Let f: R rightarrow R satisfy f(x+y)=2^{x} f( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x+y)=2^{x} f(y)+4^{y} f(x)$.Setting $y=2$,we get $f(x+2)=2^{x} f(2)+4^{2} f(x) = 3 \cdot 2^{x} + 16 f(x)$.Differentiating with respect to

Vedclass · Accepted Answer

$248$

Vedclass · Answer

(C) Given $f(x+y)=2^{x} f(y)+4^{y} f(x)$.Setting $y=2$,we get $f(x+2)=2^{x} f(2)+4^{2} f(x) = 3 \cdot 2^{x} + 16 f(x)$.Differentiating with respect to $x$,we get $f^{\prime}(x+2) = 3 \cdot 2^{x} \ln 2 + 16 f^{\prime}(x)$.For $x=2$,$f^{\prime}(4) = 3 \cdot 2^{2} \ln 2 + 16 f^{\prime}(2) = 12 \ln 2 + 16 f^{\prime}(2)$ ... $(i)$.Alternatively,setting $x=2$ in the original equation,$f(2+y)=2^{2} f(y)+4^{y} f(2) = 4 f(y) + 3 \cdot 4^{y}$.Differentiating with respect to $y$,we get $f^{\prime}(2+y) = 4 f^{\prime}(y) + 3 \cdot 4^{y} \ln 4 = 4 f^{\prime}(y) + 6 \cdot 4^{y} \ln 2$.For $y=2$,$f^{\prime}(4) = 4 f^{\prime}(2) + 6 \cdot 4^{2} \ln 2 = 4 f^{\prime}(2) + 96 \ln 2$ ... $(ii)$.Equating $(i)$ and $(ii)$,$12 \ln 2 + 16 f^{\prime}(2) = 4 f^{\prime}(2) + 96 \ln 2$.$12 f^{\prime}(2) = 84 \ln 2 \implies f^{\prime}(2) = 7 \ln 2$.Substituting into $(ii)$,$f^{\prime}(4) = 4(7 \ln 2) + 96 \ln 2 = 28 \ln 2 + 96 \ln 2 = 124 \ln 2$.Finally,$14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)} = 14 \cdot \frac{124 \ln 2}{7 \ln 2} = 2 \cdot 124 = 248$.

Let $f: R \rightarrow R$ satisfy $f(x+y)=2^{x} f(y)+4^{y} f(x)$ for all $x, y \in R$. If $f(2)=3$,then $14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)}$ is equal to

Similar Questions

If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$

If $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y), \forall x, y \in R$ and $f(1)=10$, then, $\sum_{r=1}^n(f(r))^2=$

Let $f$ be a differentiable function satisfying the relation $f(xy) = xf(y) + yf(x) - 2xy$ for all $x, y > 0$ and $f'(1) = 3$. Which of the following statements is true?

If a function $f$ satisfies $f(x+1)+f(x-1)=\sqrt{2} f(x)$,then $f(x+2)+f(x-2)=$

$A$ function $f: R \rightarrow R$ satisfies $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+f(0)}{3}$ for all $x, y \in R$. If the function $f$ is differentiable at $x=0$, then $f$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module