MCQ
If $\text{x}+\frac{\text{a}}{\text{x}}$ then the value $\frac{\text{x}^2+\text{bx}+\text{a}}{\text{bx}^2-\text{x}^3}$
- A$\frac{6\text{b}}{\text{a}}$
- B$\frac{5\text{b}}{\text{a}}$
- ✓$\frac{2\text{b}}{\text{a}}$
- D$\frac{4\text{b}}{\text{a}}$
✓
Answer
Correct option: C.
$\frac{2\text{b}}{\text{a}}$
Given $\text{x}+\frac{\text{a}}{\text{x}}=\text{b}$ Multiply by $x$ both sides
$x^2+ a = bx$
$\Rightarrow x^2 - bx + a = 0$
So,
$\frac{\text{x}^2+\text{bx}+\text{a}}{\text{bx}^2-\text{x}^3}$
$=\frac{\text{x}^2-\text{bx}+\text{a}+\text{2bx}}{\text{-x}(\text{x}^2-\text{bx})}$
$=\frac{0+\text{2bx}}{\text{-x}(\text{- a})}=\frac{\text{2bx}}{\text{ax}}$
$=\frac{\text{2b}}{\text{a}}$
$x^2+ a = bx$
$\Rightarrow x^2 - bx + a = 0$
So,
$\frac{\text{x}^2+\text{bx}+\text{a}}{\text{bx}^2-\text{x}^3}$
$=\frac{\text{x}^2-\text{bx}+\text{a}+\text{2bx}}{\text{-x}(\text{x}^2-\text{bx})}$
$=\frac{0+\text{2bx}}{\text{-x}(\text{- a})}=\frac{\text{2bx}}{\text{ax}}$
$=\frac{\text{2b}}{\text{a}}$
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