Applications of Derivative: Miscellaneous Exercise (Class 12 Maths NCERT)

Understanding the Miscellaneous Exercise

The Miscellaneous Exercise in the Applications of Derivatives chapter serves as a crucial testing ground for your comprehensive understanding. Unlike other exercises that focus on specific applications (like only rate of change or only maxima/minima), this section brings together a diverse range of problems that often combine several concepts. You might encounter questions that first ask you to find the rate of change of a quantity, then require you to determine when that quantity is increasing or decreasing, and finally, ask for its maximum or minimum value. This integrated approach demands a strong conceptual grasp and the ability to identify the correct derivative application for each part of the problem. It challenges you to think critically, break down complex problems into manageable steps, and apply your knowledge flexibly. Successfully tackling these problems demonstrates a deep mastery of the entire chapter, making it an indispensable part of your preparation for the CBSE Class 12 board exams.

Key Concepts to Remember

Rate of Change

If a quantity 'y' varies with another quantity 'x', then dy/dx represents the rate of change of y with respect to x. If y is a function of time 't', then dy/dt represents the rate of change of y with respect to time.

Increasing and Decreasing Functions

A function f(x) is increasing on an interval (a, b) if f'(x) > 0 for all x in (a, b). It is decreasing if f'(x) < 0 for all x in (a, b).

Tangents and Normals

The slope of the tangent to the curve y = f(x) at a point (x₀, y₀) is given by f'(x₀). The equation of the tangent is y - y₀ = f'(x₀)(x - x₀). The normal is perpendicular to the tangent, so its slope is -1/f'(x₀) (if f'(x₀) ≠ 0).

Maxima and Minima (Optimization)

For local maxima or minima, we find critical points where f'(x) = 0 or f'(x) is undefined. The second derivative test [f''(x)] or the first derivative test (change in sign of f'(x)) is then used to determine if it's a maximum or minimum.

Approximations

Using differentials, we can approximate the value of a quantity. Δy ≈ dy = f'(x)Δx. This is useful for estimating small changes in a function's value.

General Approach to Miscellaneous Problems

1. Understand the Problem Statement — Read the problem carefully. Identify what is given, what needs to be found, and what quantities are related. Draw a diagram if it's a geometrical problem.
2. Formulate the Function — Express the quantity to be optimized or analyzed as a function of one variable. If it involves multiple variables, use given constraints to reduce it to a single independent variable.
3. Identify the Derivative Application — Determine whether the problem requires finding a rate of change, identifying increasing/decreasing intervals, finding tangents/normals, or optimizing (maxima/minima).
4. Calculate the Derivative — Differentiate the function with respect to the relevant variable. Pay close attention to chain rule, product rule, and quotient rule if applicable.
5. Apply Tests/Conditions — For maxima/minima, set the first derivative to zero to find critical points and then use the first or second derivative test. For increasing/decreasing, check the sign of the first derivative. For tangents/normals, evaluate the derivative at the given point.
6. Interpret and Conclude — Ensure your answer addresses the original question, includes appropriate units (if applicable), and makes sense in the context of the problem. Don't forget to state your final answer clearly.

Worked Examples

• Example 1: Maxima/Minima in a Real-World Context Question: A wire of length 28 m is to be cut into two pieces. One piece is made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? Solution: Step 1: Define variables and set up equations. Let the length of one piece of wire be 'x' metres. This piece is used to form a square. So, the perimeter of the square = x. Side of the square = x/4. Area of the square (A₁) = (x/4)² = x²/16. The length of the other piece will be (28 - x) metres. This piece is used to form a circle. So, circumference of the circle = 28 - x. 2πr = 28 - x => r = (28 - x) / (2π). Area of the circle (A₂) = πr² = π [(28 - x) / (2π)]² = (28 - x)² / (4π). Step 2: Formulate the total area function. Combined Area A = A₁ + A₂ = x²/16 + (28 - x)² / (4π). Step 3: Differentiate the area function with respect to x. dA/dx = d/dx [x²/16 + (28 - x)² / (4π)] dA/dx = 2x/16 + 1/(4π) 2(28 - x) (-1) dA/dx = x/8 - (28 - x) / (2π) Step 4: Find critical points by setting dA/dx = 0. x/8 - (28 - x) / (2π) = 0 x/8 = (28 - x) / (2π) 2πx = 8(28 - x) πx = 4(28 - x) πx = 112 - 4x (π + 4)x = 112 x = 112 / (π + 4) Step 5: Use the second derivative test to confirm minimum. d²A/dx² = d/dx [x/8 - (28 - x) / (2π)] d²A/dx² = 1/8 - 1/(2π) (-1) = 1/8 + 1/(2π) Since d²A/dx² is positive (1/8 + 1/(2π) > 0), the combined area is minimum at x = 112 / (π + 4). Step 6: Calculate the lengths of the two pieces. Length of the piece for the square (x) = 112 / (π + 4) metres. Length of the piece for the circle = 28 - x = 28 - 112 / (π + 4) = [28(π + 4) - 112] / (π + 4) = (28π + 112 - 112) / (π + 4) = 28π / (π + 4) metres. Final Answer: The length of the wire for the square should be 112/(π + 4) m and for the circle should be 28π/(π + 4) m to minimize the combined area.
• Example 2: Rate of Change and Approximation Question: The radius of a sphere is measured as 7 m with an error of 0.02 m. Find the approximate error in calculating its surface area. Solution: Step 1: Identify the formula and given values. Surface area of a sphere, S = 4πr². Given radius r = 7 m. Error in radius, Δr = 0.02 m. We need to find the approximate error in surface area, ΔS. Step 2: Find the derivative of the surface area with respect to radius. dS/dr = d/dr (4πr²) = 8πr. Step 3: Use the approximation formula for error. We know that ΔS ≈ (dS/dr) Δr. Step 4: Substitute the values. ΔS = (8πr) Δr ΔS = (8 π 7) 0.02 ΔS = 56π 0.02 ΔS = 1.12π Final Answer: The approximate error in calculating its surface area is 1.12π m².

Exam Tips for Miscellaneous Problems

When attempting the Miscellaneous Exercise problems in your CBSE Class 12 Maths board exam, remember that these questions often carry higher marks due to their multi-conceptual nature. Always start by clearly defining your variables and drawing a neat diagram for geometry-based problems. This helps in visualising the scenario and establishing relationships between quantities. Show all your steps logically and neatly; even if your final answer is slightly off, correct steps can earn you partial credit. Pay extra attention to units in your final answer, especially in optimization and rate-of-change problems. For maxima and minima, explicitly state whether you are using the first derivative test or the second derivative test, and clearly show the condition for maximum or minimum. Practice identifying the 'type' of application required for each part of a problem, as sometimes a single question might combine rate of change with optimization.

Practice Questions with Solutions

• Q: A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. A: Step 1: Given the equation of the curve 6y = x³ + 2. Differentiate both sides with respect to time (t): 6(dy/dt) = 3x²(dx/dt) Step 2: Given that the y-coordinate is changing 8 times as fast as the x-coordinate, so dy/dt = 8(dx/dt). Substitute this into the differentiated equation: 6 * 8(dx/dt) = 3x²(dx/dt) 48(dx/dt) = 3x²(dx/dt) Step 3: Since dx/dt ≠ 0 (the particle is moving), we can divide by dx/dt: 48 = 3x² x² = 16 x = ±4 Step 4: Find the corresponding y-coordinates using the curve equation: If x = 4, 6y = (4)³ + 2 = 64 + 2 = 66 => y = 11. If x = -4, 6y = (-4)³ + 2 = -64 + 2 = -62 => y = -62/6 = -31/3. Final answer: The points on the curve are (4, 11) and (-4, -31/3).
• Q: Show that the function f(x) = |x - 1| + |x + 1| is not differentiable at x = 1 and x = -1. A: Step 1: Define the function f(x) piecewise. For x < -1, f(x) = -(x - 1) - (x + 1) = -x + 1 - x - 1 = -2x. For -1 ≤ x < 1, f(x) = -(x - 1) + (x + 1) = -x + 1 + x + 1 = 2. For x ≥ 1, f(x) = (x - 1) + (x + 1) = x - 1 + x + 1 = 2x. So, f(x) = { -2x, if x < -1 2, if -1 ≤ x < 1 2x, if x ≥ 1 } Step 2: Check differentiability at x = 1. Left Hand Derivative (LHD) at x = 1: lim (h→0⁻) [f(1 + h) - f(1)] / h For 1+h < 1 (h < 0), f(1+h) = 2. f(1) = 2(1) = 2. LHD = lim (h→0⁻) [2 - 2] / h = lim (h→0⁻) 0/h = 0. Right Hand Derivative (RHD) at x = 1: lim (h→0⁺) [f(1 + h) - f(1)] / h For 1+h ≥ 1 (h > 0), f(1+h) = 2(1+h). f(1) = 2. RHD = lim (h→0⁺) [2(1 + h) - 2] / h = lim (h→0⁺) [2 + 2h - 2] / h = lim (h→0⁺) 2h/h = 2. Since LHD ≠ RHD at x = 1 (0 ≠ 2), f(x) is not differentiable at x = 1. Step 3: Check differentiability at x = -1. LHD at x = -1: lim (h→0⁻) [f(-1 + h) - f(-1)] / h For -1+h < -1 (h < 0), f(-1+h) = -2(-1+h). f(-1) = 2. LHD = lim (h→0⁻) [-2(-1 + h) - 2] / h = lim (h→0⁻) [2 - 2h - 2] / h = lim (h→0⁻) -2h/h = -2. RHD at x = -1: lim (h→0⁺) [f(-1 + h) - f(-1)] / h For -1 ≤ -1+h < 1 (h > 0), f(-1+h) = 2. f(-1) = 2. RHD = lim (h→0⁺) [2 - 2] / h = lim (h→0⁺) 0/h = 0. Since LHD ≠ RHD at x = -1 (-2 ≠ 0), f(x) is not differentiable at x = -1. Final answer: The function is not differentiable at x = 1 and x = -1.
• Q: Find the interval in which the function f(x) = sin x - cos x, for 0 < x < 2π, is strictly increasing. A: Step 1: Find the first derivative of the function. f'(x) = d/dx (sin x - cos x) = cos x - (-sin x) = cos x + sin x. Step 2: For f(x) to be strictly increasing, f'(x) > 0. So, cos x + sin x > 0. Step 3: Divide by cos x (assuming cos x ≠ 0) to get tan x > -1. Alternatively, multiply by 1/√2: (1/√2)cos x + (1/√2)sin x > 0 sin(x + π/4) > 0 Step 4: For sin(θ) > 0, θ must lie in the intervals (2nπ, 2nπ + π) for integer n. So, 2nπ < x + π/4 < 2nπ + π. Since 0 < x < 2π, we have 0 + π/4 < x + π/4 < 2π + π/4, which is π/4 < x + π/4 < 9π/4. Step 5: Find the common interval for sin(x + π/4) > 0 within (π/4, 9π/4). This occurs when x + π/4 is in (π/4, π) or (2π + π/4, 3π) etc. Considering the range π/4 < x + π/4 < 9π/4, the interval where sin(x + π/4) > 0 is (π/4, π). So, π/4 < x + π/4 < π. Subtract π/4 from all parts: π/4 - π/4 < x < π - π/4 0 < x < 3π/4. Also, x + π/4 could be in (2π+π/4, 3π) which gives x > 2π, but our domain is 0 < x < 2π. We also have to consider the condition where cos x = 0. If cos x = 0, then x = π/2, 3π/2. At these points sin x = 1 or -1. So sin x + cos x is 1 or -1. So this condition does not affect the inequality. Final answer: The function is strictly increasing in the interval (0, 3π/4) and (7π/4, 2π). (Note: sin x + cos x = √2 sin(x + π/4). This is positive when x + π/4 is in (0, π) or (2π, 3π) etc. In the given domain 0 < x < 2π, x + π/4 lies in (π/4, 9π/4). So sin(x + π/4) > 0 when x + π/4 ∈ (π/4, π) or (9π/4, 2π). This means 0 < x < 3π/4 or 7π/4 < x < 2π.)
• Q: Find the equation of all lines having slope -1 that are tangent to the curve y = 1/(x - 1), x ≠ 1. A: Step 1: Find the derivative of the curve y = 1/(x - 1) with respect to x. dy/dx = d/dx [(x - 1)⁻¹] = -1(x - 1)⁻² * 1 = -1/(x - 1)². Step 2: Set the slope equal to -1 and solve for x. -1/(x - 1)² = -1 1/(x - 1)² = 1 (x - 1)² = 1 x - 1 = ±1 Case 1: x - 1 = 1 => x = 2. Case 2: x - 1 = -1 => x = 0. Step 3: Find the corresponding y-coordinates for each x-value. If x = 2, y = 1/(2 - 1) = 1. So the point is (2, 1). If x = 0, y = 1/(0 - 1) = -1. So the point is (0, -1). Step 4: Write the equations of the tangent lines using the point-slope form y - y₁ = m(x - x₁). For point (2, 1) and slope m = -1: y - 1 = -1(x - 2) y - 1 = -x + 2 y + x = 3. For point (0, -1) and slope m = -1: y - (-1) = -1(x - 0) y + 1 = -x y + x = -1. Final answer: The equations of the tangent lines are x + y = 3 and x + y = -1.

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