MHT CET · Maths · Functions
Let \(f(x+y)=f(x) \cdot f(y), \forall x, y \in R\), suppose that \(f(3)=3\) and \(f^{\prime}(0)=11\), then \(f^{\prime}(3)\) is given by
- A 22
- B 44
- C 28
- D 33
Answer & Solution
Correct Answer
(D) 33
Step-by-step Solution
Detailed explanation
\(f^{\prime}(3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}\)
\(\begin{aligned}
=\lim _{h \rightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} \\
\Rightarrow f^{\prime}(3) =3 \lim _{h \rightarrow 0} \frac{f(h)-1}{h}
\end{aligned}
\)
\([\because f(3)=3]\)
Use L'Hospital rule,
\(
\begin{aligned}
\Rightarrow f^{\prime}(3) &=3 \lim _{h \rightarrow 0} \frac{f^{\prime}(h)}{1} \\
&=3 f^{\prime}(0)
\end{aligned}
\)
\(\left[\because f^{\prime}(0)=11\right]\)
\(
=3 \times 11=33
\)
\(\begin{aligned}
=\lim _{h \rightarrow 0} \frac{f(3) \cdot f(h)-f(3)}{h} \\
\Rightarrow f^{\prime}(3) =3 \lim _{h \rightarrow 0} \frac{f(h)-1}{h}
\end{aligned}
\)
\([\because f(3)=3]\)
Use L'Hospital rule,
\(
\begin{aligned}
\Rightarrow f^{\prime}(3) &=3 \lim _{h \rightarrow 0} \frac{f^{\prime}(h)}{1} \\
&=3 f^{\prime}(0)
\end{aligned}
\)
\(\left[\because f^{\prime}(0)=11\right]\)
\(
=3 \times 11=33
\)
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