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 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. 

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 chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$

If $\displaystyle -\, \frac{d[N _2O _5]}{dt}\, =\, k _1[N _2O _5]$

$\displaystyle \frac{d[NO _2]}{dt}\, =\, k _2[N _2O _5]$

$\displaystyle \frac{d[O _2]}{dt}\, =\, k _3[N _2O _5]$
What is the relation between $k _1, k _2\, and\, k _3$ ?

  1. $k _1\, =\, k _2\, =\, k _3$
  2. $2k _1\, =\, k _2\, =\, 4k _3$
  3. $2k _1\ =\, 4k _2\, =\, k _3$

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

As we know,
for a reaction:
$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$
$\displaystyle -\, \frac{1}{2}\, \frac{d[N _2O _5]}{dt}\, =\, \frac{1}{4}\, \frac{d[NO _2]}{dt}\, =\, \frac{d[O _2]}{dt}$
So
$2k _1\, =\, k _2\, =\, 4k _3$

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

If a, b, c, are positive $\displaystyle \frac{a+c}{b+c}$ is 

  1. always smaller than $\displaystyle \frac{a}{b}$
  2. always greater than $\displaystyle \frac{a}{b}$
  3. greater than $\displaystyle \frac{a}{b}$ only if a > b
  4. greater than $\displaystyle \frac{a}{b}$ only if a < b
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since we need to compare the fraction $\displaystyle \frac{a+c}{b+c}$ with $\displaystyle \frac{a}{b}$, we cross multiply the terms and check since $a,b,c$ are all given to be positive.
We thus get $b(a + c)$ on the L.H.S. and $a(b + c)$ on R.H.S.
Thus, simplifying we are left with $ab + bc$ on the L.H.S. and $ab + ac$ on the R.H.S.
Now, which side is greater depends on $ac$ and $bc$, which in turn depends upon $a$ & $b.$
L.H.S. is greater if $b > a$, which implies $\frac{a + c}{b + c}$ is greater when $b > a.$

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
$2(3+\sqrt { 5 } )(3-\sqrt { 5 } )\\ =\quad 2(9\quad -\quad 5)\quad \\ =8$