A) a + b |
B) \( \Large \frac{2b}{a} \) |
C) \( \Large \frac{b}{a} \) |
D) ab |
B) \( \Large \frac{2b}{a} \) |
\( \Large x + \frac{a}{x} = b => \frac{x^{2}+a}{x} = b \)
=> \( \Large x^{2} + a = bx \) ...(i)
Now, \( \Large \frac{x^{2}+bx+a}{bx^{2}-x^{3}} => \frac{ \left(x^{2}+a\right) + bx} {bx^{2}-x^{3}} \) [using Eq. (i)]
= \( \Large \frac{2bx}{bx^{2} - x^{3}} = \frac{2b}{bx - x^{2}} = \frac{2b}{a} \)
\( \Large Because, x^{2} + bx \)
\( \Large bx - x^{2} = a \)