Maharashtra Board Class 12 Maths Part 1 Chapter 4 Applications of Derivatives 4.2 Solutions

Question 1. Test whether the following functions are increasing and decreasing:
(i) f(x) = x³ - 6x² + 12x - 16, x∈R Solution: \( f(x) = x^3 - 6x^2 + 12x - 16 \) \( \therefore f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 16) \) \( = 3x^2 - 6 \times 2x + 12 \times 1 - 0 \) \( = 3x^2 - 12x + 12 \) \( = 3(x^2 - 4x + 4) \) \( = 3(x - 2)^2 > 0 \text{ for all } x \in \mathbb{R}, x \neq 2 \) \( \therefore f'(x) > 0 \text{ for all } x \in \mathbb{R} - \{2\} \) \( \therefore f \text{ is increasing for all } x \in \mathbb{R} - \{2\}. \)
In simple words: To determine if a function is increasing or decreasing, we find its first derivative. If the derivative is positive, the function is increasing; if negative, it's decreasing. For this function, the derivative \( 3(x-2)^2 \) is always positive (or zero at x=2), indicating the function is increasing everywhere except at x=2.

🎯 Exam Tip: Remember to clearly state the interval(s) where the function is increasing or decreasing, and use proper set notation like \( \mathbb{R} \) or interval notation.

(ii) f(x) = x - 1/x, x ∈ R, x ≠ 0 Solution: \( f(x) = x - \frac{1}{x} \) \( \therefore f'(x) = \frac{d}{dx}(x - \frac{1}{x}) \) \( = 1 - (-\frac{1}{x^2}) \) \( = 1 + \frac{1}{x^2} > 0 \text{ for all } x \in \mathbb{R}, x \neq 0 \) \( \therefore f'(x) > 0 \text{ for all } x \in \mathbb{R}, \text{ where } x \neq 0 \) \( \therefore f \text{ is increasing for all } x > R, \text{ where } x \neq 0. \)
In simple words: The derivative of this function, \( 1 + \frac{1}{x^2} \), is always positive because \( \frac{1}{x^2} \) is always positive for any non-zero x. Thus, the function is increasing for all real numbers except x=0.

🎯 Exam Tip: Pay attention to the domain of the original function and exclude points where the derivative is undefined or zero, as these can be critical points for monotonicity.

(iii) f(x) = 7/x - 3, x ∈ R, x ≠ 0 Solution: \( f(x) = \frac{7}{x} - 3 \) \( \therefore f'(x) = \frac{d}{dx}(\frac{7}{x} - 3) = 7 (-\frac{1}{x^2}) - 0 \) \( = -\frac{7}{x^2} < 0 \text{ for all } x \in \mathbb{R}, x \neq 0 \) \( \therefore f'(x) < 0 \text{ for all } x \in \mathbb{R}, \text{ where } x \neq 0. \) \( \therefore f \text{ is decreasing for all } x \in \mathbb{R}, \text{ where } x \neq 0. \)
In simple words: The derivative of this function, \( -\frac{7}{x^2} \), is always negative because \( \frac{7}{x^2} \) is always positive for any non-zero x. This means the function is always decreasing across its domain.

🎯 Exam Tip: Remember that \( \frac{d}{dx}(c/x) = -c/x^2 \). A negative derivative means the function is strictly decreasing.

Question 2. Find the values of x, such that f(x) is increasing function:
(i) f(x) = 2x³ - 15x² + 36x + 1 Solution: \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) \( \therefore f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) \) \( = 2 \times 3x^2 - 15 \times 2x + 36 \times 1 + 0 \) \( = 6x^2 - 30x + 36 \) \( = 6(x^2 - 5x + 6) \) f is increasing, if \( f'(x) > 0 \) \( \text{i.e. if } 6(x^2 - 5x + 6) > 0 \) \( \text{i.e. if } x^2 - 5x + 6 > 0 \) \( \text{i.e. if } x^2 - 5x > -6 \) \( \text{i.e. if } x^2 - 5x + \frac{25}{4} > -6 + \frac{25}{4} \) \( \text{i.e. if } (x - \frac{5}{2})^2 > \frac{1}{4} \) \( \text{i.e. if } x - \frac{5}{2} > \frac{1}{2} \text{ or } x - \frac{5}{2} < -\frac{1}{2} \) \( \text{i.e. if } x > 3 \text{ or } x < 2 \) \( \text{i.e. if } x \in (-\infty, 2) \cup (3, \infty) \) \( \therefore f \text{ is increasing, if } x \in (-\infty, 2) \cup (3, \infty). \)
In simple words: To find where the function is increasing, we first calculate its derivative and set it greater than zero. Solving the resulting quadratic inequality \( x^2 - 5x + 6 > 0 \) (which factors to \( (x-2)(x-3) > 0 \)) reveals that the function increases when x is less than 2 or greater than 3.

🎯 Exam Tip: For quadratic inequalities, finding the roots and testing intervals is a crucial step. Remember to use appropriate interval notation for your final answer.

(ii) f(x) = x² + 2x - 5 Solution: \( f(x) = x^2 + 2x - 5 \) \( \therefore f'(x) = \frac{d}{dx}(x^2 + 2x - 5) \) \( = 2x + 2 \times 1 - 0 \) \( = 2x + 2 \) f is increasing, if \( f'(x) > 0 \) \( \text{i.e. if } 2x + 2 > 0 \) \( \text{i.e. if } 2x > -2 \) \( \text{i.e. if } x > -1, \text{ i.e. } x \in (-1, \infty) \) \( \therefore f \text{ is increasing, if } x > -1, \text{ i.e. } x \in (-1, \infty) \)
In simple words: The derivative of the function is \( 2x+2 \). For the function to be increasing, \( 2x+2 \) must be positive, which means \( x > -1 \). So, the function is increasing for all values of x greater than -1.

🎯 Exam Tip: Linear inequalities are straightforward. Isolate x to find the range for increasing/decreasing behavior. Always write your answer in interval notation.

(iii) f(x) = 2x³ - 15x² - 144x - 7 Solution: \( f(x) = 2x^3 - 15x^2 - 144x - 7 \) \( \therefore f'(x) = \frac{d}{dx}(2x^3 - 15x^2 - 144x - 7) \) \( = 2 \times 3x^2 - 15 \times 2x - 144 \times 1 - 0 \) \( = 6x^2 - 30x - 144 \) \( = 6(x^2 - 5x - 24) \) f is increasing if, \( f'(x) > 0 \) \( \text{i.e. if } 6(x^2 - 5x - 24) > 0 \) \( \text{i.e. if } x^2 - 5x - 24 > 0 \) \( \text{i.e. if } x^2 - 5x > 24 \) \( \text{i.e. if } x^2 - 5x + \frac{25}{4} > 24 + \frac{25}{4} \) \( \text{i.e. if } (x - \frac{5}{2})^2 > \frac{121}{4} \) \( \text{i.e. if } x - \frac{5}{2} > \frac{11}{2} \text{ or } x - \frac{5}{2} < -\frac{11}{2} \) \( \text{i.e. if } x > 8 \text{ or } x < -3 \) \( \text{i.e. if } x \in (-\infty, -3) \cup (8, \infty) \) \( \therefore f \text{ is increasing, if } x \in (-\infty, -3) \cup (8, \infty). \)
In simple words: First, find the derivative, \( 6(x^2 - 5x - 24) \). To find where the function increases, set this derivative greater than zero. Factoring the quadratic gives \( 6(x-8)(x+3) > 0 \), which implies the function increases when x is less than -3 or greater than 8.

🎯 Exam Tip: For complex quadratic inequalities, completing the square or factoring are useful techniques. Always remember to check the signs of the quadratic in the intervals defined by its roots.

Question 3. Find the values of x such that f(x) is decreasing function:
(i) f(x) = 2x³ - 15x² - 144x - 7 Solution: \( f(x) = 2x^3 - 15x^2 - 144x - 7 \) \( \therefore f'(x) = \frac{d}{dx}(2x^3 - 15x^2 - 144x - 7) \) \( = 2 \times 3x^2 - 15 \times 2x - 144 \times 1 - 0 \) \( = 6x^2 - 30x - 144 \) \( = 6(x^2 - 5x - 24) \) f is decreasing, if \( f'(x) < 0 \) \( \text{i.e. if } 6(x^2 - 5x - 24) < 0 \) \( \text{i.e. if } x^2 - 5x - 24 < 0 \) \( \text{i.e. if } x^2 - 5x < 24 \) \( \text{i.e. if } x^2 - 5x + \frac{25}{4} < \frac{121}{4} \) \( \text{i.e. if } (x - \frac{5}{2})^2 < \frac{121}{4} \) \( \text{i.e. if } -\frac{11}{2} < x - \frac{5}{2} < \frac{11}{2} \) \( \text{i.e. if } -\frac{11}{2} + \frac{5}{2} < x - \frac{5}{2} + \frac{5}{2} < \frac{11}{2} + \frac{5}{2} \) \( \text{i.e. if } -3 < x < 8 \) \( \therefore f \text{ is decreasing, if } -3 < x < 8. \)
In simple words: To find where the function is decreasing, we set its first derivative \( 6(x^2 - 5x - 24) \) less than zero. Solving the inequality \( (x-8)(x+3) < 0 \) shows that the function is decreasing when x is between -3 and 8.

🎯 Exam Tip: When solving \( f'(x) < 0 \), ensure you correctly determine the intervals where the quadratic expression is negative, often by sketching a parabola or testing points.

(ii) f(x) = x⁴ - 2x³ + 1 Solution: \( f(x) = x^4 - 2x^3 + 1 \) \( \therefore f'(x) = \frac{d}{dx}(x^4 - 2x^3 + 1) \) \( = 4x^3 - 2 \times 3x^2 + 0 \) \( = 4x^3 - 6x^2 \) f is decreasing, if \( f'(x) < 0 \) \( \text{i.e. if } 4x^3 - 6x^2 < 0 \) \( \text{i.e. if } x^2(4x - 6) < 0 \) \( \text{i.e. if } 4x - 6 < 0 \ldots\ldots[\because x^2 > 0] \) \( \text{i.e. if } x < \frac{3}{2} \) \( \text{i.e. } -\infty < x < \frac{3}{2} \) \( \therefore f \text{ is decreasing, if } -\infty < x < \frac{3}{2}. \)
In simple words: We find the derivative to be \( 4x^3 - 6x^2 \). Factoring it as \( x^2(4x-6) \), and setting it less than zero for decreasing behavior, we note \( x^2 \) is always non-negative. So, \( (4x-6) \) must be negative, which means \( x < \frac{3}{2} \).

🎯 Exam Tip: Always factor out common terms from the derivative. Be careful with terms like \( x^2 \), which are always non-negative, as they influence the sign of the overall derivative.

(iii) f(x) = 2x³ - 15x² - 84x - 7 Solution: \( f(x) = 2x^3 - 15x^2 - 84x - 7 \) \( \therefore f'(x) = \frac{d}{dx}(2x^3 - 15x^2 - 84x - 7) \) \( = 2 \times 3x^2 - 15 \times 2x - 84 \times 1 - 0 \) \( = 6x^2 - 30x - 84 \) \( = 6(x^2 - 5x - 14) \) f is decreasing, if \( f'(x) < 0 \) \( \text{i.e. if } 6(x^2 - 5x - 14) < 0 \) \( \text{i.e. if } x^2 - 5x - 14 < 0 \) \( \text{i.e. if } x^2 - 5x < 14 \) \( \text{i.e. if } x^2 - 5x + \frac{25}{4} < 14 + \frac{25}{4} \) \( \text{i.e. if } (x - \frac{5}{2})^2 < \frac{81}{4} \) \( \text{i.e. if } -\frac{9}{2} < x - \frac{5}{2} < \frac{9}{2} \) \( \text{i.e. if } -\frac{9}{2} + \frac{5}{2} < x - \frac{5}{2} + \frac{5}{2} < \frac{9}{2} + \frac{5}{2} \) \( \text{i.e. if } -2 < x < 7 \) \( \therefore f \text{ is decreasing, if } -2 < x < 7. \)
In simple words: To find where this function decreases, we calculate its derivative, \( 6(x^2 - 5x - 14) \), and set it less than zero. Factoring the quadratic to \( 6(x-7)(x+2) < 0 \) shows that the function decreases for x values between -2 and 7.

🎯 Exam Tip: Correctly factoring quadratic expressions is essential. The roots of the derivative define the critical points, and the sign of the derivative in the intervals between these roots determines increasing or decreasing behavior.