Class 12 Application of Derivatives Mock Test (CBSE)

Topics in Application of Derivatives

• Rate of change of quantities
• Increasing and decreasing functions
• Maxima and minima
• First derivative test
• Second derivative test
• Absolute maxima and minima in a closed interval

Application of Derivatives practice questions with answers

1. Q1Multiple choice·1 mark
   The interval in which the function $f(x) = x^2 - 4x + 6$ is strictly increasing is:

   1. (a)$(-\infty, 2)$
   2. (b)$(2, \infty)$
   3. (c)$(-\infty, \infty)$
   4. (d)$(-\infty, -2)$
   Show answer and step marks

   1. Correct option: (b) $(2, \infty)$. Find the derivative $f'(x) = 2x - 4$ and set $f'(x) > 0$ for strictly increasing function, which gives $2x - 4 > 0 \implies x > 2$ or $(2, \infty)$.+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   The radius of a circle is increasing at the rate of $0.5\text{ cm/s}$. The rate of increase of its circumference when the radius is $5\text{ cm}$ is:

   1. (a)$\pi\text{ cm/s}$
   2. (b)$2\pi\text{ cm/s}$
   3. (c)$0.5\pi\text{ cm/s}$
   4. (d)$5\pi\text{ cm/s}$
   Show answer and step marks

   1. Correct option: (a) $\pi\text{ cm/s}$. The circumference of a circle is $C = 2\pi r$. Differentiating with respect to time $t$, we get $\frac{dC}{dt} = 2\pi \frac{dr}{dt} = 2\pi(0.5) = \pi\text{ cm/s}$.+1

   Answer: (a)
3. Q3Short answer·3 marks
   Find all local maximum and local minimum values of the function $f(x) = x^3 - 3x^2 - 9x + 5$ using the second derivative test.
   Show answer and step marks

   1. Find the first derivative: $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$.+0.5
   2. Set $f'(x) = 0$: critical points $x = -1$ and $x = 3$.+0.5
   3. $f''(x) = 6x - 6$, so $f''(-1) = -12 < 0$: local maximum at $x = -1$, value $f(-1) = -1 - 3 + 9 + 5 = 10$.+1
   4. $f''(3) = 12 > 0$: local minimum at $x = 3$, value $f(3) = 27 - 27 - 27 + 5 = -22$.+1

   Answer: Local maximum value is $10$ at $x = -1$ and local minimum value is $-22$ at $x = 3$
4. Q4Short answer·3 marks
   Find the intervals in which the function $f$ given by $f(x) = 2x^3 - 9x^2 + 12x + 15$ is (a) strictly increasing, (b) strictly decreasing.
   Show answer and step marks

   1. Find the first derivative: $f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)$.+1
   2. $f'(x) = 0$ gives $x = 1$ and $x = 2$, dividing the real line into $(-\infty, 1)$, $(1, 2)$ and $(2, \infty)$.+1
   3. (a) $f'(x) > 0$ on $(-\infty, 1)$ and $(2, \infty)$, so $f$ is strictly increasing on $(-\infty, 1) \cup (2, \infty)$.+0.5
   4. (b) $f'(x) < 0$ on $(1, 2)$, so $f$ is strictly decreasing on $(1, 2)$.+0.5

   Answer: Strictly increasing on $(-\infty, 1) \cup (2, \infty)$ and strictly decreasing on $(1, 2)$

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