જો \(f\) : \(A \to B\) એ વિધેય \(f(x)\, = \frac{{x - 1}}{{x - 2}}\) કે જ્યાં \(A\, = R - \{2\}\) અને \(B\, = R - \{1\}\) હોય તો \(f\) એ . . . .

A નું વ્યસ્ત મળે અને \({f^{ - 1}}\left( y \right) = \frac{{2y + 1}}{{y - 1}}\)
B નું વ્યસ્ત મળે અને \({f^{ - 1}}\left( y \right) = \frac{{3y - 1}}{{y - 1}}\)
C વ્યસ્ત મળે નહીં
D નું વ્યસ્ત મળે અને \({f^{ - 1}}\left( y \right) = \frac{{2y - 1}}{{y - 1}}\)

Verified Solution

Answer & Solution

Correct Answer

(D) નું વ્યસ્ત મળે અને \({f^{ - 1}}\left( y \right) = \frac{{2y - 1}}{{y - 1}}\)

Step-by-step Solution

Detailed explanation

Suppose \(y = f\left( x \right)\) \( \Rightarrow y = \frac{{x - 1}}{{x - 2}}\) \( \Rightarrow yx - 2y = x - 1\) \( \Rightarrow \left( {y - 1} \right)x = 2y - 1\) \( \Rightarrow x = {f^{ - 1}}\left( y \right) = \frac{{2y - 1}}{{y - 1}}\) As the function is invertible on the given…

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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