The minimum value of the function y = 2x3 - 21x2 + 36x - 20 is:

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  1. -120
  2. -126
  3. -128
  4. None of these

Answer (Detailed Solution Below)

Option 3 : -128
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NIMCET 2020 Official Paper
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120 Questions 480 Marks 120 Mins

Detailed Solution

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Concept:

For a function y = f(x):

  • Relative (Local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
  • Relative (Local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
  • At the points of relative (local) maxima or minima, f'(x) = 0.
  • At the points of relative (local) maxima, f''(x) < 0.
  • At the points of relative (local) minima, f''(x) > 0.

 

Calculation:

Let's say that the function is y = f(x) = 2x3 - 21x2 + 36x - 20.

∴ f'(x) = \(\rm \dfrac{d}{dx}f(x)=\dfrac{d}{dx}(2x^3-21x^2+36x-20)\) = 6x2 - 42x + 36.

And, f''(x) = \(\rm \dfrac{d^2}{dx^2}f(x)=\dfrac{d}{dx}\left[\dfrac{d}{dx}f(x)\right]=\dfrac{d}{dx}(6x^2-42x+36)\) = 12x - 42.

For maxima/minima points, f'(x) = 0.

⇒ 6x2 - 42x + 36 = 0

⇒ x2 - 7x + 6 = 0

⇒ x2 - 6x - x + 6 = 0

⇒ x(x - 6) - (x - 6) = 0

⇒ (x - 6)(x - 1) = 0

⇒ x - 6 = 0 OR x - 1 = 0

⇒ x = 6 OR x = 1.

Now, let's check these points for maxima/minima by inspecting the values of f''(x) at these points.

f''(6) = 12(6) - 42 = 72 - 42 = 30.

f''(1) = 12(1) - 42 = 12 - 42 = -30.

Since, f''(6) = 30 > 0, it is the point of minimum value.

And the minimum value is f(6):

= 2(6)3 - 21(6)2 + 36(6) - 20

= 432 - 756 + 216 - 20

= -128.

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