CBSE Class 12 Mathematics HOTs Application of Derivatives Set 03

Question. The amount of pollution content added in air in a city due to \( x \)-diesel vehicles is given by \( P(x) = 0.005x^3 + 0.02x^2 + 30x \). Find the marginal increase in pollution content when 3 diesel vehicles are added and write which value is indicated in the above question.
Answer: We have, \( P(x) = 0.005x^3 + 0.02x^2 + 30x \)
\( \Rightarrow \frac{dP}{dx} = 0.015x^2 + 0.04x + 30 \)
Now, \( \left(\frac{dP}{dx}\right)_{x=3} = 0.015 \times 3^2 + 0.04 \times 3 + 30 = 30.255 \)
Value indicated in the question is that the increase in pollution is due to the increase in use of diesel vehicles.

Question. The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue). If the total revenue (in rupees) received from the sale of \( x \) units of a product is given by \( R(x) = 3x^2 + 36x + 5 \), find the marginal revenue, when \( x = 5 \), and write which value does the question indicate.
Answer: Total revenue is given by \( R(x) = 3x^2 + 36x + 5 \)
\( \therefore \text{Marginal revenue} = \frac{dR}{dx} = 6x + 36 \)
Now, \( \left(\frac{dR}{dx}\right)_{x=5} = 6 \times 5 + 36 = 66 \)
Value indicated in the question is that more amount of money is spent for the welfare of the employees with the increase in marginal revenue.

Question. The total cost \( C(x) \) associated with provision of free mid-day meals to \( x \) students of a school in primary classes is given by \( C(x) = 0.005x^3 - 0.02x^2 + 30x + 50 \). If the marginal cost is given by rate of change \( \frac{dC}{dx} \) of total cost, write the marginal cost of food for 300 students. What value is shown here?
Answer: Given, \( C(x) = 0.005x^3 - 0.02x^2 + 30x + 50 \)
Marginal cost \( = \frac{dC}{dx} = 0.015x^2 - 0.04x + 30 \)
Now, \( \left(\frac{dC}{dx}\right)_{x=300} = 0.015 \times 300^2 - 0.04 \times 300 + 30 = 1350 - 12 + 30 = 1368 \).
The value indicated here is that a kind of care and concern is shown towards the health of students of primary classes by providing free mid-day meal to them.

Question. The total expenditure (in \( \text{₹} \)) required for providing the cheap edition of a book for poor and deserving students is given by \( R(x) = 3x^2 + 36x \) where \( x \) is the number of sets of books. If the marginal expenditure is defined as \( \frac{dR}{dx} \), write the marginal expenditure required for 1200 such sets. What value is reflected in this question?
Answer: Here, \( R(x) = 3x^2 + 36x \),
\( \therefore \text{Marginal expenditure} = \frac{dR}{dx} = 6x + 36 \)
\( \left(\frac{dR}{dx}\right)_{x=1200} = 6 \times 1200 + 36 = 7236 \).
The value indicated here is that a kind of help is provided to poor and deserving students who want to study but they don’t have sources to purchase books.

Question. The side of an equilateral triangle is increasing at the rate of \( 2\text{ cm/s} \). At what rate is its area increasing when the side of the triangle is \( 20\text{ cm} \)?
Answer: Let \( a \) be the side of the equilateral triangle.
Then \( \frac{da}{dt} = 2\text{ cm/sec} \)
Let \( A \) be the area of the equilateral triangle, then
\( A = \frac{\sqrt{3}}{4} a^2 \Rightarrow \frac{dA}{dt} = 2 \times \frac{\sqrt{3}}{4} a \frac{da}{dt} = \frac{\sqrt{3}}{2} a \frac{da}{dt} \)
\( \therefore \left(\frac{dA}{dt}\right)_{a=20} = \frac{\sqrt{3}}{2} \times 20 \times 2 = 20\sqrt{3}\text{ cm}^2\text{/sec} \).

Question. The sides of an equilateral triangle are increasing at the rate of \( 2\text{ cm/sec} \). Find the rate at which the area increases, when the side is \( 10\text{ cm} \).
Answer: Let \( a \) be the side of the equilateral triangle.
Then \( \frac{da}{dt} = 2\text{ cm/sec} \)
Let \( A \) be the area of the equilateral triangle, then
\( A = \frac{\sqrt{3}}{4} a^2 \Rightarrow \frac{dA}{dt} = \frac{\sqrt{3}}{2} a \frac{da}{dt} \)
\( \therefore \left(\frac{dA}{dt}\right)_{a=10} = \frac{\sqrt{3}}{2} \times 10 \times 2 = 10\sqrt{3}\text{ cm}^2\text{/sec} \).

Question. A ladder \( 5\text{ m} \) long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of \( 2\text{ cm/s} \). How fast is its height on the wall decreasing when the foot of the ladder is \( 4\text{ m} \) away from the wall?
Answer: Let foot of the ladder is at a distance of \( x \) from the wall and height of the wall is \( y \). Here, \( x^2 + y^2 = 5^2 \)
Differentiating w.r.t. \( t \), we get
\( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \)
When \( x = 4\text{ m} \); \( y^2 = 5^2 - x^2 = 5^2 - 4^2 = 3^2 \Rightarrow y = 3\text{ m} \)
and \( \frac{dx}{dt} = 2\text{ cm/sec} \) (Given)
\( \therefore \frac{dy}{dt} = -\frac{400}{300} \times 2 = -\frac{8}{3}\text{ cm/sec} \).
Therefore, the height of the ladder on the wall is decreasing at the rate of \( \frac{8}{3}\text{ cm/sec} \).

Question. Sand is pouring from a pipe at the rate of \( 12\text{ cm}^3\text{/sec} \). The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is \( 4\text{ cm} \)?
Answer: Let at any instant of time \( t \), the radius of the base of the cone be \( r \), its height be \( h \) and the volume of cone be \( V \), then
\( h = \frac{1}{6}r \Rightarrow r = 6h \)
and \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6h)^2 h = 12\pi h^3 \) ...(i)
Differentiating (i) w.r.t. \( t \), we get
\( \frac{dV}{dt} = (12\pi)\left(3h^2 \frac{dh}{dt}\right) \Rightarrow 12 = 36\pi (4)^2 \frac{dh}{dt} \) (since \( h = 4\text{ cm} \) and \( \frac{dV}{dt} = 12\text{ cm}^3\text{/sec} \))
\( \Rightarrow \frac{dh}{dt} = \frac{12}{36\pi \times 16} = \frac{1}{48\pi}\text{ cm/sec} \).
Therefore, height of the sand cone is increasing at the rate of \( \frac{1}{48\pi}\text{ cm/sec} \).

Question. The length \( x \) of a rectangle is decreasing at the rate of \( 5\text{ cm/minute} \) and width \( y \) is increasing at the rate of \( 4\text{ cm/minute} \). When \( x = 8\text{ cm} \) and \( y = 6\text{ cm} \), find the rate of change of (a) the perimeter, (b) the area of the rectangle.
Answer: Let at any instant of time \( t \), the perimeter of the rectangle be \( P \) and the area be \( A \).
\( \frac{dx}{dt} = -5\text{ cm/min} \) and \( \frac{dy}{dt} = 4\text{ cm/min} \)
(a) \( P = 2(x + y) \)
Differentiating w.r.t. \( t \), we get
\( \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2\text{ cm/min} \).
Therefore, perimeter of the rectangle is decreasing at the rate of \( 2\text{ cm/min} \).
(b) \( A = xy \)
Differentiating w.r.t. \( t \), we get
\( \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = (8)(4) + (6)(-5) = 2\text{ cm}^2\text{/min} \).
Therefore, area of the rectangle is increasing at the rate of \( 2\text{ cm}^2\text{/min} \).

6.3 Increasing and Decreasing Functions

Question. Find the intervals in which the function \( f(x) = 3x^4 - 4x^3 - 12x^2 + 5 \) is (a) strictly increasing, (b) strictly decreasing.
Answer: We have, \( f(x) = 3x^4 - 4x^3 - 12x^2 + 5 \)
\( f'(x) = 12x^3 - 12x^2 - 24x = 12x(x^2 - x - 2) \)
\( \Rightarrow f'(x) = 12x(x + 1)(x - 2) \)
Now, \( f'(x) = 0 \Rightarrow 12x(x + 1)(x - 2) = 0 \Rightarrow x = -1, x = 0 \) or \( x = 2 \).
These points divide the real line into four disjoint open intervals namely \( (-\infty, -1), (-1, 0), (0, 2) \) and \( (2, \infty) \).

(a) \( f(x) \) is strictly increasing in \( (-1, 0) \cup (2, \infty) \).
(b) \( f(x) \) is strictly decreasing in \( (-\infty, -1) \cup (0, 2) \).

Question. Find the value(s) of \( x \) for which \( y = [x(x-2)]^2 \) is an increasing function.
Answer: Here, \( y = [x(x - 2)]^2 = x^2(x - 2)^2 \)
\( \Rightarrow \frac{dy}{dx} = 2x(x - 2)^2 + 2x^2(x - 2) = 2x(x - 2)(x - 2 + x) = 4x(x - 1)(x - 2) \)
For \( y \) to be an increasing function, \( \frac{dy}{dx} > 0 \Rightarrow x(x - 1)(x - 2) > 0 \).
Case 1: When \( -\infty < x < 0 \), \( \frac{dy}{dx} < 0 \Rightarrow y \) is a decreasing function.
Case 2: When \( 0 < x < 1 \), \( \frac{dy}{dx} > 0 \Rightarrow y \) is an increasing function.
Case 3: When \( 1 < x < 2 \), \( \frac{dy}{dx} < 0 \Rightarrow y \) is a decreasing function.
Case 4: When \( 2 < x < \infty \), \( \frac{dy}{dx} > 0 \Rightarrow y \) is an increasing function.
Therefore, \( y \) is an increasing function in \( [0, 1] \cup [2, \infty) \).

Question. Find the intervals in which the function \( f(x) = \frac{3}{2}x^4 - 4x^3 - 45x^2 + 51 \) is (i) strictly increasing, (ii) strictly decreasing.
Answer: We have, \( f(x) = \frac{3}{2}x^4 - 4x^3 - 45x^2 + 51 \)
Differentiating w.r.t. \( x \), we get
\( f'(x) = 6x^3 - 12x^2 - 90x = 6x(x^2 - 2x - 15) = 6x(x - 5)(x + 3) \).
For critical points, \( f'(x) = 0 \Rightarrow x = -3, 0, 5 \).
These points divide the real line into disjoint open intervals namely \( (-\infty, -3), (-3, 0), (0, 5) \) and \( (5, \infty) \).

(i) \( f(x) \) is strictly increasing in \( (-3, 0) \cup (5, \infty) \).
(ii) \( f(x) \) is strictly decreasing in \( (-\infty, -3) \cup (0, 5) \).

Question. Find the intervals in which the function \( f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \) is (a) strictly increasing, (b) strictly decreasing.
Answer: Here, \( f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \)
Differentiating w.r.t. \( x \), we get
\( f'(x) = \frac{3}{10} \cdot 4x^3 - \frac{4}{5} \cdot 3x^2 - 3 \cdot 2x + \frac{36}{5} \cdot 1 = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} \)
\( = \frac{6}{5}(x^3 - 2x^2 - 5x + 6) = \frac{6}{5}(x - 1)(x^2 - x - 6) = \frac{6}{5}(x - 1)(x + 2)(x - 3) \).
For critical points, \( f'(x) = 0 \Rightarrow x = -2, 1, 3 \).
These points divide the real line into disjoint intervals \( (-\infty, -2), (-2, 1), (1, 3) \) and \( (3, \infty) \).

(a) \( f(x) \) is strictly increasing in \( (-2, 1) \cup (3, \infty) \).
(b) \( f(x) \) is strictly decreasing in \( (-\infty, -2) \cup (1, 3) \).

Question. Find the intervals in which the function given by \( f(x) = \sin x + \cos x, 0 \le x \le 2\pi \) is (a) increasing, (b) decreasing.
Answer: The given function is \( f(x) = \sin x + \cos x; 0 \le x \le 2\pi \)
\( \Rightarrow f'(x) = \cos x - \sin x \)
Now \( f'(x) = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4} \).
Thus, \( f'(x) > 0 \) in \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \)
and \( f'(x) < 0 \) in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).
Therefore, the function \( f \) is decreasing in \( \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] \) and it is increasing in \( \left[0, \frac{\pi}{4}\right] \cup \left[\frac{5\pi}{4}, 2\pi\right] \).

Question. Find the intervals in which the following function is (a) increasing, (b) decreasing : \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21 \)
Answer: The given function is \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21 \)
\( \Rightarrow f'(x) = 4x^3 - 24x^2 + 44x - 24 = 4(x^3 - 6x^2 + 11x - 6) = 4(x - 1)(x^2 - 5x + 6) = 4(x - 1)(x - 2)(x - 3) \).
Thus \( f'(x) = 0 \Rightarrow x = 1, 2, 3 \).
Hence, possible disjoint intervals are \( (-\infty, 1), (1, 2), (2, 3) \) and \( (3, \infty) \).
In the interval \( (-\infty, 1) \), \( f'(x) < 0 \)
In the interval \( (1, 2) \), \( f'(x) > 0 \)
In the interval \( (2, 3) \), \( f'(x) < 0 \)
In the interval \( (3, \infty) \), \( f'(x) > 0 \).
Therefore, \( f \) is increasing in \( [1, 2] \cup [3, \infty) \) and \( f \) is decreasing in \( (-\infty, 1] \cup [2, 3] \).

Question. Prove that \( y = \frac{4\sin\theta}{2 + \cos\theta} - \theta \) is an increasing function in \( \left[0, \frac{\pi}{2}\right] \).
Answer: We have, \( y = \frac{4\sin\theta}{2 + \cos\theta} - \theta, \theta \in \left[0, \frac{\pi}{2}\right] \)
\( \Rightarrow \frac{dy}{d\theta} = 4\left[ \frac{(2 + \cos\theta)\cos\theta - \sin\theta(-\sin\theta)}{(2 + \cos\theta)^2} \right] - 1 = \frac{4(2\cos\theta + 1)}{(2 + \cos\theta)^2} - 1 = \frac{8\cos\theta + 4 - (2 + \cos\theta)^2}{(2 + \cos\theta)^2} \)
\( = \frac{8\cos\theta + 4 - (4 + 4\cos\theta + \cos^2\theta)}{(2 + \cos\theta)^2} = \frac{4\cos\theta - \cos^2\theta}{(2 + \cos\theta)^2} = \frac{\cos\theta(4 - \cos\theta)}{(2 + \cos\theta)^2} \).
Now, \( \cos\theta \ge 0 \) in \( \left[0, \frac{\pi}{2}\right] \); \( 4 - \cos\theta > 0 \) in \( \left[0, \frac{\pi}{2}\right] \) (since \( -1 \le \cos\theta \le 1 \)) and \( (2 + \cos\theta)^2 > 0 \) in \( \left[0, \frac{\pi}{2}\right] \) (being a perfect square).
\( \therefore \frac{dy}{d\theta} \ge 0 \) for all \( \theta \in \left[0, \frac{\pi}{2}\right] \).
Hence \( y \) is an increasing function in \( \left[0, \frac{\pi}{2}\right] \).

Question. Find the intervals in which the following function is (a) increasing, (b) decreasing : \( f(x) = 2x^3 - 9x^2 + 12x - 15 \)
Answer: Here, \( f(x) = 2x^3 - 9x^2 + 12x - 15 \)
\( \Rightarrow f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 2)(x - 1) \).
(a) For \( f \) to be increasing, \( f'(x) \ge 0 \Rightarrow 6(x - 2)(x - 1) \ge 0 \Rightarrow x \ge 2 \) or \( x \le 1 \).
Therefore, \( f \) is increasing in \( (-\infty, 1] \cup [2, \infty) \).
(b) For \( f \) to be decreasing, \( f'(x) \le 0 \Rightarrow (x - 2)(x - 1) \le 0 \Rightarrow 1 \le x \le 2 \).
Therefore, \( f \) is decreasing in \( [1, 2] \).

Question. Find the intervals in which the following function is (a) increasing, (b) decreasing : \( f(x) = 2x^3 + 9x^2 + 12x + 20 \)
Answer: Given: \( f(x) = 2x^3 + 9x^2 + 12x + 20 \)
\( \Rightarrow f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x + 1)(x + 2) \).
(a) For \( f \) to be increasing, \( f'(x) \ge 0 \Rightarrow 6(x + 1)(x + 2) \ge 0 \Rightarrow x \ge -1 \) or \( x \le -2 \).
Therefore, \( f \) is increasing in \( (-\infty, -2] \cup [-1, \infty) \).
(b) For \( f \) to be decreasing, \( f'(x) \le 0 \Rightarrow (x + 1)(x + 2) \le 0 \Rightarrow -2 \le x \le -1 \).
Therefore, \( f \) is decreasing in \( (-2, -1) \).

Question. Find the intervals in which the function \( f(x) = (x - 1)^3(x - 2)^2 \) is (a) increasing, (b) decreasing.
Answer: Here, \( f(x) = (x - 1)^3 \cdot (x - 2)^2 \)
\( \Rightarrow f'(x) = 3(x - 1)^2(x - 2)^2 + (x - 1)^3 \cdot 2(x - 2) = (x - 1)^2(x - 2)[3(x - 2) + 2(x - 1)] = (x - 1)^2(x - 2)(5x - 8) \).
(a) For \( f \) to be an increasing function, \( f'(x) > 0 \Rightarrow (x - 1)^2(x - 2)(5x - 8) > 0 \Rightarrow (x - 2)(5x - 8) > 0 \) [since \( (x - 1)^2 \ge 0 \) for all \( x \in \mathbb{R} \)]
\( \Rightarrow x - 2 > 0, 5x - 8 > 0 \) or \( x - 2 < 0, 5x - 8 < 0 \)
\( \Rightarrow x > 2 \) or \( x < \frac{8}{5} \Rightarrow x \in (2, \infty) \) or \( x \in \left(-\infty, \frac{8}{5}\right) \).
(b) For \( f \) to be decreasing function, \( f'(x) < 0 \Rightarrow (x - 1)^2(x - 2)(5x - 8) < 0 \Rightarrow (x - 2)(5x - 8) < 0 \)
\( \Rightarrow \frac{8}{5} < x < 2 \Rightarrow x \in \left(\frac{8}{5}, 2\right) \).

Question. Show that the function \( f \) given by \( f(x) = x^3 - 3x^2 + 4x, x \in \mathbb{R} \) is strictly increasing on \( \mathbb{R} \).
Answer: Here, \( f(x) = x^3 - 3x^2 + 4x \)
\( \Rightarrow f'(x) = 3x^2 - 6x + 4 = 3(x^2 - 2x) + 4 = 3(x^2 - 2x + 1) - 3 + 4 = 3(x - 1)^2 + 1 > 0 \) for all \( x \in \mathbb{R} \).
Therefore, \( f \) is strictly increasing on \( \mathbb{R} \).

Question. Find the intervals in which the following function is (a) increasing, (b) decreasing : \( f(x) = 2x^3 - 9x^2 + 12x + 15 \)
Answer: Refer to answer of the previous similar question.
Here, \( f(x) = 2x^3 - 9x^2 + 12x + 15 \)
\( \Rightarrow f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 2)(x - 1) \).
(a) For \( f \) to be increasing, \( f'(x) \ge 0 \Rightarrow x \ge 2 \) or \( x \le 1 \).
Therefore, \( f \) is increasing in \( (-\infty, 1] \cup [2, \infty) \).
(b) For \( f \) to be decreasing, \( f'(x) \le 0 \Rightarrow 1 \le x \le 2 \).
Therefore, \( f \) is decreasing in \( [1, 2] \).

Question. Find the intervals in which the following function is (a) increasing, (b) decreasing : \( f(x) = 2x^3 - 15x^2 + 36x + 17 \)
Answer: We have, \( f(x) = 2x^3 - 15x^2 + 36x + 17 \)
\( \Rightarrow f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 3)(x - 2) \).
Now, for critical points \( f'(x) = 0 \Rightarrow 6(x - 3)(x - 2) = 0 \Rightarrow x = 2, 3 \).
The points \( x = 2, x = 3 \) divide the real line into disjoint intervals \( (-\infty, 2), (2, 3) \) and \( (3, \infty) \).

Hence, \( f(x) \) is increasing in \( (-\infty, 2] \cup [3, \infty) \) and decreasing in \( [2, 3] \).

Question. Find the intervals in which the function \( f \) given by \( f(x) = x^3 + \frac{1}{x^3}, x \neq 0 \) is (i) increasing, (ii) decreasing.
Answer: Here, \( f(x) = x^3 + \frac{1}{x^3} \)
\( \Rightarrow f'(x) = 3x^2 - \frac{3}{x^4} = 3\left(x^2 - \frac{1}{x^4}\right) = 3\left(x - \frac{1}{x^2}\right)\left(x + \frac{1}{x^2}\right) \).
So, critical points are \( x = -1, x = 1 \).
Also, \( f(x) \) is not defined for \( x = 0 \).
So, disjoint intervals are \( (-\infty, -1), (-1, 0), (0, 1), (1, \infty) \).

Hence, \( f(x) \) is increasing in \( (-\infty, -1] \cup [1, \infty) \) and decreasing in \( [-1, 0) \cup (0, 1] \).

Question. Find the intervals in which the function \( f \) given by \( f(x) = \sin x + \cos x, 0 \le x \le 2\pi \) is strictly increasing or strictly decreasing.
Answer: Refer to the previous similar question on interval analysis for \( \sin x + \cos x \).
The given function is \( f(x) = \sin x + \cos x; 0 \le x \le 2\pi \)
\( \Rightarrow f'(x) = \cos x - \sin x \).
Now, \( f'(x) = 0 \Rightarrow \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4} \).
Thus \( f'(x) > 0 \) in \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \)
and \( f'(x) < 0 \) in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).
Therefore, the function \( f \) is strictly decreasing in \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \) and strictly increasing in \( \left[0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right] \).

Question. Find the intervals in which the following function is (a) increasing, (b) decreasing : \( f(x) = x^3 - 12x^2 + 36x + 17 \)
Answer: Here, \( f(x) = x^3 - 12x^2 + 36x + 17 \)
\( \Rightarrow f'(x) = 3x^2 - 24x + 36 = 3(x^2 - 8x + 12) = 3(x - 2)(x - 6) \).
So, critical points are \( x = 2 \) and \( x = 6 \).
Required disjoint intervals are \( (-\infty, 2), (2, 6) \) and \( (6, \infty) \).

Hence, \( f(x) \) is increasing in \( (-\infty, 2] \cup [6, \infty) \) and decreasing in \( [2, 6] \).

Question. Find the intervals in which the function \( f(x) = (x - 1)(x - 2)^2 \) is (a) increasing, (b) decreasing.
Answer: We have, \( f(x) = (x - 1)(x - 2)^2 \)
\( \Rightarrow f'(x) = 2(x - 1)(x - 2) + (x - 2)^2 = (x - 2)(2x - 2 + x - 2) = (x - 2)(3x - 4) \).
For critical points, \( f'(x) = 0 \Rightarrow (x - 2)(3x - 4) = 0 \Rightarrow x = 2, x = \frac{4}{3} \).
The points divide the real line into disjoint intervals \( \left(-\infty, \frac{4}{3}\right), \left(\frac{4}{3}, 2\right) \) and \( (2, \infty) \).

Hence, \( f(x) \) is increasing in \( \left(-\infty, \frac{4}{3}\right] \cup [2, \infty) \) and decreasing in \( \left[\frac{4}{3}, 2\right] \).

Question. Find the intervals in which \( f(x) = \sin 3x - \cos 3x, 0 < x < \pi \), is strictly increasing or strictly decreasing.
Answer: \( f(x) = \sin 3x - \cos 3x \Rightarrow f'(x) = 3\cos 3x + 3\sin 3x \).
\( f'(x) = 0 \Rightarrow 3\cos 3x = -3\sin 3x \Rightarrow \cos 3x = -\sin 3x \Rightarrow \tan 3x = -1 \),
which gives \( 3x = \frac{3\pi}{4} \) or \( \frac{7\pi}{4} \) or \( \frac{11\pi}{4} \)
\( \Rightarrow x = \frac{\pi}{4} \) or \( \frac{7\pi}{12} \) or \( \frac{11\pi}{12} \) (since \( 0 < x < \pi \)).
The points \( x = \frac{\pi}{4}, x = \frac{7\pi}{12} \) and \( x = \frac{11\pi}{12} \) divide the interval \( (0, \pi) \) into four disjoint intervals: \( \left(0, \frac{\pi}{4}\right), \left(\frac{\pi}{4}, \frac{7\pi}{12}\right), \left(\frac{7\pi}{12}, \frac{11\pi}{12}\right) \) and \( \left(\frac{11\pi}{12}, \pi\right) \).
Now, \( f'(x) > 0 \) in \( \left(0, \frac{\pi}{4}\right) \Rightarrow f \) is strictly increasing in \( \left(0, \frac{\pi}{4}\right) \).
\( f'(x) < 0 \) in \( \left(\frac{\pi}{4}, \frac{7\pi}{12}\right) \Rightarrow f \) is strictly decreasing in \( \left(\frac{\pi}{4}, \frac{7\pi}{12}\right) \).
\( f'(x) > 0 \) in \( \left(\frac{7\pi}{12}, \frac{11\pi}{12}\right) \Rightarrow f \) is strictly increasing in \( \left(\frac{7\pi}{12}, \frac{11\pi}{12}\right) \).
\( f'(x) < 0 \) in \( \left(\frac{11\pi}{12}, \pi\right) \Rightarrow f \) is strictly decreasing in \( \left(\frac{11\pi}{12}, \pi\right) \).
Hence, \( f \) is strictly increasing in \( \left(0, \frac{\pi}{4}\right) \cup \left(\frac{7\pi}{12}, \frac{11\pi}{12}\right) \) and strictly decreasing in \( \left(\frac{\pi}{4}, \frac{7\pi}{12}\right) \cup \left(\frac{11\pi}{12}, \pi\right) \).

Question. Prove that the function \( f \) defined by \( f(x) = x^2 - x + 1 \) is neither increasing nor decreasing in \( (-1, 1) \). Hence, find the intervals in which \( f(x) \) is (i) strictly increasing, (ii) strictly decreasing.
Answer: Here, \( f(x) = x^2 - x + 1; x \in (-1, 1) \)
\( \Rightarrow f'(x) = 2x - 1 \).
\( f'(x) = 0 \Rightarrow x = \frac{1}{2} \).
Now, \( f'(x) = 2\left(x - \frac{1}{2}\right) > 0 \) for \( \frac{1}{2} < x < 1 \)
\( \Rightarrow f \) is strictly increasing in \( \left(\frac{1}{2}, 1\right) \).
Also, \( f'(x) = 2\left(x - \frac{1}{2}\right) < 0 \) for \( -1 < x < \frac{1}{2} \)
\( \Rightarrow f \) is strictly decreasing in \( \left(-1, \frac{1}{2}\right) \).
Thus, \( f \) is neither increasing nor decreasing in \( (-1, 1) \).

Question. Find the intervals in which the function \( f \) given by \( f(x) = \sin x - \cos x, 0 \le x \le 2\pi \) is strictly increasing or strictly decreasing.
Answer: We have \( f(x) = \sin x - \cos x, 0 \le x \le 2\pi \)
\( f'(x) = \cos x + \sin x \).
For critical points \( f'(x) = 0 \Rightarrow \cos x + \sin x = 0 \Rightarrow \tan x = -1 \).
As \( x \in [0, 2\pi] \Rightarrow x = \frac{3\pi}{4}, \frac{7\pi}{4} \).
The points divide the intervals \( [0, 2\pi] \) into the following disjoint intervals: \( \left[0, \frac{3\pi}{4}\right), \left(\frac{3\pi}{4}, \frac{7\pi}{4}\right) \) and \( \left(\frac{7\pi}{4}, 2\pi\right] \).

Hence \( f(x) \) is strictly increasing on \( \left[0, \frac{3\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right] \) and strictly decreasing on \( \left(\frac{3\pi}{4}, \frac{7\pi}{4}\right) \).

Question. Find the intervals in which the following function \( f(x) = 20 - 9x + 6x^2 - x^3 \) is (a) strictly increasing, (b) strictly decreasing.
Answer: Here, \( f(x) = 20 - 9x + 6x^2 - x^3 \)
\( \Rightarrow f'(x) = -9 + 12x - 3x^2 = -3(3 - 4x + x^2) = -3(x - 3)(x - 1) \).
So, critical points are \( x = 1 \) and \( x = 3 \).
Required disjoint intervals are \( (-\infty, 1), (1, 3) \) and \( (3, \infty) \).

(a) \( f(x) \) is strictly increasing in \( (1, 3) \).
(b) \( f(x) \) is strictly decreasing in \( (-\infty, 1) \cup (3, \infty) \).