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Jee Main 2025 — Medium JEE math MCQ

Let \(f : \mathbb{R} \rightarrow \mathbb{R}\) 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
  1. A. \(e^2 - 1\)
  2. B. \(e^2 + 1\)
  3. C. \(e^4 + 1\)
  4. D. \(e^4 - 1\)

Solution

The correct option is **A**. (A. \(e^2 - 1\))

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
Let f:R→Rf : \mathbb{R} \rightarrow \mathbb{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 \mathbb{R}. If f′(0)=4af'(0) = 4a and ff satisfies f′′(x)−3af′(x)−f(x)=0,a>0f''(x) - 3af'(x) - f(x) = 0, a > 0, then the area of the region R={(x,y)∣0≤y≤f(ax),0≤x≤2}R = \{(x, y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2\} is
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