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UPCET / UPSEE Maxima/Minima — practice questions

Explore 2 free Maxima/Minima questions in UPCET / UPSEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Maxima/Minima questions and exercises with answers

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At a **local maximum** of a differentiable function $f$:Read the answer and worked solution →$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:Read the answer and worked solution →

Worked practice questions

At a **local maximum** of a differentiable function $f$:

  1. $f'(x) > 0$
  2. $f'(x) = 0$ and $f''(x) < 0$
  3. $f'(x) = 0$ and $f''(x) > 0$
  4. $f''(x) = 0$
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Correct answer: B. $f'(x) = 0$ and $f''(x) < 0$

**At local max:** $f'(x) = 0$ (critical/stationary point) AND $f''(x) < 0$ (curve concave down). **At local min:** $f'(x) = 0$ and $f''(x) > 0$. $f''(x) = 0$ is inconclusive — use higher-derivative test.

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$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:

  1. $x = -1$ (local max value 12)
  2. $x = 2$ (local min)
  3. $x = 0$
  4. $x = 1$
Show answer and solution

Correct answer: A. $x = -1$ (local max value 12)

We have $f'(x) = 6x^2 - 6x - 12$. That is $f'(x) = 6(x-2)(x+1)$. So $f'(x) = 0$. ⇒ $x = -1, 2$. $f''(x) = 12x - 6$. At $x = -1$: $f''(-1) = -18 < 0$ ⇒ local max. At $x = 2$: $f''(2) = 18 > 0$ ⇒ local min. $f(-1) = -2 - 3 + 12 + 5 = 12$.

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