Question

If $\frac{1}{9-x^2}=\frac{1}{k}\left[\frac{1}{3-x}+\frac{1}{3+x}\right]$, then $k=$

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is C 6

We have,

$\frac{1}{9-x^2}=\frac{1}{k}\left[\frac{1}{3-x}+\frac{1}{3+x}\right]$

$\frac{1}{9-x^2}=\frac{1}{k}\left[\frac{3+x+3-x}{(3-x)(3+x)}\right]$

$\frac{1}{9-x^2}=\frac{1}{k}\left[\frac{6}{9-x^2}\right]$

$1=\left[\frac{6}{k}\right]$

$k=6$

Hence, this is the answer.