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 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

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

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 brackets order operations and algebra using brackets in algebraic expressions order of operations

If $a$ and $ b $ are any two real numbers with opposite signs, which of the following is the greatest?

  1. $\displaystyle (a-b)^{2}$
  2. $\displaystyle (|a|-|b|)^{2}$
  3. $\displaystyle |a^{2}-b^{2}|$
  4. $\displaystyle a^{2}+b^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(a-b)^2=a^2+b^2-2ab$

as a and b are of oppsite sign ab<0 and -2ab>0,it means $(a-b)^2>a^2+b^2-2|a||b|=(|a|-|b|)^2>(|a2+b^2|)>(|a^2-b^2|)$