If f(x) = 2 tan-1 x + sin-1 ((2x/1+x2)), x 1, then f(t) is equal to :

Question

If f(x) = 2 tan-1 x + sin-1 ((2x/1+x2)), x > 1, then f(t) is equal to : (A) (π/2) (B) π (C) 4 tan-1 (5) (D)  tan-1 ((65/156))

Tardigrade Admin · Accepted Answer

f(x)=2 tan -1 x+ sin -1((2 x/1+x2)) x>1, f(5)=? We know that 2 tan -1 x= begincases sin -1((2 x/1+x2)) 1 le x le 1 -π- sin -1((2 x/1+x2)) x < -1 π- sin -1((2 x/1+x2)) x> 1 endcases ⇒ f(x)=2 tan -1 x+(π-2 tan -1 x) ⇒ f(x)=π ∀ x>1 ⇒ f(5)=π

Q. If $f (x) = 2 tan^{- 1} x + sin^{- 1} (\frac{2 x}{1 + x ^{2}}), x > 1$ , then $f (t)$ is equal to :

$\frac{π}{2}$

$π$

$4 tan^{- 1} (5)$

$tan^{- 1} (\frac{65}{156})$

Q. If f(x)=2tan−1x+sin−1(1+x22x​),x>1, then f(t) is equal to :

2π​

π

4tan−1(5)

tan−1(15665​)

f(x)=2tan−1x+sin−1(1+x22x​)x>1,f(5)=? We know that 2tan−1x=⎩⎨⎧​sin−1(1+x22x​)1≤x≤1−π−sin−1(1+x22x​)x<−1π−sin−1(1+x22x​)x>1​ ⇒f(x)=2tan−1x+(π−2tan−1x) ⇒f(x)=π∀x>1⇒f(5)=π

Q. If $f (x) = 2 tan^{- 1} x + sin^{- 1} (\frac{2 x}{1 + x ^{2}}), x > 1$ , then $f (t)$ is equal to :

$\frac{π}{2}$

$π$

$4 tan^{- 1} (5)$

$tan^{- 1} (\frac{65}{156})$