2022 SJI P2 Q9

2022 SJI P2 Q9

IB Year 5 | Grade 11
16 marks

Let \(\mathrm f(x)=b(x+1)\mathrm e^{-2x}\), where \(b>0\), \(x\in\mathbb R\).

    1. Find \(\mathrm f'(x)\) in terms of \(b\).
    2. Hence find the coordinates of the maximum point of \(y=\mathrm f(x)\) in terms of \(b\).[4]
    1. Given that \(\mathrm f(0)=3\), show that \(b=3\).
    2. Sketch the graph of \(y=\mathrm f(x)\), indicating clearly the coordinates of the turning points, axial intercepts and the equation of the asymptote.
    3. State \(\displaystyle\lim_{x\to\infty}\mathrm f(x)\).
    4. Find the coordinates of the point of inflexion and justify your answer.[9]
  1. Using \(b=3\), find the exact range of values of \(k\) for which \([\mathrm f(x)]^2=k\) has three distinct real roots.[3]

Solution:

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Answer:(a) \(\mathrm f'(x)=-b(2x+1)\mathrm e^{-2x}\); maximum \((-\dfrac12,\dfrac{b\mathrm e}2)\). (b) \(b=3\); asymptote \(y=0\); limit \(0\); inflexion \((0,3)\). (c) \(0<k<\dfrac{9\mathrm e^2}4\)

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