KCET · Maths · Functions
If \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) is defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}\), then \(\mathrm{f}^{-1}(8)\) is equal to
- A \(\{2\}\)
- B \(\left\{2,2 \omega, 2 \omega^{2}\right\}\)
- C \(\{2,-2\}\)
- D \(\{2,2\}\)
Answer & Solution
Correct Answer
(A) \(\{2\}\)
Step-by-step Solution
Detailed explanation
Let
\[
y=f(x)=x^{3}
\]
\(\therefore\)
\(\mathrm{x}=\mathrm{y}^{1 / 3}\)
\(\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^{1 / 3}\)
\(\Rightarrow \quad \mathrm{f}^{-1}(8)=(8)^{1 / 3}\)
\(=2\)
\[
y=f(x)=x^{3}
\]
\(\therefore\)
\(\mathrm{x}=\mathrm{y}^{1 / 3}\)
\(\Rightarrow \quad \mathrm{f}^{-1}(\mathrm{x})=\mathrm{x}^{1 / 3}\)
\(\Rightarrow \quad \mathrm{f}^{-1}(8)=(8)^{1 / 3}\)
\(=2\)
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