MHT CET · Maths · Functions
If \(f: R \rightarrow R\) is given by \(f(x)=7 x+8\) and \(f^{-1}(12)=\frac{k}{7}\), then the value of \(k\) is
- A 7
- B 1
- C 4
- D 8
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
(C)
We have \(f(x)=7 x+8=y\)...(let)
\(\therefore \mathrm{x}=\frac{\mathrm{y}-8}{7} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{\mathrm{y}-8}{7}\) \(\therefore \mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}-8}{7} \Rightarrow \mathrm{f}^{-1}(12)=\frac{12-8}{7}=\frac{4}{7} \Rightarrow \mathrm{k}=4\)
We have \(f(x)=7 x+8=y\)...(let)
\(\therefore \mathrm{x}=\frac{\mathrm{y}-8}{7} \Rightarrow \mathrm{f}^{-1}(\mathrm{y})=\frac{\mathrm{y}-8}{7}\) \(\therefore \mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}-8}{7} \Rightarrow \mathrm{f}^{-1}(12)=\frac{12-8}{7}=\frac{4}{7} \Rightarrow \mathrm{k}=4\)
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