Let f: R arrow R be a twice differentiable function such that f(x+y)=f(x) f(y) for all x, y ∈ 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 ≤ y ≤ f(a x), 0 ≤

Correct answer: A. e^2-1 aligned & f(x+y)=f(x) · f(y) & f(x)=e^ λ x f^ (0)=4 a & f^ (x)=λ e^ λ x λ=4 a aligned So, f(x)=e^ 4 x aligned & f^ (x)-3 af ^ (x)-f(x)=0 & λ^2-3 a λ-1=0 & 16 a^2-12 a^2-1=0 4 a^2=1 a=(1)/(2) aligned aligned & F(x)=e^ 2 x & Area =∫_0^2 e^ x dx=e^2-1 aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Area Under Curves

Let f: R arrow R be a twice differentiable function such that...

Let

f: \mathbf{R} \rightarrow \mathbf{R}

be a twice differentiable function such that

f(x+y)=f(x) f(y)

for all

x, y \in \mathbf{R}

. If

f^{\prime}(0)=4 \mathrm{a}

and

f

satisfies

f^{\prime \prime}(x)-3 \mathrm{a} f^{\prime}(x)-f(x)=0

\mathrm{a}\gt0

, then the area of the region

\mathrm{R}=\{(x, y) \mid 0 \leq y \leq f(\mathrm{a} x), 0 \leq x \leq 2\}

is:

Options

A $e^2-1$
B $\mathrm{e}^2+1$
C $e^4+1$
D $e^4-1$

Correct answer

A. $e^2-1$

Step-by-step solution

\begin{aligned} & f(x+y)=f(x) \cdot f(y) \\ & \Rightarrow f(x)=e^{\lambda x} f^{\prime}(0)=4 a \\ & \Rightarrow f^{\prime}(x)=\lambda e^{\lambda x} \Rightarrow \lambda=4 a \end{aligned}

So,

f(x)=e^{4 x}

\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{x})-3 \mathrm{af}{ }^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x})=0 \\ & \Rightarrow \lambda^2-3 \mathrm{a} \lambda-1=0 \\ & \Rightarrow 16 \mathrm{a}^2-12 \mathrm{a}^2-1=0 \Rightarrow 4 \mathrm{a}^2=1 \Rightarrow \mathrm{a}=\frac{1}{2} \end{aligned}

\begin{aligned} & \mathrm{F}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}} \\ & \text{ Area }=\int_0^2 \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^2-1\end{aligned}

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