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 business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

If $\Delta (x)=\left| \begin{matrix} 1+x+2{ x }^{ 2 } & x+3 & 1 \ x+2{ x }^{ 2 } & x & 3 \ 3x+6{ x }^{ 2 } & 3x+11 & 9 \end{matrix} \right| $ then $\displaystyle \int^{1} _{0}\Delta (x)dx$ is

  1. $\dfrac {176}{5}$
  2. $-\dfrac {176}{3}$
  3. $\dfrac {186}{3}$
  4. $-\dfrac {192}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Expanding the determinant Delta(x) and then integrating term by term from 0 to 1 yields 176/5.

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

If $\displaystyle {\sec}^{2}A\hat{i}+\hat{j}+\hat{k}$, $\displaystyle \hat{i}+{\sec}^{2}B\hat{j}+\hat{k}$,and $\displaystyle \hat{i}+\hat{j}+{\sec}^{2}C\hat{k}$, are coplanar then $\displaystyle {\cot}^{2}A+{\cot}^{2}B+{\cot}^{2}{C}$ is    

  1. $1$
  2. $2$
  3. $0$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ { { \sec   }^{ 2 } }A } & 1 & 1 \ 1 & { { { \sec   }^{ 2 } }B } & 1 \ 1 & 1 & { { { \sec   }^{ 2 } }C } \end{array} } \right| =0 \\ { C _{ 1 } }\to { C _{ 1 } }-{ C _{ 2 } }\, \, \, \, \, \, \, \, \, { C _{ 2 } }\to { C _{ 2 } }-{ C _{ 3 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ { { \tan   }^{ 2 } }A } & 0 & 1 \ { -{ { \tan   }^{ 2 } }B } & { { { \tan   }^{ 2 } }B } & 1 \ 0 & { -{ { \tan   }^{ 2 } }C } & { { { \sec   }^{ 2 } }C } \end{array} } \right| =0 \\ { \tan ^{ 2 }  }A\left[ { { { \tan   }^{ 2 } }B{ { \sec   }^{ 2 } }C+{ { \tan   }^{ 2 } }C } \right] +{ \tan ^{ 2 }  }B{ \tan ^{ 2 }  }C=0 \ \\dfrac { { { { \sec   }^{ 2 } }C } }{ { { { \tan   }^{ 2 } }C } } +{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }A=0 \ \\cos  e{ c^{ 2 } }C+{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }A=0 \\ { \cot ^{ 2 }  }A+{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }C=-1 \\ Hence,\, the\, option\, D\, is\, \, the\, correct\, answer. \end{array}$

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

If $\displaystyle \left ( 3, : \lambda, : \mu \right )$ is a point on the line then $\displaystyle 2x + y + z = 0 = x - 2y + z -1$ then

  1. $\displaystyle \lambda = \frac{-8}{3}, \: \mu = - \frac{1}{3}$
  2. $\displaystyle \lambda = \frac{-1}{3}, \: \mu = - \frac{8}{3}$
  3. $\displaystyle \lambda = \dfrac{-4}{3} \: \mu = \dfrac{-14}{3}$
  4. $\displaystyle \lambda = -5, \: \mu = -1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the point lies on the both planes we have, 

$6+\lambda+\mu=0$       ...(1)
$3-2\lambda+\mu-1=0$           ....(2) 
Subtracting equation 2 from 1 we get $\lambda=\cfrac{-4}{3}$. 

And substituting on equation 1 we get $\mu=\cfrac{-14}{3}$.

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

The sum $\displaystyle\sum _{ 0\le i }^{  }{ \sum _{ j\le 10 }^{  }{ \left( _{  }^{ 10 }{ { C } _{ j } } \right) \left( _{  }^{ j }{ { C } _{ i } } \right)  }  } $ is equal to 

  1. $2^{10}-1$
  2. $2^{10}$
  3. $3^{10}-1$
  4. $3^{10}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\displaystyle\sum _{ 0\le i }^{  }{ \sum _{ j\le 10 }^{  }{ \left( _{  }^{ 10 }{ { C } _{ j } } \right) \left( _{  }^{ j }{ { C } _{ i } } \right)  }  } $

$= ^{10}C _{0}(^0C _0) + ^{10}C _1(^1C _1+^1C _0) + ……. + ^{10}C _{10}(^{10}C _{10}+^{10}C _{9}+...…+^{10}C _{0})$

$=2^0.^{10}C _{0} + 2^1.^{10}C _{1} + …….. + 2^{10}.^{10}C _{10}$

Now,


$(1+2x)^{10} = ^{10}C _{0}(1)^{10} + ^{10}C _1(1)^9(2x)^1 + ………….+^{10}C _{10}(2x)^{10} $

Put x = 1,

$(3)^{10} = ^{10}C _{0}(1)^{10} + ^{10}C _1(1)^9(2)^1 + ………….+^{10}C _{10}(2)^{10} $

Thus the required sum is $3^{10}$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $\displaystyle a\times b=a\times c,a\neq 0,$ then

  1. $\displaystyle b=c+\lambda a$
  2. $\displaystyle c=a+\lambda b$
  3. $\displaystyle a=b+\lambda c$

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

$\displaystyle a\times b=a\times c\Rightarrow a\times b-a\times c=0$
$\Rightarrow a\times (b-c)=0$ $\Rightarrow a \parallel (b-c)$
$\Rightarrow b-c = \lambda a\Rightarrow b= c+\lambda a$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

$\displaystyle a\times \left ( b+c \right )+b\times \left ( c+a \right )+c\times \left ( a+b \right )$ is equal to

  1. $\displaystyle 2\left [ a\:b\:c \right ]$
  2. $0$
  3. $3$

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

$\displaystyle a\times \left ( b+c \right )+b\times \left ( c+a \right )+c\times \left ( a+b \right )$
$=a\times b+a\times c+b\times c+b\times a+c\times a+c\times b$
$=a\times b-c\times a+b\times c-a\times b+c\times a-b\times c = 0$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

Let $\displaystyle a=i+j$ and $\displaystyle b=2i-k,$ the point of intersection of the lines $\displaystyle r\times a=b\times a $ and $\displaystyle r\times b=a\times b $ is

  1. $\displaystyle -i+j+k$
  2. $\displaystyle 3i-j+k$
  3. $\displaystyle 3i+j-k$
  4. $\displaystyle i-j-k$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\vec r \times \vec a = \vec b \times \vec a\Rightarrow \vec{r} = \vec{b}+\lambda \vec{a}$
and $\vec r \times \vec b = \vec a \times \vec b\Rightarrow \vec{r} = \vec{a}+\mu \vec{b}$
For intersection of both the lines, $\vec{b}+\lambda \vec{a}=\vec{a}+\mu \vec{b}$
Comparing coefficients, $\lambda=\mu = 1$
Hence point of intersection is $\vec{r}=\vec{a}+\vec{b} = 3\hat{i}+\hat{j}-\hat{k}$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $\displaystyle \bar{a}+p\bar{b}+q\bar{c}=0 $ then

  1. $\displaystyle p(\bar{a}\times\bar{b})=pq(\bar{b}\times\bar{c})=q(\bar{c}\times\bar{a})$
  2. $\displaystyle \bar{a}\times\bar{b}=pq(\bar{c}\times\bar{a})$
  3. $\displaystyle \bar{c}\times\bar{a}=p(\bar{a}\times\bar{b})$
  4. $\displaystyle \bar{a}\times\bar{c}=q(\bar{b}\times\bar{c})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\bar{a}+p\bar{b}+q\bar{c}=0$
$\bar{b}=\dfrac{-q\bar{c}-\bar{a}}{p}$

Then
$p(\bar{a}\times b)$
$=p(\bar{a}\times (\dfrac{-q\bar{c}-\bar{a}}{p}))$

$=((q\bar{c}+\bar{a})\times \bar{a})$
$=q(\bar{c}\times \bar{a})$
$=q(\bar{c}\times (-p\bar{b}-q\bar{c}))$
$=q((p\bar{b}+q\bar{c})\times \bar{c})$
$=pq(\bar{b}\times \bar{c})$.

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $\displaystyle a\cdot b=a\cdot c$ and $\displaystyle a\times b=a\times c,$ then

  1. either $\displaystyle a=0$ or $\displaystyle b=c$
  2. $a$ is parallel to $\displaystyle \left ( b-c \right )$
  3. $a$ is perpendicular to $\displaystyle \left ( b-c \right )$

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

Given,  $\displaystyle a\cdot b=a\cdot c$ and $\displaystyle a\times b=a\times c,$
$\Rightarrow a\cdot (b-c) = 0$ and $a\times (b-c) = 0$
Hence  either $a=0$ or $b=c$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If in $\displaystyle \left[ \begin{matrix} { a } _{ 1 } \ { a } _{ 2 } \ { a } _{ 3 } \end{matrix}\begin{matrix} { b } _{ 1 } \ { b } _{ 2 } \ { b } _{ 3 } \end{matrix}\begin{matrix} { c } _{ 1 } \ { c } _{ 2 } \ { c } _{ 3 } \end{matrix} \right] $, the cofactor of $\displaystyle { a } _{ r }$ is $\displaystyle { A } _{ r }$, then $\displaystyle { c } _{ 1 }{ A } _{ 1 }+{ c } _{ 2 }{ A } _{ 2 }+{ c } _{ 3 }{ A } _{ 3 }$ is 

  1. $\displaystyle 0$
  2. $\displaystyle -D$
  3. $\displaystyle D$
  4. $\displaystyle { D }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A\begin{bmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \\ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \\ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{bmatrix}$
Co-factor$\begin{bmatrix} { A } _{ 1 } & { B } _{ 1 } & { C } _{ 1 } \\ { A } _{ 2 } & { B } _{ 2 } & { C } _{ 2 } \\ { A } _{ 3 } & { B } _{ 3 } & { C } _{ 3 } \end{bmatrix}$
${ A } _{ 1 }={ b } _{ 2 }{ c } _{ 3 }-{ c } _{ 2 }{ b } _{ 3 }$
${ A } _{ 2 }=-\left( { b } _{ 1 }{ c } _{ 3 }-{ c } _{ 1 }{ b } _{ 3 } \right) $
${ A } _{ 3 }={ b } _{ 1 }{ c } _{ 2 }-{ c } _{ 1 }{ b } _{ 2 }$
${ C } _{ 1 }{ A } _{ 1 }+{ C } _{ 2 }{ A } _{ 2 }+{ C } _{ 3 }{ A } _{ 3 }$
${ c } _{ 1 }{ c } _{ 3 }{ b } _{ 2 }-{ c } _{ 1 }{ c } _{ 2 }{ b } _{ 3 }-{ c } _{ 2 }{ c } _{ 3 }{ b } _{ 1 }+{ c } _{ 1 }{ c } _{ 2 }{ b } _{ 3 }$
${ c } _{ 2 }{ c } _{ 3 }{ b } _{ 1 }-{ c } _{ 1 }{ c } _{ 3 }{ b } _{ 2 }=0$
Option A
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ then $\displaystyle \frac{dy}{dx}$ is equal to

  1. $\dfrac{y}{x}$
  2. $\dfrac{x}{y}$
  3. $\displaystyle\dfrac{ x^{2}}{y^{2}}$
  4. $\displaystyle\dfrac{ y^{2}}{x^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ 
$\Rightarrow \displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\cos \log a=A$ (say)
Putting $u=\dfrac{y}{x}$ and applying componendo and dividendo, we have
$\left ( \dfrac{y}{x} \right )^{2}=u^{2}=\left ( 1-A \right )\left ( 1+A \right )$
$\Rightarrow \dfrac{y}{x}=\sqrt{\left ( 1-A \right )\left ( 1+A \right )}\Rightarrow x\dfrac{dy}{dx}-y=0$
$\Rightarrow $   $\dfrac{dy}{dx}=\dfrac{y}{x}$
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $\displaystyle y=\sec(\tan^{-1}x)$, then $\displaystyle \frac {dy}{dx}$ at $x=1$ is equal to

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

Given, $\displaystyle y=\sec(\tan^{-1}x)=\sqrt {1+x^{2}}$
$\Rightarrow \displaystyle \frac {dy}{dx}=\frac {x}{\sqrt {1+x^{2}}}$
$\displaystyle\therefore  \left.\begin{matrix}\frac {dy}{dx}\end{matrix}\right| _{x=1}=\frac {1}{\sqrt {2}}$

Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

If 'a' is  x % more than 'b' and 'b' is y % less than 'a'. then the relation between x and y is

  1. $\displaystyle \frac{1}{x}\, +\, \displaystyle \frac{1}{y}\, =\, \displaystyle \frac{1}{100}$
  2. $\displaystyle \frac{1}{y}\, -\, \displaystyle \frac{1}{x}\, =\, \displaystyle \frac{1}{100}$
  3. $\displaystyle \frac{1}{x}\, -\, \displaystyle \frac{1}{y}\, =\, 100$
  4. $\displaystyle \frac{1}{y}\, -\, \displaystyle \frac{1}{x}\, =\, 100$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$y\, \%\, =\, \displaystyle \frac{100\, \times\, x}{100\, +\, x} \%$


$\Rightarrow y\, =\, \displaystyle \frac{100\, \times\, x}{100\, +\, x}$

$\Rightarrow \displaystyle \frac{1}{y}\, =\, \displaystyle \frac{100\, +\, x}{100\, \times\, x}\, =\, \displaystyle \frac{1}{x}\, =\, \displaystyle \frac{1}{100}$

$\Rightarrow \displaystyle \frac{1}{y}\, -\, \displaystyle \frac{1}{x}\, =\, \displaystyle \frac{1}{100}$