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 logarithm and its uses basic mathematical concepts physics

The value of $\displaystyle\sum _{r=1}^{n}log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$ is

  1. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n}} \right )$
  2. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n+1}} \right )$
  3. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n+1}} \right )$
  4. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n-1}} \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Now,
$\displaystyle\sum _{r=1}^{n}\log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$
$=\displaystyle\sum _{r=1}^{n}\left(\log a^{r}-\log b^{r-1}\right)$
$=\displaystyle\sum _{r=1}^{n}\left(r\log a-(r-1)\log b\right)$
$=(\log a)\times \dfrac{n(n+1)}{2}-\log b\times\dfrac{(n-1)n}{2}$
$=\dfrac{n}{2}\left(\log a^{n+1}-\log b^{n-1}\right)$
$=\dfrac{n}{2}\log\left(\dfrac{a^{n+1}}{b^{n-1}}\right)$
Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If (x + 6) is the HCF of $\displaystyle p\left ( x \right )=x^{2}-a$ and $\displaystyle q\left ( x \right )=x^{2}-bx+6$ then $\displaystyle \frac{p\left ( x \right )}{q\left ( x \right )}$ in its lowest terms is______

  1. $\displaystyle \frac{x-6}{x-2}$
  2. $\displaystyle \frac{x+6}{x+1}$
  3. $\displaystyle \frac{x-6}{x-1}$
  4. $\displaystyle \frac{x-6}{x+1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since (x + 6) is the HCF, (x + 6) must be a factor of both polynomials. For p(x) = x^2 - a, x = -6 makes p(-6) = 36 - a = 0, so a = 36. Thus p(x) = (x - 6)(x + 6). For q(x) = x^2 - bx + 6, x = -6 makes 36 + 6b + 6 = 0, so 6b = -42, b = -7. Thus q(x) = x^2 + 7x + 6 = (x + 6)(x + 1). The ratio p(x)/q(x) simplifies to (x - 6)/(x + 1).

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

$\displaystyle 9x^{2}+2hxy+4y^{2}+6x+2fy-3=0$ represents two parallel lines if

  1. $\displaystyle h=6, f=2 $
  2. $\displaystyle h=-6, f=-2 $
  3. $\displaystyle h=-6, f=2 $
  4. $\displaystyle h=6, f=-2 $
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Since the given equation represents a pair of parallel lines, we have
$\displaystyle h^{2}=9 \times 4\Rightarrow h= \pm 6$
and $\displaystyle \begin{vmatrix}
9 & h & 3\ 
 h& 4 & f\ 
 3& f & -3
\end{vmatrix}=0$
$\displaystyle \Rightarrow 9\left ( -12-f^{2} \right )-h\left ( -3h-3f \right )+3\left ( hf-12 \right )=0$
$\displaystyle \Rightarrow 3h^{2}+6hf-9f^{2}-144=0$
$\displaystyle \Rightarrow 108 \pm 36f-9f^{2}-144=0 \ \ \ \left ( \because h= \pm 6 \right )$
$\displaystyle \Rightarrow 9f^{2} \mp 36f+36=0 \ \ \ \ \  (if \ \  h= \pm 6)$
$\displaystyle \Rightarrow f=2 \ \ \ \ if \ \ \ \ \ ( h=6)$
and $\displaystyle \Rightarrow f=-2 \ \ \ \ \ if \ \ \ \ \ \ (h=-6)$

Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following is logically equivalent to $\displaystyle \sim \left (\sim p\rightarrow q\right )$?

  1. $\displaystyle p\wedge q$
  2. $\displaystyle p\wedge \sim q$
  3. $\displaystyle \sim p\wedge q$
  4. $\displaystyle \sim p\wedge \sim q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
p  q  $\sim p$  $\sim q$  $\sim p \rightarrow q$  $\sim (\sim p \rightarrow q)$  $p \wedge q$  $p \wedge \sim q$   $\sim p \wedge q$   $\sim p \wedge \sim q$  
T  T  F  F  T  F  T  F  F  F 
T F  F  T  T  F  F  T  F  F 
F T  T  F  T  F  F  F  T  F 
F F  T  T  F  T  F  F  F  T 

The values in column 6 and column 10 are same.

Hence, option D is correct.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Which of the following formula is wrong?

  1. $\displaystyle{C _{v} = \dfrac{R}{\gamma - 1}}$
  2. $\displaystyle{C _{p} = \dfrac{\gamma R}{\gamma - 1}}$
  3. $\displaystyle \dfrac{C _{p}}{ C _{v}} = \gamma$
  4. $C _{p} - C _{v} = 2R$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The different formula for specific heats is given by:

  • $\dfrac{C _{p}}{C _{v}} = \gamma$
  • $C _{p} - C _{v} = R$
Upon further simplification, we get:
  • $C _{p} = \dfrac{\gamma R}{\gamma -1}$
  • $C _{v} = \dfrac{R}{\gamma -1}$
The incorrect formula is
$C _{p} - C _{v} = 2R$
Hence option D is the answer.

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Simplify: 
$\displaystyle x-\left[ y-{ x-\left( y-1 \right) -2x}  \right] $

  1. $2y+1$
  2. $-2y+1$
  3. $2x+y-1$
  4. $2x-y-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

On simplified, we have

$\displaystyle x-\left[ y-{ x-\left( y-1 \right) -2x}  \right] $
$=x-\left[ y-{ x-y+1-2x}  \right] $
=$\displaystyle x-\left[ y-{ -x-y+1}  \right] =x-\left[ y+x+y-1 \right] $
=$\displaystyle x-\left[ 2y+x-1 \right] =x-2y-x+1=-2y+1$
Hence, simplified form of the given expression is $-2y+1$.

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $\displaystyle M=a\left ( m+n \right )$ and $\displaystyle N=b(m-n)$ then the value of  $\displaystyle \left ( \frac{M}{a}+\frac{N}{b} \right )\div \left ( \frac{M}{a}-\frac{N}{b} \right )$ is :

  1. $\displaystyle \frac{m}{n}$
  2. $\frac{n}{m}$

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

$\displaystyle \frac{M}{a}=m+n;\frac{N}{b}=m-n$
$\displaystyle \therefore \left ( \frac{M}{a}+\frac{N}{b} \right )\div \left ( \frac{M}{a}-\frac{N}{b} \right )=2m\div 2n=\frac{m}{n}$

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $A\,\colon\,B=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3},\,B\,\colon\,C=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}$, then $A\,\colon\,B\,\colon\,C$ is equal to:

  1. $\;2\,\colon\,3\,\colon\,3$
  2. $\;1\,\colon\,2\,\colon\,6$
  3. $\;3\,\colon\,2\,\colon\,6$
  4. $\;9\,\colon\,6\,\colon\,4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\;A\,\colon\,B=\displaystyle\frac{1}{2} \colon\displaystyle\frac{1}{3}=\displaystyle\frac{1}{2}\times6\ \colon\displaystyle\frac{1}{3}\times6=3\,\colon\,2$


$\;\;\;\;\;\;\;\;B\,\colon\,C=\displaystyle\frac{1}{2}\colon\displaystyle\frac{1}{3}=3\,\colon\,2$

By taking the L.C.M. of $2$ and $3$, i.e., $6$, we can make the value of $B$ equal in both the ratio.

$\;\;\;\;\;\;\;\;\displaystyle\frac{A}{B}=\displaystyle\frac{3}{2}=\displaystyle\frac{9}{6}$ and $\displaystyle\frac{B}{C}=\displaystyle\frac{3}{2}=\displaystyle\frac{6}{4}$

$\;\;\;\;\;\;\;\therefore\,A\,\colon\,B\,\colon\,C=9\,\colon\,6\,\colon\,4$.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Determine the order relation between the following pairs of ratios.

$\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}}, \frac{2\sqrt{2}}{3\sqrt{3}}$

  1. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} > \frac{2\sqrt{2}}{3\sqrt{3}}$
  2. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} < \frac{2\sqrt{2}}{3\sqrt{3}}$

  3. Cannot be determined

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

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle A=\pi \left ( R^{2}-r^{2} \right )$, then $R$ is equal to

  1. $\displaystyle \sqrt{\frac{A-\pi r^{2}}{\pi }}$
  2. $\displaystyle \sqrt{\frac{A+\pi r^{2}}{\pi }}$
  3. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{\pi }}$
  4. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{r}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, $A=\pi(R^2-r^2)$
Therefore, $A =$ $\displaystyle \pi R^{2}-\pi r^{2}$
$\Rightarrow  A+\pi r^{2}=\pi R^{2}$
$\displaystyle \Rightarrow R^{2}=\frac{A+\pi r^{2}}{\pi }$
$\displaystyle \Rightarrow $ $\displaystyle R=\sqrt{\frac{A+\pi r^{2}}{\pi }}$
Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right) $ is equal to

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

$\displaystyle \left( { 3x }^{ 2 }-x \right) \div \left( -x \right)$

By separating denominators, we get
$  =\dfrac { 3{ x }^{ 2 } }{ -x } +\dfrac { \left( -x \right)  }{ \left( -x \right)  } =-3x+1$
Hence, final result after given operation is $-3x+1$.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Assume that, $\Delta RST \sim \Delta XYZ$. Complete the following statement.


$\displaystyle \frac{RT}{XY} = \frac{- -}{YZ}, \frac{RS}{XY} = \frac{ST}{- -}, \frac{XY}{ - -} = \frac{YZ}{ST}$

  1. ST, YZ, RT

  2. ST, YZ, RS

  3. YT, YS, RZ

  4. ST, YZ, RZ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given triangle RST similar to triangle XYZ, the ratios of corresponding sides are equal: RS/XY = ST/YZ = RT/XZ. The provided option B correctly completes the ratios.