If f:[1,∞ )→ [2,∞ ] is given by f(x)=x+(1/x), then f-1(x) is equal to

Question

If f:[1,∞ )→ [2,∞ ] is given by f(x)=x+(1/x), then f-1(x) is equal to (A)  (x+√x2-4/2)  (B)  (x/1+x2)  (C)  (x-√x2-4/2)  (D)  1+√x-4

Tardigrade Admin · Accepted Answer

We have, f(x)=x(1/x) let f(x)=y=x+(1/x) ⇒ xy=x2+1 ⇒ x2-yx+1=0 ⇒ x=(y± √y2-4/2)=f-1(y) ⇒ f-1(x)=(x± √x2-4/2) ⇒ f-1(x)=(x+√x2-4/2) (neglecting -ve as x>1 )

Q. If $f : [1, \infty) \to [2, \infty]$ is given by $f (x) = x + \frac{1}{x},$ then $f^{- 1} (x)$ is equal to

$\frac{x + x ^{2} - 4}{2}$

$\frac{x}{1 + x ^{2}}$

$\frac{x - x ^{2} - 4}{2}$

$1 + x - 4$

Q. If f:[1,∞)→[2,∞] is given by f(x)=x+x1​, then f−1(x) is equal to

2x+x2−4​​

1+x2x​

2x−x2−4​​

1+x−4​

We have, f(x)=xx1​ let f(x)=y=x+x1​ ⇒ xy=x2+1 ⇒ x2−yx+1=0 ⇒ x=2y±y2−4​​=f−1(y) ⇒ f−1(x)=2x±x2−4​​ ⇒ f−1(x)=2x+x2−4​​ (neglecting −ve as x>1 )

Q. If $f : [1, \infty) \to [2, \infty]$ is given by $f (x) = x + \frac{1}{x},$ then $f^{- 1} (x)$ is equal to

$\frac{x + x ^{2} - 4}{2}$

$\frac{x}{1 + x ^{2}}$

$\frac{x - x ^{2} - 4}{2}$

$1 + x - 4$