Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

The value of $\cfrac { { 2 }^{ 2n-2 } }{ { 2 }^{ n(n-1) } }-\cfrac { { 8 }^{ n-1 } }{ { 2 }^{ (n-1)(n+1) } } $ will be

  1. $2$
  2. $0$
  3. $\dfrac {1}{2}$
  4. $\dfrac {1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$\dfrac{2^{2n-2}}{2^{n(n-1)}} - \dfrac{8^{n-1}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^{2(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^{3(n-1)}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^2\times 2^{(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^3\times 2^{(n-1)}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^2\times 2^{(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^3\times 2^{(n-1)}}{2^{(n-1)}2^{(n+1)}}$

$=\dfrac{2^2}{2^n}-\dfrac{2^3}{2^{n+1}}$

$=\dfrac{2^2}{2^n}-\dfrac{2^3}{2^n \times 2^1}$

$=\dfrac{2^2}{2^n}-\dfrac{2^2}{2^n}$

$=0$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Find the sum of all values of $x$, so that $16^{\left(x^{2}+3x-1\right)}=8^{\left(x^{2}+3x+2\right)}$.

  1. $0$
  2. $3$
  3. $-3$
  4. $-5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $16^{(x^2 + 3x -1)} = 8^{(x^2 + 3x + 2)}$


As, 16 = $2^4 and \ \ \ 8 = 2^3$


$2^{4(x^2 + 3x -1)} = 2^{3(x^2 + 3x + 2)}$

So, we can write 

${4(x^2 + 3x -1)} = {3(x^2 + 3x + 2)}$

${4x^2 + 12x - 4} = {3x^2 + 9x + 6}$

${4x^2 + 12x - 4} - {(3x^2 + 9x + 6)} = 0$
$4x^2 + 12x - 4 - 3x^2 - 9x - 6 = 0$

$x^2 + 3x - 10  = 0$

$x^2 + 5x - 2x - 10  = 0$

$x(x + 5) - 2(x + 5)  = 0$

$(x - 2)(x + 5)  = 0$

So, $x = 2, - 5$

Sum of values of $x = 2 + (-5) = -3$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If the normal at any point $P$ of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ meets the axes in $G$ and $g$ respectively, then $|PG| : |Pg|$ is equal to 

  1. $a:b$
  2. $a^2:b^2$
  3. $b^2:a^2$
  4. $b:a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1, the normal at P(x1, y1) meets the axes at G and g. The ratio PG:Pg is a:b.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

One foot of normal of the ellipse $4x^2$ $+$ 9$y^2$ $= 36 $, that is parallel to the line $2x + y = 3 $, is

  1. $\left ( \dfrac{9}{8}, \dfrac{5}{8} \right )$
  2. $\left ( \dfrac{9}{8}, \dfrac{8}{5} \right )$
  3. $\left ( \dfrac{8}{9}, \dfrac{8}{5} \right )$

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

The normal to 4x^2 + 9y^2 = 36 (x^2/9 + y^2/4 = 1) parallel to 2x + y = 3 (slope -2). The normal slope is -2. Using the normal equation y = mx - (a^2-b^2)m/sqrt(a^2+b^2m^2), we find the point.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The equation of normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

  1. $y- 3 = 0$
  2. $y + 3 = 0$
  3. $x$-axis
  4. $y$-axis
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$9x^2+5y^2=45$


Differentiating with respect to $x$, we get,

$18x+10y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{18x}{10y}$

Therefore, Slope of tangent at $(0,3)=\dfrac{-0}{30}=0$

Slope of normal = $\dfrac{1}{0}$

The equation of normal is :
$y-3=\dfrac{1}{0}(x-0)$
$x-0=0$
$x=0$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Find the equation of the normal to the ellipse $9x^2 + 16y^2 = 288$ at the point $(4, 3).$

  1. $4x-3y=7$
  2. $3x-4y=7$
  3. $4x+3y=7$
  4. $3x+4y=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ellipse may be written as, $\displaystyle \frac{x^2}{32}+\frac{y^2}{18}=1$
$\Rightarrow a^2 = 32, b^2 = 18$
Hence required normal at $(4,3)$ is given by,
$\displaystyle y-3=\frac{3\times 32}{4\times 18}(x-4)\Rightarrow y-3=\frac{4}{3}(x-4)\Rightarrow 4x-3y=7$