Previous year questions: Application of Derivatives
Application of Derivatives28 min readLesson 6 of 6
Q1 (CBSE Board 2023, 2 marks)
The product of two positive numbers is 9. Find the numbers for which the sum of their squares is minimum. Answer: Let the numbers be x and 9 over x, both positive. The sum of squares is S equal to x squared plus 81 over x squared. Differentiating gives 2x minus 162 over x cubed, which is zero when x to the fourth equals 81, so x equals 3. The second derivative is 2 plus 486 over x to the fifth, positive at 3, so this is a minimum. Both numbers are 3 and the minimum sum of squares is 18. The second derivative test confirms the minimum before the final numeric answer is written. The negative root is rejected because the problem states the numbers are positive. The second derivative is positive at the critical point, so the sum of squares indeed has a minimum there. The second derivative test is written before the final values because it closes the argument.
Q2 (CBSE Board 2024, 1 mark)
The function f(x) = x cubed minus 3x squared plus 12x minus 18 is which of strictly decreasing on R, strictly increasing on R, neither, or strictly decreasing on the negative side only? Answer: Differentiate: f prime of x equals 3x squared minus 6x plus 12. Complete the square inside the bracket to get 3 times (x minus 1) squared plus 9 over 3, which is always positive. Since the derivative is positive for every real x, the function is strictly increasing on the whole real line. The answer is the second option. Completing the square inside the derivative shows the positivity without testing intervals. The completed square is non-negative, so the derivative is positive everywhere. The squared term is non-negative for every real x, so the whole derivative stays positive.
Q3 (CBSE Board 2024, 2 marks)
Find the local maximum and local minimum values of f(x) = 4x squared plus 1 over x for x not zero. Answer: The derivative is 8x minus 1 over x squared. Setting it to zero gives 8x cubed equal to 1, so x equals 1 by 2. The second derivative is 8 plus 2 over x cubed, which is positive at 1 by 2, so this point is a local minimum. The minimum value is f(1/2) = 1 plus 2, which is 3. There is no local maximum because the function grows without bound as x approaches zero from either side. The function has no local maximum, so only the minimum value is reported. The minimum is reported with its point, and the absence of a maximum is stated explicitly.
Q4 (CBSE Board 2024, 2 marks)
On the interval from 0 to 1, the function f(x) = 12x to the power four thirds minus 6x to the power one third is to be checked for its largest and smallest values. Answer: The derivative is 16x to the power one third minus 2x to the power minus two thirds, which factors as 2x to the power minus two thirds times (8x minus 1). The only critical point inside the interval is x equal to 1 by 8. Evaluate at the critical point and the endpoints: f(0) = 0, f(1/8) = minus 9 by 4, and f(1) = 6. The absolute maximum is 6 at x equal to 1, and the absolute minimum is minus 9 by 4 at x equal to 1 by 8. The endpoint values are part of the comparison and not an optional extra. Writing the three values in one line makes the comparison explicit and prevents an endpoint from being missed. Both the critical point and the two endpoints appear in the final comparison. All three values are compared before the largest and smallest are named.
Q5 (CBSE Board 2026, 1 mark)
Find the rate of change of the volume of a sphere with respect to its diameter when the radius is 5 cm. Answer: Write the volume as four thirds pi r cubed and the diameter as D equal to 2r, so the volume is pi D cubed over 6. Differentiating with respect to D gives pi D squared over 2. When the radius is 5 cm the diameter is 10 cm, so the rate is pi times 100 over 2, which is 50 pi cubic centimetres per centimetre. The answer is the option 50 pi. Writing the volume in terms of the diameter avoids a chain through the radius. Substituting the diameter before differentiating would give a constant and lose the relation, so the order matters. Writing the volume in terms of the diameter is the step that makes the derivative simple.
Q6 (CBSE Board 2026, 1 mark)
The distance travelled by a particle in t seconds is S = 72t plus 3t squared minus t cubed. Find the time at which the particle comes to rest. Answer: Velocity is the derivative of distance, so v equals 72 plus 6t minus 3t squared. The particle rests when velocity is zero. Dividing by 3 gives t squared minus 2t minus 24 equal to zero, which factors as (t minus 6)(t plus 4) equal to zero. Time is positive, so t equals 6 seconds. The other root is rejected. Rejecting the negative root is part of the answer because time cannot be negative. The velocity expression is a downward parabola in t, so the positive root is the only physical time. The negative root is rejected on physical grounds, and the positive root is the answer.
Q7 (CBSE Board 2026, 2 marks)
Find the values of x for which f(x) = (x minus 3) divided by (x plus 1), with x not equal to minus 1, is increasing. Answer: Differentiate using the quotient rule. The derivative is ((x plus 1) minus (x minus 3)) divided by (x plus 1) squared, which simplifies to 4 over (x plus 1) squared. A square is positive, so the derivative is positive for every allowed x. The function is therefore increasing on both intervals (minus infinity, minus 1) and (minus 1, infinity), and the answer is written as the union of the two. The derivative being a positive square makes the union of the two intervals the full answer. The derivative is a positive square, so both intervals of the domain are included in the answer.
Q8 (CBSE Board 2026, 2 marks)
Find the absolute maximum value of f(x) = cos x plus sin squared x on the interval [0, pi]. Answer: The derivative is minus sin x plus 2 sin x cos x, which is sin x times (2 cos x minus 1). It vanishes when sin x is zero, giving x equal to 0 or pi, or when cos x equals 1 by 2, giving x equal to pi by 3. Evaluate: f(0) = 1, f(pi) = minus 1, and f(pi/3) = 1 by 2 plus 3 by 4, which is 5 by 4. The largest value is 5 by 4, so the absolute maximum is 5 by 4 at x equal to pi by 3. The table of three values covers both critical points and the two endpoints of the interval. Three candidate values are enough because the derivative factors completely. The three candidate values cover the two critical points and the interval endpoints.
Q9 (CBSE SQP 2021-22, 1 mark)
Find the point on the curve y = x plus 1 over x, x positive, where the normal is perpendicular to the line 3x minus 4y minus 7 = 0. Answer: The slope of the given line is 3 by 4. A normal perpendicular to it has slope minus 4 by 3, so the tangent slope must be 3 by 4. The derivative of the curve is 1 minus 1 over x squared. Setting it equal to 3 by 4 gives 1 over x squared equal to 1 by 4, so x equals 2 for positive x. Then y equals 2 plus 1 by 2, which is 5 by 2. The point is (2, 5/2). The normal slope is the negative reciprocal of the tangent slope, and the substitution gives the point.
Q10 (CBSE SQP 2021-22, 1 mark)
Find the point on the curve y = x cubed minus 11x plus 5 at which the tangent is parallel to the line y = x minus 11. Answer: The tangent slope is the derivative 3x squared minus 11, and the line has slope 1. Setting 3x squared minus 11 equal to 1 gives x squared equal to 4, so x equals 2 or minus 2. At x equal to 2, y equals 8 minus 22 plus 5, which is minus 9, and the point (2, minus 9) lies on the line y equal to x minus 11. At x equal to minus 2, y equals 19, and the tangent there is parallel but the point is not on the given line. The required point is (2, minus 9). The second candidate point fails the line test, so only one point is reported.