Let f :[(1/2), ∞) arrow[(3/4), ∞) defined by f(x)=x2-x+1 and x0 be the value of x satisfying f(x)=f-1(x). Then the value of sin -1(x0)+ cot -1(-x0) equals

Question

Let f :[(1/2), ∞) arrow[(3/4), ∞) defined by f(x)=x2-x+1 and x0 be the value of x satisfying f(x)=f-1(x). Then the value of sin -1(x0)+ cot -1(-x0) equals (A)  (5 π/4) (B) (π/4) (C) (3 π/4) (D) none

Tardigrade Admin · Accepted Answer

f ( x )= f -1( x ) ⇒ x 0=1 . text Now sin -1(1)+ cot -1(-1)=(π/2)+(3 π/4)=(5 π/4)

Q. Let $f : [\frac{1}{2}, \infty) \to [\frac{3}{4}, \infty)$ defined by $f (x) = x^{2} - x + 1$ and $x_{0}$ be the value of $x$ satisfying $f (x) = f^{- 1} (x)$ . Then the value of $sin^{- 1} (x_{0}) + cot^{- 1} (- x_{0})$ equals

$\frac{5 π}{4}$

$\frac{π}{4}$

$\frac{3 π}{4}$

none

$f (x) = f^{- 1} (x)$
$\Rightarrow x_{0} = 1. Now sin^{- 1} (1) + cot^{- 1} (- 1) = \frac{π}{2} + \frac{3 π}{4} = \frac{5 π}{4}$

Q. Let f:[21​,∞)→[43​,∞) defined by f(x)=x2−x+1 and x0​ be the value of x satisfying f(x)=f−1(x). Then the value of sin−1(x0​)+cot−1(−x0​) equals

45π​

4π​

43π​