MHT CET · Maths · Functions

Let \(A=\{x \in R / x\) is not a positive integer). Let a function \(f\) be defined as \(f: A \rightarrow R\) so that \(f(x)=\frac{2 x}{x-1}\), then \(f\) is

A Not injective.
B surjective but not injective.
C neither injective nor surjective.
D injective but not surjective.

Verified Solution

Answer & Solution

Correct Answer

(D) injective but not surjective.

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & f(x)=\frac{2 x}{x-1} \\ & \Rightarrow f^{\prime}(x)=\frac{(x-1) 2-2 x(1-0)}{(x-1)^2}=\frac{-2}{(x-1)^2}<0\end{aligned}\)
\(f(x)\) is strictly decreasing so \(f(x)\) is injective
But \(f(x)=4\)
\(\Rightarrow \frac{2 x}{x-1}=4\)
\(\Rightarrow x=2\) (which is a positive integer)
i.e., \(4 \in R\) (co-domain of \(f\) ) has no pre-image in \(A\) (domain of \(f\) )
So, ' \(f\) ' is not surjective

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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