Set, Relations and FunctionsPYQ July 21Question 1967 of 217
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Let A=R−{3}\displaystyle A = R - \{3\} and B=R−{1}\displaystyle B = R - \{1\}. Let f:A→B\displaystyle f: A \to B defined by f(x)=x−2x−3\displaystyle f(x) = \frac{x-2}{x-3}. What is the value of f−1(12)\displaystyle f^{-1}(\frac{1}{2})?

Options

A1/2\displaystyle 1/2
B2/3\displaystyle 2/3
C1\displaystyle 1
D3/4\displaystyle 3/4
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Correct Answer

✅ Option c — 1\displaystyle 1

All Options:

  • A1/2\displaystyle 1/2
  • B2/3\displaystyle 2/3
  • C1\displaystyle 1
  • D3/4\displaystyle 3/4

Detailed Solution & Explanation

To find the value of f−1(12)\displaystyle f^{-1}\left(\frac{1}{2}\right), we need to find the value of x∈A\displaystyle x \in A such that f(x)=12\displaystyle f(x) = \frac{1}{2}.

Given the function:
f(x)=x−2x−3f(x) = \frac{x-2}{x-3}
We set f(x)=12\displaystyle f(x) = \frac{1}{2}:
x−2x−3=12\frac{x-2}{x-3} = \frac{1}{2}
Cross-multiplying both sides to solve for x\displaystyle x:
2(x−2)=1(x−3)2(x - 2) = 1(x - 3)
2x−4=x−32x - 4 = x - 3
Subtracting x\displaystyle x from both sides:
x−4=−3x - 4 = -3
Adding 4\displaystyle 4 to both sides:
x=−3+4=1x = -3 + 4 = 1
Since 1∈A\displaystyle 1 \in A (as A=R−{3}\displaystyle A = \mathbb{R} - \{3\}), the value exists and is equal to 1\displaystyle 1.

Hence, **Option C** is the correct answer.

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