Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
If $y = f(x); f'(x) \geq 0$ & $f(0) = 0$ bounding a curvilinear trapezoid with base $[0, x]$ whose area is proportional to $3^{rd}$ power of $f(x)$. If $f(1) = 3$, then:
Range of $f(\sin^2 x)$ is $[-3, 3]$
Domain of $f(\ln(2^x - 3))$ is $[2, \infty)$
Range of $f(\sec^2 x)$ is $[3, \infty)$
Area bounded by $y = f(x)$, line $x = 0, y = 1$ & $y = 2$ is $\frac{7}{9}$
Step-by-Step Solution
Key Concept: Differentiate the integral condition to obtain a tractable differential equation, then solve by recognizing the functional relationship.
From the integral condition $\int_0^x f(t)dt = kf^3(x) = \frac{1}{2}f^2(x) = \frac{x}{3k}$, we deduce $f(x) = 3\sqrt{x}$. Differentiating the integral equation $\int_0^x f(t)dt = kf^3(x)$ gives $f(x) = 3kf^2(x)f'(x)$, confirming that $f(\sin^2 x) = 3\sin x$. Therefore $\int_0^x f(t)dt = kf^3(x)$ yields the functional form $f(\sin^2 x) = 3\sin x$.
Correct Answer: 2,3,4