Area Under the Curve
Area Under Curves
nta_pyq_2025_apr
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that $f(x+y) = f(x)f(y)$ for all $x,y\in\mathbb{R}$. If $f'(0) = 4a$ and $f$ satisfies $f''(x)-3af'(x)-f(x)=0$, $a>0$, then the area of the region $R = \{(x,y)\mid 0\leq y\leq f(ax),\; 0\leq x\leq 2\}$ is:
$e^2-1$
$e^2+1$
$e^4+1$
$e^4-1$
Step-by-Step Solution
Key Concept: From $f(x+y)=f(x)f(y)$ deduce $f(x)=e^{\lambda x}$; use $f'(0)=\lambda=4a$ and the ODE $\lambda^2-3a\lambda-1=0$ to find $a=\tfrac{1}{2}$, giving $f(x)=e^{2x}$ and $f(ax)=e^x$.
$f(x)=e^{\lambda x}$, $f'(0)=\lambda=4a$. Substituting in ODE: $\lambda^2-3a\lambda-1=0$.
$(4a)^2-3a(4a)-1 = 16a^2-12a^2-1 = 4a^2-1 = 0 \Rightarrow a = \dfrac{1}{2}$.
So $\lambda=2$, $f(x)=e^{2x}$, $f(ax)=f\!\left(\tfrac{x}{2}\right)=e^x$.
$$\text{Area} = \int_0^2 e^x\,dx = e^2-1.$$
Correct Answer: 1