Question 1

Find the rate of change of the area of a circle with respect to its radius when the radius is 5 cm.

Accepted Answer

Let the area of the circle be A = πr^2. The rate of change of area with respect to radius is dA/dr = d(πr^2)/dr = 2πr. When r = 5 cm, dA/dr = 2π × 5 = 10π cm^2/cm. — We differentiate A = πr^2 with respect to r to get dA/dr = 2πr. Substituting r = 5 gives dA/dr = 10π.

Question 2

The volume of a cube is increasing at the rate of 8 cm^3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Accepted Answer

Let the edge of the cube be x cm. Volume V = x^3, Surface area S = 6x^2. Given dV/dt = 8 cm^3/s. dV/dt = 3x^2 dx/dt ⇒ dx/dt = dV/dt / (3x^2) = 8 / (3 × 12^2) = 8 / 432 = 1/54 cm/s. Now, dS/dt = 12x dx/dt = 12 × 12 × (1/54) = 144 / 54 = 8/3 cm^2/s. — We relate the rates using chain rule: dS/dt = dS/dx × dx/dt. First, find dx/dt from dV/dt, then substitute into dS/dt.

Question 3

Find the intervals in which the function f(x) = x^2 - 4x + 6 is increasing or decreasing.

Accepted Answer

First, find f'(x): f'(x) = 2x - 4. Set f'(x) = 0 ⇒ 2x - 4 = 0 ⇒ x = 2. For x < 2, f'(x) < 0 ⇒ f(x) is decreasing on (-∞, 2). For x > 2, f'(x) > 0 ⇒ f(x) is increasing on (2, ∞). — The sign of the derivative determines increasing/decreasing. Compute f'(x), find where it is positive/negative.

Question 4

Find the equation of the tangent and normal to the curve y = x^2 + 2x at the point (1, 3).

Accepted Answer

First, find dy/dx: dy/dx = 2x + 2. At x = 1, dy/dx = 2 × 1 + 2 = 4. Equation of tangent: y - 3 = 4(x - 1) ⇒ y = 4x - 1. Slope of normal = -1/4. Equation of normal: y - 3 = -1/4(x - 1) ⇒ y = -1/4x + 3.25. — Differentiate y = x^2 + 2x to get slope at x = 1. Use point-slope form for tangent and normal.

Question 5

Using derivatives, find the approximate value of √(26).

Accepted Answer

Let f(x) = √x. Choose x = 25, so f(25) = 5. f'(x) = 1/(2√x). At x = 25, f'(25) = 1/(2 × 5) = 0.1. Using linear approximation: f(26) ≈ f(25) + f'(25) × (26 - 25) = 5 + 0.1 × 1 = 5.1. — Linear approximation: f(a + h) ≈ f(a) + f'(a)h. Substitute a = 25, h = 1.

Question 6

Find the maximum and minimum values of the function f(x) = x^2 - 6x + 9.

Accepted Answer

First, f'(x) = 2x - 6. Set f'(x) = 0 ⇒ 2x - 6 = 0 ⇒ x = 3. f''(x) = 2 > 0, so x = 3 is a minimum. Minimum value: f(3) = 3^2 - 6 × 3 + 9 = 9 - 18 + 9 = 0. Since the function is a parabola opening upwards, there is no maximum value (as x → ±∞, f(x) → ∞). — Find critical points by setting derivative to zero. Use second derivative test to classify minima/maxima.

Question 7

If $y = f(x)$, then the derivative $\frac{dy}{dx}$ represents:

Accepted Answer

The rate of change of $y$ with respect to $x$ — The derivative $\frac{dy}{dx}$ gives the instantaneous rate at which $y$ changes as $x$ changes. It does not represent the average value, sum, or area.

Question 8

The derivative of a function at a point gives the slope of the tangent to the curve at that point.

Accepted Answer

True — The derivative $f'(x_0)$ at $x_0$ gives the slope of the tangent line to the curve $y = f(x)$ at the point $(x_0, f(x_0))$.

Question 9

If the radius of a circle is increasing at a rate of $2$ cm/s, what is the rate of change of its area when the radius is $5$ cm?

Accepted Answer

20\pi 	ext{ cm}^2/	ext{s} — Given: $\frac{dr}{dt} = 2$ cm/s, $r = 5$ cm
Find: $\frac{dA}{dt}$
Formula: $A = \pi r^2 \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt}$
Solution:
Step 1: Substitute values: $\frac{dA}{dt} = 2\pi 	imes 5 	imes 2$
Step 2: $= 20\pi$
Step 3: $= 20\pi$ cm$^2$/s
Answer: $20\pi$ cm$^2$/s
Note: Students sometimes forget to multiply by $\frac{dr}{dt}$ or use wrong formula.

Question 10

A function $f(x)$ is said to be increasing on an interval if:

Accepted Answer

$f'(x) > 0$ for all $x$ in the interval — A function is increasing on an interval if its derivative is positive throughout that interval.

Question 11

Assertion (A): If $f'(x) = 0$ at $x = c$, then $f(x)$ always has a maximum or minimum at $x = c$.
Reason (R): Critical points occur where the derivative is zero or does not exist.
A) Both A and R are true and R is the correct explanation of A
B) Both A and R are true but R is NOT the correct explanation of A
C) A is true but R is false
D) A is false but R is true

Accepted Answer

D) A is false but R is true. Explanation: The assertion is false because $f'(x) = 0$ is a necessary but not sufficient condition for maxima or minima; it could also be a point of inflection. The reason is true, as critical points are where the derivative is zero or does not exist. — A is false: $f'(x) = 0$ does not guarantee a maximum or minimum (e.g., $f(x) = x^3$ at $x=0$).
R is true: Critical points are where $f'(x) = 0$ or $f'(x)$ does not exist.
R does not explain A because A is false.

Question 12

The normal to the curve $y = f(x)$ at $x = a$ has a slope of _____, provided $f'(a) 
eq 0$.

Accepted Answer

-1/f'(a) — The slope of the normal is the negative reciprocal of the tangent's slope, which is $-1/f'(a)$.

Question 13

Explain the difference between the average rate of change and the instantaneous rate of change of a function. Give one example.

Accepted Answer

— The average rate of change of a function between two points is the total change in the function value divided by the change in the variable, while the instantaneous rate of change is given by the derivative at a specific point. For example, the average velocity over a time interval is the total distance divided by time, while the instantaneous velocity at a moment is the derivative of distance with respect to time.

Question 14

For the function $f(x) = x^3 - 3x + 1$, find the equation of the tangent at $x = 2$.

Accepted Answer

y = 9x - 23 — Given: $f(x) = x^3 - 3x + 1$, $x_0 = 2$
Find: Equation of tangent at $x = 2$
Formula: Slope $m = f'(x_0)$; Equation: $y - y_0 = m(x - x_0)$
Solution:
Step 1: $f'(x) = 3x^2 - 3$; $f'(2) = 3	imes4 - 3 = 9$
Step 2: $y_0 = f(2) = 8 - 6 + 1 = 3$
Step 3: $y - 3 = 9(x - 2) \implies y = 9x - 15$
Correction: $y = 9x - 15$ (corrected from above)
Answer: $y = 9x - 15$
Note: Students may forget to compute $y_0$ correctly.

Question 15

If $f(x) = x^2$, use linear approximation to estimate $f(2.1)$.

Accepted Answer

4.2 — Given: $f(x) = x^2$, $a = 2$, $x = 2.1$
Find: Approximate $f(2.1)$ using linear approximation
Formula: $f(x) \approx f(a) + f'(a)(x - a)$
Solution:
Step 1: $f(2) = 4$, $f'(x) = 2x$, $f'(2) = 4$
Step 2: $f(2.1) \approx 4 + 4 	imes (2.1 - 2) = 4 + 4 	imes 0.1 = 4.4$
Correction: $4 + 0.4 = 4.4$
Answer: $4.4$
Note: Students may forget to use $x - a$ or use wrong $f'(a)$.

Application of Derivatives

Application of Derivatives — Study Notes

Introduction

Rate of Change of Quantities

Increasing and Decreasing Functions

Practice Questions — Application of Derivatives

All 6 Chapters in Mathematics Part-I