Let f(x) be a differentiable function such th | vedclass

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

Question

(D) Given $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$. Substitute $x=1$ and $y=1$ into the functional equation: $f(1+1) = f(1) + k(1)(1) +

Vedclass · Accepted Answer

$\frac{4}{3}x^2+\frac{2}{3}$

Vedclass · Answer

(D) Given $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$. Substitute $x=1$ and $y=1$ into the functional equation: $f(1+1) = f(1) + k(1)(1) + \frac{4}{3}(1)^2$ $f(2) = f(1) + k + \frac{4}{3}$ $6 = 2 + k + \frac{4}{3}$ $4 = k + \frac{4}{3}$ $k = 4 - \frac{4}{3} = \frac{8}{3}$. Now,substitute $k = \frac{8}{3}$ into the original equation: $f(x+y) = f(x) + \frac{8}{3}xy + \frac{4}{3}y^2$. To find $f(x)$,we use the definition of the derivative: $f'(x) = \lim_{h 	o 0} \frac{f(x+h)-f(x)}{h} = \lim_{h 	o 0} \frac{\frac{8}{3}xh + \frac{4}{3}h^2}{h} = \lim_{h 	o 0} (\frac{8}{3}x + \frac{4}{3}h) = \frac{8}{3}x$. Integrating $f'(x) = \frac{8}{3}x$ with respect to $x$: $f(x) = \int \frac{8}{3}x \, dx = \frac{8}{3} \cdot \frac{x^2}{2} + C = \frac{4}{3}x^2 + C$. Using $f(1)=2$: $2 = \frac{4}{3}(1)^2 + C \Rightarrow C = 2 - \frac{4}{3} = \frac{2}{3}$. Thus,$f(x) = \frac{4}{3}x^2 + \frac{2}{3}$.

Let $f(x)$ be a differentiable function such that $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$ for all $x, y \in R$. Then $f(x)$ is:

Similar Questions

If $y = \tan^{-1} \left\{ \frac{ax - b}{bx + a} \right\}$,then $y' = $

Differentiate the function with respect to $x$: $\cos(x^{3}) \cdot \sin^{2}(x^{5})$

If $f'(x)$ is zero in the interval $(a, b)$,then in this interval it is

If $f(x)=\cosh ^{-1}\left(\frac{1-x}{1+x}\right)$ is well defined, then $f^{\prime}(x)=$

$\mathop {\text{Limit}}\limits_{h \to 0} \frac{{\int\limits_a^{x + h} {\ln^2 t \, dt} - \int\limits_a^x {\ln^2 t \, dt} }}{h} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module