Solveeit Logo
Login

Question

Question: f(x) = 1 + x log\((x + \sqrt{x^{2} + 1})\)– \(\sqrt{1 + x^{2}}\) where x is any real number then f(x...

f(x) = 1 + x log(x+x2+1)(x + \sqrt{x^{2} + 1})– 1+x2\sqrt{1 + x^{2}} where x is any real number then f(x) is

A

Increasing in (–∞, ∞)

B

Increasing in (–∞, 0)

C

Increasing in (0, ∞)

D

None of these

Answer

Increasing in (0, ∞)

Explanation

Solution

f(x) = 1 + x log (x+x2+1)(x + \sqrt{x^{2} + 1}) – 1+x2\sqrt{1 + x^{2}}

f '(x) = 1+xx+x2+11 + \frac{x}{x + \sqrt{x^{2} + 1}} × (1+2x2x2+1)\left( 1 + \frac{2x}{2\sqrt{x^{2} + 1}} \right) + log

(x+x2+1)(x + \sqrt{x^{2} + 1}).1 – 2x21+x2\frac{2x}{2\sqrt{1 + x^{2}}}

= 1 + xx2+1\frac{x}{\sqrt{x^{2} + 1}} + log (x+x2+1)(x + \sqrt{x^{2} + 1})– x1+x2\frac{x}{\sqrt{1 + x^{2}}}

= 1 + log(x+x2+1)(x + \sqrt{x^{2} + 1})

When x > 0 f '(x) is > 0 ⇒ f(x) is increasing