AP EAMCET · Maths · Differentiation
A function \(f: R \rightarrow R\) satisfies the relation \(f(x+y)=f(x) \cdot f(y), \forall x, y \in R\) and \(f(x) \neq 0\), \(\forall x \in R\). If \(f\) is differentiable at \(x=0\) and \(f^{\prime}(0)=4\) and \(f(6)=3\), then \(f^{\prime}(6)\) is equal to
- A 0
- B 12
- C 3
- D 6
Answer & Solution
Correct Answer
(B) 12
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } f: R \rightarrow R \\ & f(x+y)=f(x) \cdot f(y), \forall x, y \in R \\ & f(x) \neq 0, \forall x \in R \\ & \text { and } f^{\prime}(0)=4 \text { and } f(6)=3 \\ & \because \quad f(x+y)=f(x) \cdot f(y) ...(i)\\ & \text { Put } x=0 \quad y=0 \\ & \quad…
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