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Application of Derivatives question

2021 · 20 Jul · Shift 2 · Q32
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  5. /2021 · 20 Jul · Shift 2 · Q32

Application of Derivatives question

2021 · 20 Jul · Shift 2 · Q32

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
The sum of all the local minimum values of the twice differentiable function f : R →\to→ R defined by f(x)=x3−3x2−3f′′(2)2x+f′′(1)f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)f(x)=x3−3x2−23f′′(2)​x+f′′(1) is :
  1. A
    −-− 22
  2. B
    5
  3. C
    −-− 27
  4. D
    0
View written solutionFree

Correct answer: C

  1. Write the function and compute its derivatives

Given

f(x)=x3−3x2−3f′′(2)2x+f′′(1).f(x)=x^3-3x^2-\frac{3f''(2)}{2}x+f''(1).f(x)=x3−3x2−23f′′(2)​x+f′′(1).

Since the function itself contains f′′(2)f''(2)f′′(2) and f′′(1)f''(1)f′′(1), first compute f′′(x)f''(x)f′′(x) in terms of the polynomial.

Differentiate:

f′(x)=3x2−6x−3f′′(2)2.f'(x)=3x^2-6x-\frac{3f''(2)}{2}.f′(x)=3x2−6x−23f′′(2)​.

Again differentiating,

f′′(x)=6x−6.f''(x)=6x-6.f′′(x)=6x−6.
  1. Find f′′(2)f''(2)f′′(2) and f′′(1)f''(1)f′′(1)

Using

f′′(x)=6x−6,f''(x)=6x-6,f′′(x)=6x−6,

we get

f′′(2)=12−6=6,f''(2)=12-6=6,f′′(2)=12−6=6, f′′(1)=6−6=0.f''(1)=6-6=0.f′′(1)=6−6=0.

So the given function becomes

f(x)=x3−3x2−3⋅62x+0=x3−3x2−9x.f(x)=x^3-3x^2-\frac{3\cdot 6}{2}x+0 = x^3-3x^2-9x.f(x)=x3−3x2−23⋅6​x+0=x3−3x2−9x.
  1. Find critical points

Differentiate:

f′(x)=3x2−6x−9=3(x2−2x−3)=3(x−3)(x+1).f'(x)=3x^2-6x-9=3(x^2-2x-3)=3(x-3)(x+1).f′(x)=3x2−6x−9=3(x2−2x−3)=3(x−3)(x+1).

Hence critical points are

x=−1,x=3.x=-1,\quad x=3.x=−1,x=3.
  1. Determine local minima

Second derivative:

f′′(x)=6x−6.f''(x)=6x-6.f′′(x)=6x−6.

At x=−1x=-1x=−1:

f′′(−1)=−12<0,f''(-1)=-12<0,f′′(−1)=−12<0,

so x=−1x=-1x=−1 is a local maximum.

At x=3x=3x=3:

f′′(3)=12>0,f''(3)=12>0,f′′(3)=12>0,

so x=3x=3x=3 is a local minimum.

Thus there is only one local minimum value, namely f(3)f(3)f(3).

  1. Compute the local minimum value
f(3)=33−3(32)−9(3)=27−27−27=−27.f(3)=3^3-3(3^2)-9(3)=27-27-27=-27.f(3)=33−3(32)−9(3)=27−27−27=−27.

Therefore, the sum of all local minimum values is

−27.-27.−27.
  1. Compare with stored correct answer

Stored correct answer: C

Option C corresponds to

−27,-27,−27,

which matches our result.

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