Application of Derivatives — Class 12 MCQs with Answers

Application of Derivatives — important questions & MCQs with answers (Class 12 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Slope of tangent to y = f(x) at x = a:

   Af'(a)Bf(a)C1/f(a)Da
   Show answerHide answer

   Answer: f'(a)

   Explanation: The slope of the tangent to y = f(x) at x = x₀ is the derivative evaluated there, f'(x₀) — not f(x₀), which is just the height of the curve, a common mix-up.

2. Q2 of 6Easy

   At a local extremum (interior, differentiable):

   Af(x) = 0Bf''(x) = 0Cf'(x) = 0Dx = 0
   Show answerHide answer

   Answer: f'(x) = 0

   Explanation: At an interior differentiable extremum, the tangent must be horizontal, so f'(x) = 0. f''(x) = 0 only tests concavity and can fail even at a genuine extremum (e.g. f(x) = x⁴ at x = 0).

3. Q3 of 6Medium

   A balloon's volume increases at 100 cm³/s. Find dr/dt when r = 5 cm:

   A1/π cm/sB5/πC2/πD10/π
   Show answerHide answer

   Answer: 1/π cm/s

   Explanation: V = (4/3)πr³ gives dV/dt = 4πr²·(dr/dt). Substituting dV/dt = 100 and r = 5: 100 = 4π(25)(dr/dt) = 100π(dr/dt), so dr/dt = 1/π cm/s — always plug in r before solving, not after.

4. Q4 of 6Medium

   Assertion (A): The function f(x) = x² − 4x + 6 is decreasing on the interval (−∞, 2). Reason (R): For f(x) = x² − 4x + 6, f′(x) = 2x − 4, which is negative for every x < 2.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
   Show answerHide answer

   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: f′(x) = 2x − 4 < 0 whenever x < 2, so R is true. A function whose derivative is negative throughout an interval is decreasing there, so A follows directly from R.

5. Q5 of 6Hard

   f(x) = (x − 1)² · (x + 2) is increasing on:

   Ax > 0BAll xC(−∞, −1) ∪ (1, ∞)DCannot determine
   Show answerHide answer

   Answer: (−∞, −1) ∪ (1, ∞)

   Explanation: Expanding gives f(x) = x³ − 3x + 2, so f'(x) = 3x² − 3 = 3(x−1)(x+1), positive when x < −1 or x > 1 — the same sign pattern as x³ − 3x itself, since adding a constant never changes where the derivative is positive.

6. Q6 of 6Hard

   Assertion (A): Of all pairs of positive numbers whose sum is 15, the sum of their squares is least when each number is 7.5. Reason (R): S(x) = x² + (15 − x)² satisfies S′(7.5) = 0, and S″(x) = 4 > 0 for all x.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
   Show answerHide answer

   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: S′(x) = 4x − 30 vanishes only at x = 7.5 and S″(x) = 4 > 0, so R is true. Hence x = 7.5 is the only critical point and it is a minimum, giving the numbers 7.5 and 15 − 7.5 = 7.5; R is exactly the argument for A.

Application of Derivatives notes: key concepts (Class 12 Mathematics)

Derivatives are put to work: rates of change and related rates, slopes of tangents and normals, testing where a function increases or decreases by the sign of f′, locating critical points, and classifying local and absolute maxima and minima by the first- and second-derivative tests, plus optimisation and approximation.

Rate of change

dy/dt measures how y changes with time; the derivative interprets any instantaneous rate, e.g. dV/dr = 4πr² for a sphere.

Related rates

Link two changing quantities through the chain rule, dy/dt = (dy/dx)(dx/dt); the ladder, balloon and conical-tank problems use this.

Slope of tangent and normal

The tangent's slope at x = a is f′(a); the normal is perpendicular, with slope −1/f′(a).

Equation of tangent/normal

Tangent at (a, f(a)): y − f(a) = f′(a)(x − a); normal uses slope −1/f′(a) through the same point.

Increasing/decreasing functions

f is strictly increasing where f′(x) > 0 and strictly decreasing where f′(x) < 0 on an interval.

Critical points

Interior points where f′(x) = 0 or is undefined; candidates for local maxima or minima.

First-derivative test

At a critical point, f′ changing + to − gives a local maximum, − to + a local minimum, no sign change neither.

Second-derivative test

If f′(c) = 0 and f″(c) > 0 then c is a local minimum; f″(c) < 0 a local maximum; f″(c) = 0 is inconclusive.

Absolute maxima and minima

On a closed interval [a, b], compare f at all critical points and at both endpoints to find the global extremes.

Optimisation problems

Express the quantity as a function of one variable, then use derivatives; e.g. the largest rectangle in a circle of radius r has area 2r².

Approximation by differentials

For small Δx, f(x + Δx) ≈ f(x) + f′(x)Δx and dy = f′(x)dx, used to estimate values like √4.1.