The function f(x)= cot -1 x+x increases in the interval :

Question

The function f(x)= cot -1 x+x increases in the interval : (A) (1, ∞) (B) (-1, ∞) (C) (-∞, ∞) (D) (0, ∞)

Tardigrade Admin · Accepted Answer

Key Idea: A function f(x) is said to be increasing function, if f prime(x)>0 ∵ f(x)= cot -1 x+x ∴ f'(x)=-(1/1+x2)+1=(x2/1+x2) Hence, f(x) is increasing function since f'(x)>0 for all x

Q. The function $f (x) = cot^{- 1} x + x$ increases in the interval :

$(1, \infty)$

$(- 1, \infty)$

$(- \infty, \infty)$

$(0, \infty)$

Key Idea : A function $f (x)$ is said to be increasing function, if $f^{'} (x) > 0$
$∵ f (x) = cot^{- 1} x + x$
$∴ f^{'} (x) = - \frac{1}{1 + x ^{2}} + 1 = \frac{x ^{2}}{1 + x ^{2}}$
Hence, $f (x)$ is increasing function since $f^{'} (x) > 0$ for all $x$

Q. The function f(x)=cot−1x+x increases in the interval :

(1,∞)

(−1,∞)

(−∞,∞)

(0,∞)