Search any question & find its solution
Question:
Answered & Verified by Expert
Let $f:(-1,1) \rightarrow$ IR be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. If $g(x)=(f(2 f(x)+2))^2$, then $g^{\prime}(0)=$
Options:
Solution:
1868 Upvotes
Verified Answer
The correct answer is:
-4
We have,
$$
\begin{aligned}
& g(x)=(f(2 f(x)+2))^2 \\
& g^{\prime}(x)=2\left(f(2 f(x)+2) \cdot f^{\prime}(2 f(x)+2) f^{\prime}(x) \cdot 2\right. \\
& \Rightarrow g^{\prime}(0)=2\left(f(2 f(0)+2) \cdot f^{\prime}(2 f(0)+2) \cdot 2 f^{\prime}(0)\right. \\
& \left.\Rightarrow g^{\prime}(0)=2(f(2)-1)+2\right) \cdot f^{\prime}(2(-1)+2 \cdot 2(1) \\
& {\left[\because f(0)=-1, f^{\prime}(0)=1\right]} \\
& \Rightarrow g^{\prime}(0)=2\left[f(-2+2) \cdot f^{\prime}(-2+2)\right] \cdot 2 \\
& \Rightarrow g^{\prime}(0)=4 f(0) \cdot f^{\prime}\langle 0\rangle \\
& \Rightarrow g^{\prime}(0)=4 \times(-1)(1) \\
& =-4 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& g(x)=(f(2 f(x)+2))^2 \\
& g^{\prime}(x)=2\left(f(2 f(x)+2) \cdot f^{\prime}(2 f(x)+2) f^{\prime}(x) \cdot 2\right. \\
& \Rightarrow g^{\prime}(0)=2\left(f(2 f(0)+2) \cdot f^{\prime}(2 f(0)+2) \cdot 2 f^{\prime}(0)\right. \\
& \left.\Rightarrow g^{\prime}(0)=2(f(2)-1)+2\right) \cdot f^{\prime}(2(-1)+2 \cdot 2(1) \\
& {\left[\because f(0)=-1, f^{\prime}(0)=1\right]} \\
& \Rightarrow g^{\prime}(0)=2\left[f(-2+2) \cdot f^{\prime}(-2+2)\right] \cdot 2 \\
& \Rightarrow g^{\prime}(0)=4 f(0) \cdot f^{\prime}\langle 0\rangle \\
& \Rightarrow g^{\prime}(0)=4 \times(-1)(1) \\
& =-4 \\
&
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.