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JEE Advanced 2023 Paper 2 · Mathematics · Differential Equations · medium

Let f:mathbb R→mathbb R be a differentiable function satisfying x f'(x)-f(x)=x^2e^x for all real x, and f(1)=e.

One or more options correct

Let f:R→Rf:\mathbb R\to\mathbb R be a differentiable function satisfying xf′(x)−f(x)=x2exx f'(x)-f(x)=x^2e^{x} for all real xx, and f(1)=ef(1)=e. Which of the following is/are correct?
  1. A. f(2)=2e2f(2)=2e^{2}
  2. B. ff has a local minimum at x=−1x=-1 and the minimum value is −1e-\dfrac1e
  3. C. ∫01f(x) dx=e−1\displaystyle\int_0^1 f(x)\,dx=e-1
  4. D. ff is concave on (−∞,0)(-\infty,0)
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Answer: A, B

On x≠0x\neq0, (fx)′=ex\left(\frac{f}{x}\right)'=e^x, so f=x(ex+c)f=x(e^x+c) on each side of 00. Differentiability at 00 forces the same cc on both sides, and f(1)=ef(1)=e gives c=0c=0: f(x)=xexf(x)=xe^x. (A) f(2)=2e2f(2)=2e^2: true. (B) f′(x)=(1+x)exf'(x)=(1+x)e^x changes sign from −- to ++ at x=−1x=-1, minimum value −1/e-1/e: true. (C) ∫01xexdx=[(x−1)ex]01=1≠e−1\int_0^1xe^xdx=[(x-1)e^x]_0^1=1\ne e-1: false. (D) f′′(x)=(x+2)ex>0f''(x)=(x+2)e^x>0 for −2<x<0-2<x<0, so ff is not concave on (−∞,0)(-\infty,0): false.
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