KCET · Maths · Functions
Let \(A=\{x: x \in R, x\) is not a positive integer \()\) Define \(f: A \rightarrow R\) as \(f(x)=\frac{2 x}{x-1}\), then \(f\) is
- A injective but not surjective.
- B surjective but not injective.
- C bijective.
- D neither injective nor surjective.
Answer & Solution
Correct Answer
(A) injective but not surjective.
Step-by-step Solution
Detailed explanation
Given function,
\(f(x)=\frac{2 x}{x-1}\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned}
f^{\prime}(x) &=\frac{(x-1)(2)-2 x(1-0)}{(x-1)^{2}} \\
&=\frac{2 x-2-2 x}{(x-1)^{2}}=\frac{-2}{(x-1)^{2}} < 0
\end{aligned}\)
Function is strictly decreasing.
Function is injective
\(\frac{2 x}{x-1}=y \Rightarrow 2 x=y x-y\)
\(\Rightarrow \quad y=x(y-2)\)
Let \(\quad x=\frac{y}{y-2} \notin \pi\) for \(y=2\)
\(f\) is not surjective.
\(f(x)=\frac{2 x}{x-1}\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned}
f^{\prime}(x) &=\frac{(x-1)(2)-2 x(1-0)}{(x-1)^{2}} \\
&=\frac{2 x-2-2 x}{(x-1)^{2}}=\frac{-2}{(x-1)^{2}} < 0
\end{aligned}\)
Function is strictly decreasing.
Function is injective
\(\frac{2 x}{x-1}=y \Rightarrow 2 x=y x-y\)
\(\Rightarrow \quad y=x(y-2)\)
Let \(\quad x=\frac{y}{y-2} \notin \pi\) for \(y=2\)
\(f\) is not surjective.
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