Mathematics

Differentiation and Algebraic Fractions

43 Questions

Differentiation combined with algebraic fractions involves solving implicit functions and evaluating complex mathematical ratios. Test items require calculating second derivatives, inverse trigonometric functions, and variable fraction equivalencies. This specialized mathematics topic appears regularly in high level aptitude screenings.

Implicit function derivativesAlgebraic fraction ratiosSecond derivative calculationInverse trigonometric differentiationVariable fraction equivalencies

Differentiation and Algebraic Fractions Questions

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $\quad y={ log } _{ x }({ log } _{ e }x)({ log } _{ e }x)\quad then\quad \dfrac { dy }{ dx } \quad equals$ to 

  1. $\dfrac { 1 }{ x{ log } _{ x }{ log } _{ x }x } $
  2. $\dfrac { 1 }{ x{ log } _{ e }x } $

  3. 0

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

$y=\log _x\left(\log _e\left(x\right)\right)\left(\log _e\left(x\right)\right)$

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\log _x\left(\log _e\left(x\right)\right)\log _e\left(x\right)\right)$

$\left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\rightarrow f=\log _x\left(\log _e\left(x\right)\right),\:g=\log _e\left(x\right)$

$=\dfrac{d}{dx}\left(\log _x\left(\log _e\left(x\right)\right)\right)\log _e\left(x\right)+\dfrac{d}{dx}\left(\log _e\left(x\right)\right)\log _x\left(\log _e\left(x\right)\right)$

$=\dfrac{1-\log _e \left(\log \left(x\right)\right)}{x\log ^2\left(x\right)}\log _e\left(x\right)+\dfrac{1}{x}\log _x\left(\log _e\left(x\right)\right)$

$=\dfrac{1}{x\log _e \left(x\right)}$
Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

Prove that $\dfrac{a^{-1}}{(a^{-1}+b^{-1})}$ is equal to $\dfrac{b}{(a+b)}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\cfrac{{ a }^{ -1 }}{{ a }^{ -1 }+{ b }^{ -1 }}\Leftrightarrow \cfrac{{ a }^{ -1 }}{\cfrac{1}{a}+\cfrac{1}{b}}$
$\Rightarrow$ $\cfrac{{ a }^{ -1 }}{\cfrac{b+a}{a.b}}$
$\Rightarrow$ $\cfrac{a.b}{a(b+a)}$
$\Rightarrow$ $\cfrac{b}{b+a}$
$\Rightarrow$ $\cfrac{b}{a+b}$
$\therefore$ $\cfrac{{ a }^{ -1 }}{{ a }^{ -1 }+{ b }^{ -1 }}=\cfrac{b}{a+b}$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x = 1\, + a + {a^2} + ......\infty $, $y = 1\, + b + {b^2}\,\, + ......\infty $ where $\left| a \right| < 1$ and $\left| b \right| < 1$, then $\left( {1 + ab + {a^2}{b^2} + ........\infty } \right) = ?$

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

Given x = 1/(1-a) and y = 1/(1-b), we have a = (x-1)/x and b = (y-1)/y. The series 1 + ab + a^2b^2 + ... is a geometric series with sum 1/(1-ab). Substituting a and b: 1 / (1 - ((x-1)/x)((y-1)/y)) = 1 / (1 - (xy - x - y + 1)/(xy)) = xy / (xy - xy + x + y - 1) = xy / (x + y - 1).

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $y=x-x^2+x^3-x^4+....\infty$, then value of x will be?

  1. $y+\dfrac{1}{y}$
  2. $\dfrac{y}{1+y}$
  3. $y-\dfrac{1}{y}$
  4. $\dfrac{y}{1-y}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series y = x - x^2 + x^3 - x^4 + ... is a geometric series with first term a = x and common ratio r = -x. The sum is y = x / (1 - (-x)) = x / (1 + x). Solving for x: y(1 + x) = x, so y + xy = x, which means y = x - xy = x(1 - y). Thus, x = y / (1 - y).

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $, then $y =$

  1. $x^{1/3}$
  2. $x^{2/3}$
  3. $x^{1/2}$
  4. $x$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $
 $ =x^{\cfrac{1}{3}+\cfrac{1}{3^2}+\cfrac{1}{3^3}+........\infty }=x^{\cfrac{1/3}{1-1/3}}=x^{1/2}$
Hence, option 'C' is correct.

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

If $\dfrac{1}{x} + y = 3$ and $x + \dfrac{1}{y} = 2$ then $x:y$ is 

  1. $3:2$
  2. $2:3$
  3. $1:2$
  4. $2:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The question states '...then x: is' 
It should state '..then x:y is'

Given
$\dfrac { 1 }{ x } +y=3$ --- Eqn (1)
$x+\dfrac { 1 }{ y } =2$ ---Eqn (2)

Multiplying Eqn (1) by x and Eqn (2) by y, we get:

$x\left( \dfrac { 1 }{ x } +y \right) =3x\quad \Rightarrow 1+yx=3x$ --- Eqn (3)
$y\left( x+\dfrac { 1 }{ y }  \right) =2y\quad \Rightarrow xy+1=2y$ --- Eqn (4)

Subtracting Eqn (4) and Eqn (5), we get:

$1+yx-1-yx=3x-2y$
$\Rightarrow 0=3x-2y$
$\Rightarrow 3x=2y$
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2 }{ 3 } $
$\therefore x:y=2:3$

Hence the answer is B
Multiple choice

The sum of 1/ \frac{1}{x+y}x + y and \frac{1}{x-y} is

  1. <latex>\frac{2x}{x^2-y^2}</latex>

  2. <latex>\frac{2}{2x}</latex>

  3. <latex>\frac{2y}{x^2-y^2}</latex>

  4. <latex>\frac{-2x}{x^2-y^2}</latex>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum is 1/(x + y) + 1/(x - y). Find a common denominator: (x - y)/((x + y)(x - y)) + (x + y)/((x + y)(x - y)) = (x - y + x + y) / (x^2 - y^2) = 2x / (x^2 - y^2).

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $\dfrac {y}{x-z}=\dfrac{y+x}{z}=\dfrac{x}{y}$ then find $x:y:z$

  1. $1:2:3$
  2. $3:2:1$
  3. $4:2:3$
  4. $2:4:7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


$ \dfrac{y}{x-z}=\dfrac{y+x}{z}=\dfrac{x}{y} $

 

Now,

$ \dfrac{y}{x-z}=\dfrac{x}{y} $

$ {{y}^{2}}={{x}^{2}}-xz\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(1) $

 

And

$ \dfrac{y+x}{z}=\dfrac{x}{y} $

$ {{y}^{2}}+xy=xz\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(2) $

$ {{x}^{2}}-xz+xy=xz $

$ x-z+y=z $

$ 2z=x+y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......(3) $

 

$ And $

$ \dfrac{y}{x-z}=\dfrac{y+x}{z} $

$ yz=xy-yz+{{x}^{2}}-xz $

$ 2yz=xy+{{x}^{2}}-xz $

$ 2yz=x\left( y+x \right)-xz $                    [From equation (3)]

$ 2yz=2xz-xz $

$ 2yz=xz $

$ 2y=x $

$ \dfrac{x}{y}=\dfrac{2}{1}\,\,\,\,\,\,\,\,......\,\,\left( 4 \right) $


Substituting this value in equation (3), we get

$ 2z=2y+y $

$ 2z=3y $

$ \dfrac{y}{z}=\dfrac{2}{3}\,\,\,\,\,......\,\,\left( 5 \right) $


By equation (4) and (5), we get

$ x:y:z=4:2:3 .$


Hence, this is the answer.

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

$\left(\dfrac{1}{x^{a-b}}\right)^{\tfrac{1}{(a-c)}}. \left(\dfrac{1}{x^{b-c}}\right)^{\tfrac{1}{(b-a)}}. \left(\dfrac{1}{x^{c-a}}\right)^{\tfrac{1}{(c-b)}}=$

  1. $0$
  2. $1$
  3. $a+b+c$
  4. $(a-b+c)^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can write the given equation as, 

$(x^{b-a})^{\frac{1}{a-c}}\cdot (x^{c-b})^{\frac{1}{b-a}}\cdot (x^{a-c})^{\frac{1}{c-b}}$

$=x^{\cfrac{b-a}{a-c}}\cdot x^{\cfrac{c-b}{b-a}}\cdot x^{\cfrac{a-c}{c-b}}$
On adding all the powers of $x$, We get
$x^{\Bigl(\cfrac{(b-a)^2(c-b)+(c-b)^2(a-c)+(a-c)^2(b-a)}{(a-c)(b-c)(c-b)}\Bigr)}\ =x^0=1$

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $ x+\dfrac{1}{x}=2\cos \theta \   and \ y+\dfrac{1}{y}=2\cos \phi$  then which of the following is not correct?

  1. $\displaystyle \frac{x}{y} +\frac{y}{x}=2\cos \left ( \theta -\phi \right )$
  2. $x^{m}y^{n}=\cos \left ( m\theta +n\phi \right )+i\sin \left ( m\theta +n\phi \right )$
  3. $x^{m}y^{n}+x^{-m}y^{-n}=2\cos \left ( m\theta +n\phi \right )$

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

$x+\dfrac { 1 }{ x } =2\cos { \theta  } \quad &amp; \quad y+\dfrac { 1 }{ y } =2\cos { \phi  } \ $

$\Rightarrow x=\cos { \theta  } +i\sin { \theta  } =cis\theta \ \quad &amp; \quad y=\cos { \phi  } +i\sin { \phi  } =cis\phi \ $

$\dfrac { x }{ y } +\dfrac { y }{ x } =\dfrac { cis\theta  }{ cis\phi  } +\dfrac { cis\phi  }{ cis\theta  } =cis\left( \theta -\phi  \right) +cis\left( -\theta +\phi  \right) $

$\therefore \quad \dfrac { x }{ y } +\dfrac { y }{ x } =2\cos { \left( \theta -\phi  \right)  } $

${ x }^{ m }{ y }^{ n }={ \left( cis\theta  \right)  }^{ m }{ \left( cis\phi  \right)  }^{ n }=\left( cism\theta  \right) \left( cisn\phi  \right) $          ...{ De Moivre's Theorem}

$\therefore \quad { x }^{ m }{ y }^{ n }=cis\left( m\theta +n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } +i\sin { \left( m\theta +n\phi  \right)  } $

${ x }^{ -m }{ y }^{ -n }={ \left( cis\theta  \right)  }^{ -m }{ \left( cis\phi  \right)  }^{ -n }=\left( cis\left( -m\theta  \right)  \right) \left( cis\left( -n\phi  \right)  \right) \ \therefore \quad { x }^{ -m }{ y }^{ -n }=cis\left( -m\theta -n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } -i\sin { \left( m\theta +n\phi  \right)  } \ $

$\therefore \quad { x }^{ m }{ y }^{ n }+{ x }^{ -m }{ y }^{ -n }=2\cos { \left( m\theta +n\phi  \right)  } $