MHT CET · Maths · Differentiation
If \(\mathrm{f}\) and \(\mathrm{g}\) are differentiable functions satisfying \(\mathrm{g}^{\prime}(\mathrm{a})=2, \mathrm{~g}(\mathrm{a})=\mathrm{b}\) and \(\mathrm{fog}=\mathrm{I}\), where is an identity function, then \(\mathrm{f}^{\prime}(\mathrm{b})\) is equal to
- A \(\frac{1}{2}\)
- B \(\frac{3}{2}\)
- C \(\frac{2}{3}\)
- D 2
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given
\(
g(a)=b, g^{\prime}(a)=2, f[g(x)]=x
\)
Now \(f^{\prime}[g(x)] g^{\prime}(x)=1 \Rightarrow f^{\prime}(g(x))=\frac{1}{g^{\prime}(x)}\)
Put \(x=a\), we get
\(
f^{\prime}[g(a)]=\frac{1}{g^{\prime}(a)} \Rightarrow f^{\prime}(b)=\frac{1}{2}
\)
\(
g(a)=b, g^{\prime}(a)=2, f[g(x)]=x
\)
Now \(f^{\prime}[g(x)] g^{\prime}(x)=1 \Rightarrow f^{\prime}(g(x))=\frac{1}{g^{\prime}(x)}\)
Put \(x=a\), we get
\(
f^{\prime}[g(a)]=\frac{1}{g^{\prime}(a)} \Rightarrow f^{\prime}(b)=\frac{1}{2}
\)
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