Question

If y = 2x +cot^–1 x + log ($\sqrt{1 + x^{2}}$– x), then y –

• A. Decreases on (–∞, ∞)
• B. Decreases on [0, ∞)
• C. Neither decreases nor increases on [0, ∞)
• D. Increases on (–∞, ∞)

Answer 1

Answer: (D)

Increases on (–∞, ∞)

Answer 2

y′(x) = 2 – $\frac{1}{1 + x^{2}}$ + $\frac{1}{\sqrt{1 + x^{2}} - x}$$\left( \sqrt{1 + x^{2}} - x \right)$

= 2 – $\frac{1}{1 + x^{2}}$ + $\frac{1}{\sqrt{1 + x^{2}} - x}$×$\left( \frac{x}{\sqrt{1 + x^{2}}} - 1 \right)$

= 2 – $\frac{1}{1 + x^{2}}$ – $\frac{1}{\sqrt{1 + x^{2}}}$ ≥ 0

Since 1/(1 + x²) and 1/$\sqrt{1 + x^{2}}$ are less than or equal to 1 for all x. So ƒ(x) increases on (–∞, ∞).