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QJEEM25Ma6048: JEE Main 2025 Mathematics - Class 12 JEEM25Ma6048 Solution: Let be a twice differentiable function such that for all . If…

Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} be a twice differentiable function such that f(x+y)=f(x)f(y)f(x+y)=f(x) f(y) for all x,y∈Rx, y \in \mathbf{R} . If f′(0)=4af^{\prime}(0)=4 \mathrm{a} and ff satisfies f^{\prime \prime}(x)-3 \mathrm{a} f^{\prime}(x)-f(x)=0, \mathrm{a}>0 , then the area of the region R={(x,y)∣0≤y≤f(ax),0≤x≤2}\mathrm{R}=\{(x, y) \mid 0 \leq y \leq f(a x), 0 \leq x \leq 2\} is :

Multiple Choice Options

  • A e2−1\mathrm{e}^2-1
  • B e4+1e^4+1
  • C e2+1\mathrm{e}^2+1
  • D e4−1e^4-1

Step-by-Step Verified Solution

amp;f(x+y)=f(x)f(y)amp;⇒f(x)=eλxf′(0)=4aamp;⇒f′(x)=λeλx⇒λ=4a\begin{aligned} & \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \\ & \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{\lambda \mathrm{x}} \mathrm{f}^{\prime}(0)=4 \mathrm{a} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\lambda \mathrm{e}^{\lambda \mathrm{x}} \Rightarrow \lambda=4 \mathrm{a} \end{aligned}

So, f(x)=e4ax\mathrm{f}(\mathrm{x})=\mathrm{e}^{4 \mathrm{ax}}

amp;f′′(x)−3af′(x)−f(x)=0amp;⇒λ2−3aλ−1=0amp;⇒16a2−12a2−1=0⇒4a2=1⇒a=12\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}

JEE Main 2025 (Online) 22nd January Morning Shift Mathematics - Differential Equations Question 12 English Explanation amp;F(x)=e2xamp; Area =∫02exdx=e2−1\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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