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Application of Derivatives question

2019 · 8 Apr · Shift 1 · Q31
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Application of Derivatives question

2019 · 8 Apr · Shift 1 · Q31

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x ∈\in∈ R, then :
  1. A
    S1 = {–1}; S2 = {0, 2}
  2. B
    S1 = {–2}; S2 = {0, 1}
  3. C
    S1 = {–2, 0}; S2 = {1}
  4. D
    S1 = {–2, 1}; S2 = {0}
View written solutionFree

Correct answer: D

  1. We are given f(x)=9x4+12x3−36x2+25.f(x)=9x^4+12x^3-36x^2+25.f(x)=9x4+12x3−36x2+25. We need to find the sets of local minimum points and local maximum points.

  2. First derivative: f′(x)=36x3+36x2−72x.f'(x)=36x^3+36x^2-72x.f′(x)=36x3+36x2−72x. Factor it: f′(x)=36x(x2+x−2)=36x(x+2)(x−1).f'(x)=36x(x^2+x-2)=36x(x+2)(x-1).f′(x)=36x(x2+x−2)=36x(x+2)(x−1).

So the critical points are: x=−2,  0,  1.x=-2,\;0,\;1.x=−2,0,1.

  1. Use the second derivative test.

Compute: f′′(x)=108x2+72x−72.f''(x)=108x^2+72x-72.f′′(x)=108x2+72x−72.

Now evaluate at each critical point.

  • At x=−2x=-2x=−2: f′′(−2)=108(4)+72(−2)−72=432−144−72=216>0.f''(-2)=108(4)+72(-2)-72=432-144-72=216>0.f′′(−2)=108(4)+72(−2)−72=432−144−72=216>0. Hence, x=−2x=-2x=−2 is a local minimum.

  • At x=0x=0x=0: f′′(0)=−72<0.f''(0)=-72<0.f′′(0)=−72<0. Hence, x=0x=0x=0 is a local maximum.

  • At x=1x=1x=1: f′′(1)=108+72−72=108>0.f''(1)=108+72-72=108>0.f′′(1)=108+72−72=108>0. Hence, x=1x=1x=1 is a local minimum.

  1. Therefore, S1={−2,1}S_1=\{-2,1\}S1​={−2,1} (the set of local minimum points), and S2={0}S_2=\{0\}S2​={0} (the set of local maximum points).

  2. Compare with the options:

  • Option A: incorrect
  • Option B: incorrect
  • Option C: incorrect
  • Option D: correct

So the correct answer is D.

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