Integral Calculus-2
Integral Calculus-2
Allen Star Batch
Grade 12
Question:
Let $'x'$ be a real valued differentiable function satisfying $f\left(\frac{x}{y}\right) = f(x) - f(y)$ and $\lim_{x \to 0} \frac{f(1+x)}{x} = 3$. If the area bounded by the curve $y = f(x)$, the Y-axis and the line $y = 3$, where $x,y \in \mathbb{R}$ is $K$, then $K$ is____.
Step-by-Step Solution
Key Concept: Solve the functional equation f(x/y) = f(x) - f(y) using the given limit condition lim_{x→0} f(1+x)/x = 3 to determine f'(x) = 3/x, then integrate to find f(x) = 3ln(x), and finally compute the area between the curve and the line y = 3 using the integral ∫[f(x)=3] 3ln(x) dx with appropriate bounds.
Given the functional equation $f\left(\frac{x}{y}\right) = f(x) - f(y)$, we use the definition of derivative to find $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{f\left(1 + \frac{h}{x}\right)}{h}$. Using the functional equation property, this becomes $f'(x) = \frac{3}{x}$. Integrating gives $f(x) = 3\ln x + c$. With the condition $f(1) = 0$, we get $c = 0$, so $f(x) = 3\ln x$. The required area is $\int_1^e xdy = \int_0^1 e^{y/3} dy = 3[e^{y/3}]_0^1 = 3(e^{1/3} - 1)$ square units.
Correct Answer: 8.15