Question

For any real x the expression 2(k – x) [x +$\sqrt{x^{2} + k^{2}}$] cannot exceed –

• A. k²
• B. 2 k²
• C. 3 k²
• D. None of these

Answer 1

Answer: (B)

2 k²

Answer 2

y = 2 (k – x) (x + $\sqrt{x^{2} + k^{2}}$) or x + $\sqrt{x^{2} + k^{2}}$

= $\frac{y}{2(k - x)}$ …(1)

$\frac{(x + \sqrt{x^{2} + k^{2}})(\sqrt{x^{2} + k^{2}} - x)}{\lbrack(\sqrt{x^{2} + k^{2}}) - x\rbrack}$ = $\frac{y}{2(k - x)}$

$\frac{k^{2}}{\sqrt{x^{2} + k^{2}} - x}$ = $\frac{y}{2(k - x)}$

⇒ $\sqrt{(x^{2} + k^{2})}$ – x = $\frac{2k^{2}(k - x)}{y}$ … (2)

(1) – (2) ⇒ 2x = $\frac{y}{2(k - x)}$ – $\frac{2k^{2}(k - x)}{y}$

4 (y – k²) x² + 4k (2k² – y) x + y² – 4k⁴ = 0

But x is real D ≥ 0 y² (y – 2k²) ≤ 0

∴ y ≤ 2k²

Q y² ≥ 0

max. value of y is 2k².