Let be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, . If and f satisfies , a 0, then the area of the region is : [2025]

Question

Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

Smart Achievers · Accepted Answer

(2)

We have, f(x + y) = f(x)f(y)

Let $f (x) = e^{λ x}$

Differentiating w.r.t. x, we get

$f' (x) = λ \cdot e^{λ x}$

Put x = 0, we get

$f' (0) = λ$ [ $∵ f' (0) = 4 a$ ]

$\Rightarrow λ = 4 a$

So, $f (x) = e^{4 a x}$

Now, $f'' (x) - 3 a f' (x) - f (x) = 0$

$\Rightarrow \frac{d}{d x} (λ e^{λ x}) - 3 a (λ e^{λ x}) - e^{λ x} = 0$

$\Rightarrow λ^{2} - 3 a λ - 1 = 0$

$\Rightarrow 16 a^{2} - 12 a^{2} - 1 = 0 \Rightarrow 4 a^{2} = 1$

$\Rightarrow a = \frac{1}{2}$ [ $∵ a > 0$ ]

$∴ f (x) = e^{2 x}$

Now, $f (x) = f (a x) = e^{x}$

$∴$ Area of region $\int_{0}^{2} e^{x} d x = {[e^{x}]}_{0}^{2} = e^{2} - 1$ .

Solution

Q.
Let $f : R \to R$ be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, $y \in R$ . If $f' (0) = 4 a$ and f satisfies $f'' (x) = 3 a f' (x) - f (x) = 0$ , a > 0, then the area of the region $R = {(x, y) | 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is : [2025]

1 $e^{4} + 1$

2 $e^{2} - 1$

3 $e^{4} - 1$

4 $e^{2} + 1$

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Solution

Q. Let f : R→R be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, y∈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) | 0≤y≤f(ax), 0≤x≤2} is : [2025]

1 e4+1

2 e2–1

3 e4–1