MHT CET · Maths · Differentiation
If \(\mathrm{g}(x)=[\mathrm{f}(2 \mathrm{f}(x)+2)]^2\) and \(\mathrm{f}(0)=-1, \mathrm{f}^{\prime}(0)=1\), then \(g^{\prime}(0)\) is
- A -4
- B 4
- C -3
- D 3
Answer & Solution
Correct Answer
(A) -4
Step-by-step Solution
Detailed explanation
1. (A). Std. \(12 \mid\) Part-2 \(|\mathrm{Ch}-1|\) Exercise-1.1
\(\mathrm{g}(x) =\{\mathrm{f}[2 \mathrm{f}(x)+2]\}^2 \)
\( \therefore \mathrm{~g}^{\prime}(x) =2 \mathrm{f}[2 \mathrm{f}(x)+2] \cdot \mathrm{f}^{\prime}[2 \mathrm{f}(x)+2] \cdot 2 \mathrm{f}^{\prime}(x) \)
\( \therefore \mathrm{g}^{\prime}(0) =2 \mathrm{f}[2 \mathrm{f}(0)+2] \cdot \mathrm{f}^{\prime}[2 \mathrm{f}(0)+2] \cdot 2 \mathrm{f}^{\prime}(0) \)
\( =2 \mathrm{f}[2(-1)+2] \cdot \mathrm{f}^{\prime}(2(-1)+2) \cdot 2(1) \)
\( \Rightarrow \mathrm{g}^{\prime}(0) =4 \mathrm{f}(0) \cdot \mathrm{f}^{\prime}(0) \)
\( =4(-1)(1) \)
\( =-4\)
\(\mathrm{g}(x) =\{\mathrm{f}[2 \mathrm{f}(x)+2]\}^2 \)
\( \therefore \mathrm{~g}^{\prime}(x) =2 \mathrm{f}[2 \mathrm{f}(x)+2] \cdot \mathrm{f}^{\prime}[2 \mathrm{f}(x)+2] \cdot 2 \mathrm{f}^{\prime}(x) \)
\( \therefore \mathrm{g}^{\prime}(0) =2 \mathrm{f}[2 \mathrm{f}(0)+2] \cdot \mathrm{f}^{\prime}[2 \mathrm{f}(0)+2] \cdot 2 \mathrm{f}^{\prime}(0) \)
\( =2 \mathrm{f}[2(-1)+2] \cdot \mathrm{f}^{\prime}(2(-1)+2) \cdot 2(1) \)
\( \Rightarrow \mathrm{g}^{\prime}(0) =4 \mathrm{f}(0) \cdot \mathrm{f}^{\prime}(0) \)
\( =4(-1)(1) \)
\( =-4\)
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