The harmonic mean of frac{a}{1 - ab} and frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The harmonic mean $(H)$ of two numbers $x$ and $y$ is given by $H = \frac{2xy}{x + y}$.Let $x = \frac{a}{1 - ab}$ and $y = \frac{a}{1 + ab}$.Then

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(C) The harmonic mean $(H)$ of two numbers $x$ and $y$ is given by $H = \frac{2xy}{x + y}$.Let $x = \frac{a}{1 - ab}$ and $y = \frac{a}{1 + ab}$.Then $xy = \frac{a^2}{(1 - ab)(1 + ab)} = \frac{a^2}{1 - a^2b^2}$.And $x + y = \frac{a}{1 - ab} + \frac{a}{1 + ab} = \frac{a(1 + ab) + a(1 - ab)}{(1 - ab)(1 + ab)} = \frac{a + a^2b + a - a^2b}{1 - a^2b^2} = \frac{2a}{1 - a^2b^2}$.Therefore,$H = \frac{2 	imes \left( \frac{a^2}{1 - a^2b^2} ight)}{\frac{2a}{1 - a^2b^2}} = \frac{2a^2}{2a} = a$.

The harmonic mean of $\frac{a}{1 - ab}$ and $\frac{a}{1 + ab}$ is

Similar Questions

$A$ sound absorber attenuates the sound level by $20\, dB$. The intensity decreases by a factor of

The centre of mass of a body

If $\vec{a}, \vec{b}$ and $\vec{c}$ are vectors with magnitudes $2, 3$ and $4$ respectively,then the best upper bound of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ among the given values is

The perimeter of a sector is a constant. If its area is to be maximum,the sectorial angle is:

The two pairs of straight lines $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ constitute a

Vedclass Test Series

Exam Paper Generator

Online Exam Module