The function f(x) = log(1+ x ) - (2x/2+x) is increasing on

Question

The function f(x) = log(1+ x ) - (2x/2+x) is increasing on (A) (-1, ∞) (B) (-∞, 0) (C) (-∞, ∞) (D) None of these

Tardigrade Admin · Accepted Answer

Given, f(x) = log(1+ x ) - (2x/2+x) ⇒ f'(x) = (1/1+x) - ((2+x)2-2x/(2+x)2) ⇒ f'(x) = (x2/(x+1)(x+2)2) Clearly, f'(x) > 0 for all x > -1, hence f(x) is increasing on (-1, ∞).

Q. The function $f (x) = l o g (1 + x) - \frac{2 x}{2 + x}$ is increasing on

$(- 1, \infty)$

$(- \infty, 0)$

$(- \infty, \infty)$

None of these

Given, $f (x) = l o g (1 + x) - \frac{2 x}{2 + x}$
$\Rightarrow f^{'} (x) = \frac{1}{1 + x} - \frac{( 2 + x ) 2 - 2 x}{( 2 + x ) ^{2}}$
$\Rightarrow f^{'} (x) = \frac{x ^{2}}{( x + 1 ) ( x + 2 ) ^{2}}$
Clearly, $f^{'} (x) > 0$ for all $x > - 1$ , hence $f (x)$ is increasing on $(- 1, \infty)$ .

Q. The function f(x)=log(1+x)−2+x2x​ is increasing on

(−1,∞)

(−∞,0)

(−∞,∞)