A function f :[(5/2), ∞) arrow[(11/4), ∞) defined as f ( x )= x 2-5 x +9. Number of solution of the equation f(x)=f-1(x) will be

Question

A function f :[(5/2), ∞) arrow[(11/4), ∞) defined as f ( x )= x 2-5 x +9. Number of solution of the equation f(x)=f-1(x) will be (A) 0 (B) 1 (C) 2 (D) ∞

Tardigrade Admin · Accepted Answer

∵ f ( x )=( x -(5/2))2+(11/4)R ∴ f text is one one and onto f ( x )= f -1( x ) text will be same as f ( x )= x ⇒ x 2-5 x +9= x ⇒ x 2-6 x +9=0 ⇒ x =3

Q. A function $f : [\frac{5}{2}, \infty) \to [\frac{11}{4}, \infty)$ defined as $f (x) = x^{2} - 5 x + 9$ . Number of solution of the equation $f (x) = f^{- 1} (x)$ will be

0

1

2

$\infty$

$∵ f (x) = (x - \frac{5}{2})^{2} + \frac{11}{4} R$
$∴ f is one one and onto$
$f (x) = f^{- 1} (x) will be same as f (x) = x$
$\Rightarrow x^{2} - 5 x + 9 = x \Rightarrow x^{2} - 6 x + 9 = 0$
$\Rightarrow x = 3$

Q. A function f:[25​,∞)→[411​,∞) defined as f(x)=x2−5x+9. Number of solution of the equation f(x)=f−1(x) will be

0

1

2

∞

∵f(x)=(x−25​)2+411​R ∴f is one one and onto f(x)=f−1(x) will be same as f(x)=x ⇒x2−5x+9=x⇒x2−6x+9=0 ⇒x=3

Q. A function $f : [\frac{5}{2}, \infty) \to [\frac{11}{4}, \infty)$ defined as $f (x) = x^{2} - 5 x + 9$ . Number of solution of the equation $f (x) = f^{- 1} (x)$ will be

$\infty$

$∵ f (x) = (x - \frac{5}{2})^{2} + \frac{11}{4} R$
$∴ f is one one and onto$
$f (x) = f^{- 1} (x) will be same as f (x) = x$
$\Rightarrow x^{2} - 5 x + 9 = x \Rightarrow x^{2} - 6 x + 9 = 0$
$\Rightarrow x = 3$