MHT CET · Maths · Functions
\(f: R \rightarrow R ; g: R \rightarrow R\) are two functions such that \(\mathrm{f}(x)=2 x-3, \mathrm{~g}(x)=x^3+5\), then \((f \circ g)^{-1}(-9)\) is
- A \(-2\)
- B \(2\)
- C \(-\sqrt{2}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(-2\)
Step-by-step Solution
Detailed explanation
We have, \(\mathrm{f}(x)=2 x-3, \mathrm{~g}(x)=x^3+5\)
\(\begin{aligned}
\operatorname{fog}(x) & =\mathrm{f}(\mathrm{g}(x)) \\
& =2 \mathrm{~g}(x)-3 \\
& =2\left(x^3+5\right)-3=2 x^3+7
\end{aligned}\)
Let \((f \circ g)(x)=y=2 x^3+7\)
\(\begin{aligned}
& y=2 x^3+7 \\
& \Rightarrow y-7=2 x^3 \\
& \Rightarrow x^3=\frac{y-7}{2} \\
& \Rightarrow x=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\
\therefore \quad & (\text { fog })^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\
\therefore \quad & (\text { fog })^{-1}(-9)=\left(\frac{-9-7}{2}\right)^{\frac{1}{3}}=(-8)^{\frac{1}{3}}=-2
\end{aligned}\)
\(\begin{aligned}
\operatorname{fog}(x) & =\mathrm{f}(\mathrm{g}(x)) \\
& =2 \mathrm{~g}(x)-3 \\
& =2\left(x^3+5\right)-3=2 x^3+7
\end{aligned}\)
Let \((f \circ g)(x)=y=2 x^3+7\)
\(\begin{aligned}
& y=2 x^3+7 \\
& \Rightarrow y-7=2 x^3 \\
& \Rightarrow x^3=\frac{y-7}{2} \\
& \Rightarrow x=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\
\therefore \quad & (\text { fog })^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\
\therefore \quad & (\text { fog })^{-1}(-9)=\left(\frac{-9-7}{2}\right)^{\frac{1}{3}}=(-8)^{\frac{1}{3}}=-2
\end{aligned}\)
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