$ lies between

  1. $-6$ and $3$
  2. $-3$ and $6$
  3. $-6$ and $6$
  4. $-3$ and $9$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\left| w-\left( 2+i \right)  \right| <3\Rightarrow \left| w \right| -\left| 2+i \right| <3\ \Rightarrow -3+\sqrt { 5 } <\left| w \right| <3+\sqrt { 5 } $
$\Rightarrow -3-\sqrt { 5 } <-\left| w \right| <3-\sqrt { 5 } $   ...(1)
Also, $\left| z-\left( 2+i \right)  \right| =3$
$\Rightarrow -3+\sqrt { 5 } <-\left| z \right| \le 3+\sqrt { 5 } $   ...(2)
$\therefore -3<\left| z \right| -\left| w \right| +3<9$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $z=a+ib$ where $a>0,b>0$, then

  1. $\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 } } \left( a-b \right) $
  2. $\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 } } \left( a+b \right) $
  3. $\displaystyle \left| z \right| < \frac { 1 }{ \sqrt { 2 } } \left( a+b \right) $

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

As ${ \left( a-b \right)  }^{ 2 }\ge 0,{ a }^{ 2 }+{ b }^{ 2 }\ge 2ab$   ...(1)

But $\left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ;$ si from (1), ${ \left| z \right|  }^{ 2 }\ge 2ab$
$\therefore { \left| z \right|  }^{ 2 }+{ a }^{ 2 }+{ b }^{ 2 }\ge { a }^{ 2 }+{ b }^{ 2 }+2ab\ \Rightarrow { \left| z \right|  }^{ 2 }+{ \left| z \right|  }^{ 2 }\ge { \left( a+b \right)  }^{ 2 }\Rightarrow 2{ \left| z \right|  }^{ 2 }\ge { \left( a+b \right)  }^{ 2 }$
$\Rightarrow \sqrt { 2 } \left| z \right| \ge a+b$ as $\left| z \right| $ is positive
$\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\displaystyle \left | z-\frac{2}{z} \right |=1$, then the greatest value of $\left | z \right |$ is 

  1. 2

  2. 1

  3. 4

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\left| z-\frac { 2 }{ z }  \right| =1$        ...(1)
Let $ \dfrac{2}{z}=w$

$\left| z \right| =\left| \left( z-w \right) +w \right| \le \left| z-w \right| +\left| w \right| $      ..(Triangle inequality)

$\Rightarrow \left| z \right| -\left| w \right| \le \left| z-w \right| $

$\Rightarrow \left| z \right| -\left| \frac { 2 }{ z }  \right| \le \left| z-\frac { 2 }{ z }  \right| $

$\Rightarrow \left| z \right| -\left| \frac { 2 }{ z }  \right| \le 1$         ...{ from 1 }

$\Rightarrow { \left| z \right|  }^{ 2 }-\left| z \right| -2\le 0\ \Rightarrow -1\le \left| z \right| \le 2\ \Rightarrow 0\le \left| z \right| \le 2$

Therefore,  maximum value of $\left| z \right| $ is 2
Hence, option A is correct.