Application of Derivatives — Class 12 MCQs with Answers

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

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

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

   A.f'(a)✓ Correct

   B.f(a)

   C.1/f(a)

   D.a

   Answer: A. 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. Q2Easy

   At a local extremum (interior, differentiable):

   A.f(x) = 0

   B.f''(x) = 0

   C.f'(x) = 0✓ Correct

   D.x = 0

   Answer: C. 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. Q3Easy

   Rate of change of y w.r.t. x is:

   A.Δy

   B.dy/dx✓ Correct

   C.y/x

   D.dy

   Answer: B. dy/dx

   Explanation: The rate of change of y with respect to x is precisely the derivative dy/dx, the limit of the average rate Δy/Δx as Δx → 0 — not the ratio y/x, which only works for a line through the origin.

4. Q4Easy

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

   A.f'(a)✓ Correct

   B.f(a)

   C.1/f'(a)

   D.f''(a)

   Answer: A. f'(a)

   Explanation: Slope of the tangent at x = x₀ is f'(x₀), the instantaneous rate of change — squaring or inverting it (1/f'(x₀), f''(x₀)) describes a different line entirely.

5. Q5Easy

   f is increasing on an interval if:

   A.f(x) > 0

   B.f'(x) < 0

   C.f'(x) = 0

   D.f'(x) > 0✓ Correct

   Answer: D. f'(x) > 0

   Explanation: f is increasing on an interval exactly where its derivative is positive there, f'(x) > 0 — the function's own sign, f(x) > 0, says nothing about whether it's rising or falling.

6. Q6Easy

   A critical point is where:

   A.f(x) is max

   B.f(x) = 0

   C.f''(x) = 0

   D.f'(x) = 0 or f'(x) doesn't exist✓ Correct

   Answer: D. f'(x) = 0 or f'(x) doesn't exist

   Explanation: A critical point is any x where f'(x) = 0 OR f' fails to exist (a corner or cusp) — restricting it to just f'(x) = 0 misses sharp points like |x| at x = 0, which is still a critical point.

7. Q7Medium

   f is strictly increasing on (a,b) if:

   A.f''(x) > 0

   B.f'(x) = 0

   C.f'(x) < 0

   D.f'(x) > 0 for all x in (a,b)✓ Correct

   Answer: D. f'(x) > 0 for all x in (a,b)

   Explanation: f is strictly increasing on (a,b) only when f'(x) > 0 for EVERY x in that interval — a single point where f'(x) ≤ 0 breaks strict increase, so f'(x)=0 at one point alone isn't enough.

8. Q8Medium

   If f'(c) = 0 and f''(c) > 0:

   A.Inflection

   B.Local maximum

   C.Local minimum at c✓ Correct

   D.Discontinuity

   Answer: C. Local minimum at c

   Explanation: Second derivative test: f'(x₀) = 0 locates a stationary point, and f''(x₀) > 0 means the curve bends upward (concave up) there, so x₀ is a local minimum — not a maximum.

9. Q9Medium

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

   A.1/π cm/s✓ Correct

   B.5/π

   C.2/π

   D.10/π

   Answer: A. 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.

10. Q10Medium

    Tangent to y = x³ at x = 1:

    A.y = x

    B.y = 3x + 2

    C.y = 3x

    D.y = 3x − 2✓ Correct

    Answer: D. y = 3x − 2

    Explanation: For y = x³ at x = 1: f(1) = 1 and f'(1) = 3(1)² = 3. The tangent is y − 1 = 3(x − 1), which simplifies to y = 3x − 2 — keep the constant term; dropping it gives the wrong line y = 3x through the origin.

11. Q11Medium

    f(x) = x³ − 3x is decreasing on:

    A.All x

    B.(−∞, −1) ∪ (1, ∞)

    C.(−1, 1)✓ Correct

    D.x > 0

    Answer: C. (−1, 1)

    Explanation: f'(x) = 3x² − 3 for f(x) = x³ − 3x, and this is negative exactly when x² < 1, i.e. −1 < x < 1 — the interval where the cubic dips between its two turning points, not outside it.

12. Q12Hard

    Approximate increase in y when x increases by dx (linear):

    A.f(x)·dx

    B.f'(x)·dx✓ Correct

    C.dx²

    D.dx/f(x)

    Answer: B. f'(x)·dx

    Explanation: The linear approximation says a small change dx produces dy ≈ f'(x)·dx — multiplying by the SLOPE, not the function value f(x), which is the trap in the wrong option.

13. Q13Hard

    Maximum area of rectangle inscribed in circle of radius R:

    A.4R²

    B.R²

    C.2R² (square inscribed)✓ Correct

    D.πR²

    Answer: C. 2R² (square inscribed)

    Explanation: Among all rectangles inscribed in a circle of radius R, the diagonal is fixed at 2R, and area xy is maximised when x = y (a square), giving side R√2 and area 2R² — not 4R² or πR² (that's the circle's own area).

14. Q14Hard

    Water flows into a conical tank at 3 m³/min. Tank radius 4 m, height 8 m. Water rises at:

    A.2

    B.6/π

    C.Cannot determine

    D.3/(4π) m/min when water is at height 4✓ Correct

    Answer: D. 3/(4π) m/min when water is at height 4

    Explanation: From r/h = 4/8 we get r = h/2 and V = πh³/12, so dV/dh = πh²/4. At h = 4 that is 4π, and dV/dt = 3 gives dh/dt = 3/(4π) ≈ 0.24 m/min. Dropping the factor 4 from h² is what turns the answer into 3/π.

Application of Derivatives — FAQs

More Class 12 Mathematics chapters

• 01Relations and Functions
• 02Inverse Trigonometric Functions
• 03Matrices
• 04Determinants
• 05Continuity and Differentiability
• 07Integrals

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