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 congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\displaystyle \left | z \right |< \sqrt{3}-1 $ then $\displaystyle \left | z^{2}+2z\cos\alpha  \right | $ is

  1. less than $2$
  2. $\displaystyle \sqrt{3}+1$
  3. $\displaystyle \sqrt{3}-1 $

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

$\displaystyle \left | z^{2}+2z \cos \alpha  \right |\leq \left | z \right |^{2}+2\left | z \right |\left | \cos \alpha  \right |  \leq \left | z \right |^{2}+2\left | z \right |$

$|z^2+2z \cos \alpha|  < \left ( \sqrt{3}-1 \right )^{2}+2\left ( \sqrt{3}-1 \right ) = 3+1-2\sqrt{3}+2\sqrt{3}-2=2$

$\displaystyle \therefore \left | z^{2}+2z \cos  \alpha  \right |< 2$

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

If $\left| z  - \displaystyle \frac{1}{z}\right| = 1$ then

  1. $|z| _{max} = \displaystyle \frac {1+\sqrt 5}{2}$
  2. $|z| _{min} = \displaystyle \frac {1+\sqrt 5}{2}$
  3. $|z| _{max} =\displaystyle \frac {-1+\sqrt 5}{2}$

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

$\left| z  - \displaystyle \frac{1}{z}\right| = 1$
$|z|-\displaystyle\frac{1}{|z|}\leq |z-\displaystyle\frac{1}{z}|$
$\Rightarrow |z|-\displaystyle\frac{1}{|z|}\leq 1$
$\Rightarrow |z|^2-|z|-1\leq 0$
$\Rightarrow\displaystyle \frac {1-\sqrt 5}{2}\leq |z|\leq \frac {1+\sqrt 5}{2} $
$\therefore |z| _{max}=\displaystyle \frac {1+\sqrt 5}{2}$
Hence, option A.

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

Let $\left| { z } _{ r }-r \right| \le r$, for all $ r = 1, 2, 3, ..., n.$ Then $\left| \sum _{ r=1 }^{ n }{ { z } _{ r } }  \right| $ is less than

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

$\left| { z } _{ 1 }-1 \right| \le 1,\quad \left| { z } _{ 2 }-2 \right| \le 2,\quad \left| { z } _{ 3 }-3 \right| \le 3...\left| { z } _{ n }-n \right| \le n$
Adding these and using triangle inequality:
$\left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n }-(1+2+...n) \right| \le 1+2+...n\quad =>\quad \left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n }-\left(\dfrac { n(n+1) }{ 2 } \right) \right| \le \dfrac { n(n+1) }{ 2 } $
Thus, $\left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n } \right| -\left(\dfrac { n(n+1) }{ 2 } \right)\le \dfrac { n(n+1) }{ 2 } \quad =>\quad \left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n } \right| \le n(n+1)$
Hence, (c) is correct.

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y=\displaystyle\dfrac{1}{a-z}$, then $\displaystyle\dfrac{dz}{dy}$ is:

  1. $(a-z)^2$
  2. $-(z-a)^2$
  3. $(z+a)^2$
  4. $-(z+a)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$y=\dfrac{1}{a-z}$
$\dfrac{{d} y}{{d} z}=\dfrac{-1}{(a-z)^{2}}(-1)$   (By differentiating w.r.t z)
$\dfrac{{d} y}{{d} z}=\dfrac{1}{(a-z)^{2}}$
   $\therefore \dfrac{{d} z}{{d} y}=(a-z)^{2}$
Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $\displaystyle [A]\neq 0 $ then which of the following is not true?

  1. $\displaystyle (A^{2})^{-1}= (A^{-1})^{2}$
  2. $\displaystyle (A')^{-1}= (A^{-1})^{'}$
  3. $\displaystyle A^{-1}= \left | A \right |^{-1}$

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

We know, $(A^{n})^{-1}=(A^{-1})^{n}$
So, $(A^{2})^{-1}=(A^{-1})^{2}$
Hence, option A is correct.

We know that inverse of transpose of matrix is equal to transpose of inverse of matrix
$(A^{-1})' =(A')^{-1}$
Hence, option B is correct

For option C,
In the LHS, there is a matrix and in RHS , its a determinant i.e. a single value.
So, option C is incorrect.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ satisfies the equation $\displaystyle x^{3}-5x^{2}+4x+\lambda =0$, then $\displaystyle A^{-1}$ exists if

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

Since, A satisfies the equation
$\displaystyle x^{3}-5x^{2}+4x+\lambda =0$
$\Rightarrow A^{3}-5A^{2}+4A+\lambda=O$
$\Rightarrow A^{3}A^{-1}-5A^{2}A^{-1}+4AA^{-1}+\lambda A^{-1}=O$
$\Rightarrow A^{2}-5A+4I+\lambda A^{-1}=O$
So, $A^{-1}$ exists if $\lambda\ne 0$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

If $ \displaystyle a+b+c=0$ then value of $ \displaystyle (s) $ of $x$ which makes $\displaystyle \begin{vmatrix}
a-x &c  &b \
 c&b-x  &a \
b & a &c-x
\end{vmatrix}$ zero is (are)

  1. $\displaystyle x=0 $
  2. $\displaystyle x=\sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$
  3. $\displaystyle x=- \sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

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

Applying $\displaystyle R _{1}\rightarrow R _{1}+R _{2}+R _{3}$ and taking $\displaystyle a+b+c-x $ common from the first row, we obtain
$\displaystyle \Delta =\left ( a+b+c-x \right )\begin{vmatrix}
1 &1  &1 \ 
c &b-x  &a \ 
b &a  &c-x 
\end{vmatrix}$
Applying  $\displaystyle C _{2}\rightarrow C _{2}-C _{1}$ and $\displaystyle C _{3}\rightarrow C _{3}-C _{1}$ we obtain
$\displaystyle \Delta =\begin{vmatrix}
1 &0  &0 \ 
c &b-c-x  &a-c \ 
b &a-b  &c-b-x 
\end{vmatrix}\left [ \because a+b+c=0 \right ]$
Expanding along $\displaystyle R _{1}$ we get
$\displaystyle \Delta =x\left [ \left ( b-c-x \right )\left (c-b-x  \right )-\left ( a-b \right )\left ( a-c \right ) \right ] $
$\displaystyle \Delta =x\left [ \left ( a-b \right )\left (a-c  \right )-\left ( x+b-c\right )\left ( x-b+c \right ) \right ] $
$\displaystyle  =x\left [ a^{2}-ab-ac+bc-x^{2}+b^{2}+c^{2}-2bc \right ]$
$\displaystyle \Delta =x\left [ a^{2}+b^{2}+c^{2}-bc-ab-ac-x^{2} \right ]$
$\displaystyle \Delta =0$ implies $\displaystyle x =0$ or $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
Now $\displaystyle x^{2}=a^{2}+b^{2}+c^{2}-bc-ab-ac$
$\displaystyle =a^{2}+b^{2}+c^{2}-\frac{1}{2}\left [ \left ( a+b+c \right )^{2}- a^{2}-b^{2}-c^{2} \right ]$
$\displaystyle =\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )\left [ \because a+b+c=0 \right ]$
$\displaystyle \Rightarrow x= \pm \sqrt{\frac{3}{2}\left ( a^{2}+b^{2}+c^{2} \right )}$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $\displaystyle \alpha ,\beta $ are the roots of $\displaystyle x^{2}+x+1=0 $ and $\displaystyle \gamma ,\delta  $ are the roots of $\displaystyle x^{2}+3x+1=0 $ then $\displaystyle \left ( \alpha -\gamma  \right )\left ( \beta +\delta  \right )\left ( \alpha +\delta  \right )\left ( \beta -\gamma  \right )$ = 

  1. 2

  2. 4

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ x }^{ 2 }+x+1=0$
If $\alpha ,\beta $ are the roots, then $\alpha +\beta =-1\quad ,\quad \alpha \beta =1$
Solving the above equation, we get the $\alpha =\dfrac { -1+i\sqrt { 3 }  }{ 2 } ,\quad \beta =\dfrac { -1-i\sqrt { 3 }  }{ 2 } \\ $

Similarly, ${ x }^{ 2 }+3x+1=0$
If $\gamma ,\delta  $ are the roots, then $\gamma +\delta =-3\quad ,\quad \gamma \delta =1$
Solving the above equation, we get the $\gamma =\dfrac { -3+\sqrt { 5 }  }{ 2 } ,\quad \delta =\dfrac { -3-\sqrt { 5 }  }{ 2 } $

$(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )$ is..
$(\alpha -\gamma )(\beta -\gamma )$
$=\alpha \beta -\gamma (\alpha +\beta )+{ \gamma  }^{ 2 }\\ =1-(-1)(\dfrac { -3+\sqrt { 5 }  }{ 2 } )+(\dfrac { 14-6\sqrt { 5 }  }{ 4 } )\\ =1+(\dfrac { -3+\sqrt { 5 }  }{ 2 } )+(\dfrac { 14-6\sqrt { 5 }  }{ 4 } )\\ =\dfrac { 12-4\sqrt { 5 }  }{ 4 } \\ =3-\sqrt { 5 } $

$(\alpha +\delta )(\beta +\delta )\\ =\alpha \beta +\delta (\alpha +\beta )+\delta ^{ 2 }\\ =1+(-1)(\dfrac { -3-\sqrt { 5 }  }{ 2 } )+(\dfrac { 14+6\sqrt { 5 }  }{ 4 } )\\ =1+(\dfrac { 3+\sqrt { 5 }  }{ 2 } )+(\dfrac { 14+6\sqrt { 5 }  }{ 4 } )\\ =\dfrac { 24+8\sqrt { 5 }  }{ 4 } \\ =6+2\sqrt { 5 } =\quad 2(3+\sqrt { 5 } )$

Multiplying the above two results, we get
$2(3+\sqrt { 5 } )(3-\sqrt { 5 } )\\ =\quad 2(9\quad -\quad 5)\quad \\ =8$


Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\displaystyle y= a^{\left(\frac{1}{1-\log _{a}x}\right)}$ and $\displaystyle z= a^{\left(\frac{1}{1-\log _{a}y}\right)}$, then relation between $x$ and $z$ is

  1. $\displaystyle x= a^{\left(\frac{1}{1-\log _{a}z}\right)}$
  2. $\displaystyle x= a^{\left(\frac{1}{1+\log _{a}z}\right)}$
  3. $\displaystyle x= a\left(\frac{1}{1-\log _{a}z}\right)$
  4. $\displaystyle x= a\left(\frac{1}{1+\log _{a}z}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle y={ a }^{ \frac { 1 }{ 1-\log _{ a }{ x }  }  }\Rightarrow \log _{ a }{ y } =\frac { 1 }{ 1-\log _{ a }{ x }  } $, taking log both sides on base 'a'

$\displaystyle z={ a }^{ \frac { 1 }{ 1-\log _{ a }{ y }  }  }\Rightarrow \log _{ a }{ z } =\frac { 1 }{ 1-\log _{ a }{ y }  } =\frac { 1 }{ 1-\frac { 1 }{ 1-\log _{ a }{ x }  }  } $

$\displaystyle \Rightarrow \log _{ a }{ z } =\frac { 1-\log _{ a }{ x }  }{ 1-\log _{ a }{ x } -1 } \Rightarrow -\log _{ a }{ x } \log _{ a }{ z } =1-\log _{ a }{ x } $

$\displaystyle \Rightarrow \log _{ a }{ x } =\frac { 1 }{ 1-\log _{ a }{ z }  } \Rightarrow x={ a }^{ \frac { 1 }{ 1-\log _{ a }{ z }  }  }$

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $a, b, c$ are positive numbers such that $a^{\log _37}=27, b^{\log _711}=49, c^{\log _{11}25}=\sqrt{11}$, then the sum of digits of $S=a^{(\log _37)^2}+b^{(\log _711)^2}+c^{(\log _{11}25)^2}$ is

  1. 15

  2. 17

  3. 19

  4. 21
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ a }^{ \log _{ 3 }{ 7 }  }=27,\quad { b }^{ \log _{ 7 }{ 11 }  }=49,\quad { c }^{ \log _{ 11 }{ 25 }  }=\sqrt { 11 } $
$S=({ a }^{ \log _{ 3 }{ 7 } })^{\log _{ 3 }{ 7 }  }+{ \left( { b }^{ \log _{ 7 }{ 11 }  } \right)  }^{ \log _{ 7 }{ 11 }  }+{ \left( { c }^{ \log _{ 11 }{ 25 }  } \right)  }^{ \log _{ 11 }{ 25 }  }$
$={ \left( 27 \right)  }^{ \log _{ 3 }{ 7 }  }+{ \left( 49 \right)  }^{ \log _{ 7 }{ 11 }  }+{ \left( \sqrt { 11 }  \right)  }^{ \log _{ 11 }{ 25 }  }$
$={ 3 }^{ 3\log _{ 3 }{ 7 }  }+{ 7 }^{ 2\log _{ 7 }{ 11 }  }+{ 11 }^{ { 1 }/{ 2 }\log _{ 11 }{ 25 }  }$
$={ 7 }^{ 3 }+{ 11 }^{ 2 }+{ 25 }^{ { 1 }/{ 2 } }$
$=469$
Sum of digits $=19$
Hence, C is correct.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

If $\displaystyle \overset{n-r}{\underset{k=1}{\sum }}\ ^{n-k}C _r=^{x}C _y$ then-

  1. $x=n+1\ ;\ y=r$
  2. $x=n\ ;\ y=r+1$
  3. $x=n\ ;\ y=r$
  4. $x=n+1\ ;\ y=r+1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the hockey stick identity (or repeated application of Pascal's identity C(n,r) + C(n,r+1) = C(n+1,r+1)), the sum of combinations simplifies to C(n, r+1). Therefore, x = n and y = r + 1.

Multiple choice exponent of a prime in n! factorial notation combinatorics and mathematical induction permutations and combinations maths

If $\displaystyle \frac{^{n}P _{r-1}}{a}=\frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$,then which of the following holds good 

  1. $c^{2}=a(b+c)$
  2. $a^{2}=c(a+b)$
  3. $b^{2}=a(b+c)$
  4. $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \frac{^{n}p _{r-1}}{a}=\frac{^{n}P _{r}}{b}$

$\displaystyle \Rightarrow n-r=\frac{b}{a}-1$

and $\displaystyle \frac{^{n}P _{r}}{b}=\frac{^{n}P _{r+1}}{c}$

$\displaystyle \Rightarrow n-r=\frac{c}{b}$
On dividing (i) and (ii) we get
$b^{2}=a(b+c)$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $\displaystyle x^{3}-mx^{2}-3x+2=0$ has two roots equal in magnitude but opposite in sign, then $m$ is:

  1. $\displaystyle \frac{3}{2}$
  2. $\displaystyle \frac{2}{3}$
  3. $\displaystyle -\frac{2}{3}$

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

Let $\alpha ,-\alpha ,\beta $ be the roots of $x^{ 3 }-mx^{ 2 }-3x+2=0$
Then
${ s } _{ 1 }=\alpha -\alpha +\beta =m\ \Rightarrow \beta =m$
Substituting $x=m$ in equation, we get
$m^{ 3 }-m.m^{ 2 }-3.m+2=0\ \Rightarrow m=\cfrac { 2 }{ 3 } $
Hence, option 'B' is correct.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

The value of $\displaystyle\ \alpha^{4n-1}+\alpha^{4n-3}, n\epsilon\mathbb{N}$ and $\displaystyle\ \alpha$ is a nonreal fourth root of unity is 

  1. $0$
  2. $-1$
  3. $3$

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

$x^{4}=1$
$x^{2}=\pm1$
$x=\pm i$ and $x=\pm 1$
Hence
$\alpha^{4n-1}+\alpha^{4n-3}$
$=\alpha^{4n}[\alpha^{-1}+\alpha^{-3}]$
$=[\alpha^{-1}+\alpha^{-3}]$
$=\alpha^{-1}[1+\alpha^{-2}]$
$=\alpha^{-3}[\alpha^{2}+1]$
$=\alpha^{-3}[(\pm i)^{2}+1]$
$=0$

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $\displaystyle \alpha$ is a non-real root of $\displaystyle x^{5}+1=0$ then $\displaystyle \alpha ^{10n+2}+\alpha ^{5n+2}+\alpha ^{5n}$, where n is an odd positive integer,has the value

  1. $1$
  2. $0$
  3. $-1$

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

$x^{5}=-1$
Hence
$\alpha^{5}=-1$
Therefore
$\alpha^{10n+2}+\alpha^{5n+2}+\alpha^{5n}$
$=(\alpha^{5n})^{2}\alpha^{2}+(\alpha^{5n}).\alpha^{2}+\alpha^{5n}$
$=(-1)^{2}\alpha^{2}+(-1)\alpha^{2}+\alpha^{5n}$ .... Since n is odd
$=-\alpha^{2}+\alpha^{2}-1$
$=-1$