The function f(x)= tan -1((1-x2/1+x2)) is

Question

The function f(x)= tan -1((1-x2/1+x2)) is (A) increasing in its domain (B) decreasing in its domain (C) decreasing in (-∞, 0) and increasing in (0, ∞) (D) increasing in (-∞, 0) and decreasing in (0, ∞)

Tardigrade Admin · Accepted Answer

put mathrmx2= tan θ to get f(x)=(π/4)- tan -1(x2) ⇒ mathrmf prime( mathrmx)=-(2 mathrmx/1+ mathrmx4) which is greater than zero for mathrmx<0 and less than zero for mathrmx>0 ⇒ ( mathrmD)]

Q. The function $f (x) = tan^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})$ is

increasing in its domain

decreasing in its domain

decreasing in $(- \infty, 0)$ and increasing in $(0, \infty)$

increasing in $(- \infty, 0)$ and decreasing in $(0, \infty)$

put $x^{2} = tan θ$ to get
$f (x) = \frac{π}{4} - tan^{- 1} (x^{2})$
$\Rightarrow f^{'} (x) = - \frac{2 x}{1 + x ^{4}}$ which is greater than zero for $x < 0$ and less than zero for $x > 0$
$\Rightarrow (D)]$

Q. The function f(x)=tan−1(1+x21−x2​) is

increasing in its domain

decreasing in its domain

decreasing in (−∞,0) and increasing in (0,∞)

increasing in (−∞,0) and decreasing in (0,∞)

put x2=tanθ to get f(x)=4π​−tan−1(x2) ⇒f′(x)=−1+x42x​ which is greater than zero for x<0 and less than zero for x>0 ⇒(D)]

Q. The function $f (x) = tan^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})$ is

decreasing in $(- \infty, 0)$ and increasing in $(0, \infty)$

increasing in $(- \infty, 0)$ and decreasing in $(0, \infty)$

put $x^{2} = tan θ$ to get
$f (x) = \frac{π}{4} - tan^{- 1} (x^{2})$
$\Rightarrow f^{'} (x) = - \frac{2 x}{1 + x ^{4}}$ which is greater than zero for $x < 0$ and less than zero for $x > 0$
$\Rightarrow (D)]$