Let f:(1$$,∞$$)→(1$$,∞$$) be defined by $$f(x)=x+2x$$−$$1$$.Then

Correct option is 1f is 1 – 1 and onto

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Let $f : (1, \infty) \to (1, \infty)$ be defined by $f (x) = \frac{x + 2}{x - 1} .$
Then

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detailed solution

Correct option is A

If x1+2x1−1=x2+2x2−1 ⇒x1x2+2x2−x1−2 =x1x2+2x1−x2−2⇒3x2=3x1 ⇒ x1=x2So f in one - one If y=f(x)=x+2x−1 ⇒ yx−y=x+2⇒ (y−1)x=y+2⇒ x=y+2y−1∈(1,∞)Thus f is onto

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Let f:(1,∞)→(1,∞) be defined by f(x)=x+2x−1.Then

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Let $f : (1, \infty) \to (1, \infty)$ be defined by $f (x) = \frac{x + 2}{x - 1} .$
Then