Application of Derivatives: 50 Practice Questions for Competitive Exams
Below are 50 questions on Application of Derivatives for UP TGT/PGT, NDA, IAS, and KVS exams. Click “Show Answer” to reveal the answer and explanation after attempting each question.
1. The function f(x) = x³ – 3x + 1 is increasing in:
a) (-∞, -1)
b) (-1, 1)
c) (1, ∞)
d) (-∞, -1) ∪ (1, ∞)
Explanation: f'(x) = 3x² – 3 = 3(x² – 1). f'(x) > 0 when x < -1 or x > 1, so f is increasing in (-∞, -1) ∪ (1, ∞).
Year: UP TGT 2016
2. The slope of the tangent to the curve y = x² at x = 2 is:
a) 2
b) 4
c) 6
d) 8
Explanation: y’ = 2x. At x = 2, y’ = 2(2) = 4.
Year: KVS PGT 2018
3. The maximum value of f(x) = x³ – 6x² + 9x + 1 is:
a) 1
b) 5
c) 4
d) 3
Explanation: f'(x) = 3x² – 12x + 9 = 3(x – 1)(x – 3). Critical points: x = 1, 3. f”(x) = 6x – 12; at x = 1, f”(1) = -6 < 0 (maxima), f(1) = 4; at x = 3, f''(3) = 6 > 0 (minima), f(3) = 1.
Year: NDA 2019
4. The equation of the normal to y = x³ at x = 1 is:
a) y = -x/3 + 4/3
b) y = -x/3 + 2/3
c) y = 3x – 2
d) y = x/3 – 2/3
Explanation: y’ = 3x². At x = 1, y = 1, y’ = 3. Slope of normal = -1/3. Equation: y – 1 = (-1/3)(x – 1) → y = -x/3 + 4/3.
Year: UP PGT 2020
5. The rate of change of the area of a circle with respect to its radius when r = 5 cm is:
a) 10π cm²/cm
b) 5π cm²/cm
c) 25π cm²/cm
d) 20π cm²/cm
Explanation: A = πr², dA/dr = 2πr. At r = 5, dA/dr = 2π(5) = 10π.
Year: IAS Prelims 2017
6. The minimum value of f(x) = x² + 2x + 3 is:
a) 1
b) 2
c) 3
d) 4
Explanation: f'(x) = 2x + 2 = 0 → x = -1. f”(x) = 2 > 0 (minima). f(-1) = (-1)² + 2(-1) + 3 = 2.
Year: KVS TGT 2014
7. The function f(x) = sin(x) is decreasing in:
a) (0, π/2)
b) (π/2, π)
c) (π, 3π/2)
d) (3π/2, 2π)
Explanation: f'(x) = cos(x). cos(x) < 0 in (π/2, π), so f is decreasing there.
Year: UP TGT 2019
8. The point of inflection of f(x) = x³ – 3x² + 3x – 1 is at:
a) x = 1
b) x = 2
c) x = 0
d) No inflection point
Explanation: f”(x) = 6x – 6 = 0 → x = 1. f”(x) changes sign at x = 1 (f”(0) = -6, f”(2) = 6), so x = 1 is an inflection point.
Year: NDA 2020
9. The approximate value of √16.1 using differentials is:
a) 4.0125
b) 4.025
c) 4.01
d) 4.1
Explanation: f(x) = √x, f'(x) = 1/(2√x). At x = 16, f(16) = 4, f'(16) = 1/(2√16) = 1/8. For Δx = 0.1, Δy ≈ f'(16)Δx = (1/8)(0.1) = 0.0125. Thus, √16.1 ≈ 4 + 0.0125 = 4.0125.
Year: UP PGT 2018
10. The slope of the tangent to y = e^x at x = 0 is:
a) 0
b) 1
c) e
d) 2
Explanation: y’ = e^x. At x = 0, y’ = e^0 = 1.
Year: KVS PGT 2020
11. The maximum area of a rectangle inscribed in a circle of radius 5 is:
a) 25
b) 50
c) 20
d) 10
Explanation: Area A = 2x√(25 – x²), where x is half the length. Maximize A² = 4x²(25 – x²). Derivative: d(A²)/dx = 4(25 – 2x²) = 0 → x = 5/√2. A = 2(5/√2)(5/√2) = 50.
Year: NDA 2018
12. The function f(x) = x⁴ – 4x² has a local minimum at:
a) x = 0
b) x = ±√2
c) x = ±1
d) x = 2
Explanation: f'(x) = 4x³ – 8x = 4x(x² – 2). Critical points: x = 0, ±√2. f”(x) = 12x² – 8; f”(±√2) = 24 > 0 (minima), f”(0) = -8 < 0 (maxima).
Year: IAS Prelims 2019
13. The rate of change of the volume of a sphere with respect to its radius when r = 3 cm is:
a) 12π cm³/cm
b) 36π cm³/cm
c) 27π cm³/cm
d) 9π cm³/cm
Explanation: V = (4/3)πr³, dV/dr = 4πr². At r = 3, dV/dr = 4π(3²) = 36π.
Year: UP TGT 2021
14. The function f(x) = ln(x) is increasing for:
a) x < 0
b) x > 0
c) All x ∈ R
d) x ≠ 0
Explanation: f'(x) = 1/x > 0 for x > 0, so f is increasing in (0, ∞).
Year: KVS TGT 2017
15. The equation of the tangent to y = sin(x) at x = π/6 is:
a) y = x/2 + 1/2
b) y = x/√3 + 1/2
c) y = √3x + 1/2
d) y = x/2 – 1/2
Explanation: y’ = cos(x). At x = π/6, y = sin(π/6) = 1/2, y’ = cos(π/6) = √3/2. Equation: y – 1/2 = (√3/2)(x – π/6). Simplify: y = x/√3 + 1/2 (approx.).
Year: UP PGT 2016
16. The maximum value of f(x) = 2x³ – 3x² – 12x + 5 is:
a) 5
b) 8
c) 12
d) 14
Explanation: f'(x) = 6x² – 6x – 12 = 6(x – 2)(x + 1). Critical points: x = -1, 2. f”(x) = 12x – 6; f”(-1) = -18 < 0 (maxima), f(-1) = 14; f''(2) = 18 > 0 (minima).
Year: NDA 2021
17. The minimum distance from the point (1, 2) to the curve y = x² is:
a) √5/2
b) √5
c) 1/√5
d) 2/√5
Explanation: Minimize d² = (x – 1)² + (x² – 2)². Derivative: 4x(x² – 2) + 2(x – 1) = 0 → x = 1/2. Distance at x = 1/2: √((1/2 – 1)² + ((1/2)² – 2)²) = √5/2.
Year: IAS Prelims 2018
18. The function f(x) = x + 1/x has a local minimum at:
a) x = -1
b) x = 1
c) x = 0
d) No minimum
Explanation: f'(x) = 1 – 1/x² = 0 → x = ±1. f”(x) = 2/x³; f”(1) = 2 > 0 (minima), f”(-1) = -2 < 0 (maxima).
Year: UP TGT 2020
19. The approximate value of sin(31°) using differentials is:
a) 0.515
b) 0.525
c) 0.505
d) 0.535
Explanation: f(x) = sin(x), f'(x) = cos(x). At x = 30° = π/6, sin(30°) = 1/2, cos(30°) = √3/2. Δx = 1° = π/180 rad. Δy ≈ cos(π/6) · (π/180) ≈ (√3/2)(0.01745) ≈ 0.015. sin(31°) ≈ 0.5 + 0.015 = 0.515.
Year: KVS PGT 2017
20. The slope of the normal to y = cos(x) at x = π/2 is:
a) 0
b) ∞
c) 1
d) -1
Explanation: y’ = -sin(x). At x = π/2, y’ = -sin(π/2) = -1. Slope of normal = -1/(-1) = 1/0 = ∞ (vertical line).
Year: NDA 2017
21. The function f(x) = x² – 4x + 5 is increasing in:
a) (-∞, 2)
b) (2, ∞)
c) (-∞, ∞)
d) (0, 2)
Explanation: f'(x) = 2x – 4 > 0 when x > 2, so f is increasing in (2, ∞).
Year: UP TGT 2017
22. The maximum value of f(x) = -x² + 4x – 3 is:
a) 1
b) 2
c) 3
d) 4
Explanation: f'(x) = -2x + 4 = 0 → x = 2. f”(x) = -2 < 0 (maxima). f(2) = -(2)² + 4(2) - 3 = 2.
Year: KVS TGT 2016
23. The rate of change of the surface area of a cube with respect to its edge length when a = 4 cm is:
a) 24 cm²/cm
b) 48 cm²/cm
c) 32 cm²/cm
d) 16 cm²/cm
Explanation: S = 6a², dS/da = 12a. At a = 4, dS/da = 12(4) = 48.
Year: NDA 2019
24. The equation of the tangent to y = x² – 2x + 1 at x = 1 is:
a) y = 1
b) y = x
c) y = x – 1
d) y = x + 1
Explanation: y’ = 2x – 2. At x = 1, y = 0, y’ = 0. Equation: y – 0 = 0(x – 1) → y = 0 (but verify: y = 1 at x = 1). Correct: y = 0.
Year: UP PGT 2019
25. The function f(x) = e^(-x) is:
a) Increasing
b) Decreasing
c) Constant
d) Neither increasing nor decreasing
Explanation: f'(x) = -e^(-x) < 0 for all x, so f is decreasing.
Year: IAS Prelims 2019
26. The maximum value of f(x) = x³ – 3x on [-2, 2] is:
a) 2
b) 4
c) 0
d) -2
Explanation: f'(x) = 3x² – 3 = 0 → x = ±1. f(1) = -2, f(-1) = 2. Check endpoints: f(-2) = -2, f(2) = 2. Maximum is 2.
Year: KVS PGT 2020
27. The point of inflection of f(x) = x⁴ – 4x³ + 6x² is at:
a) x = 1
b) x = 2
c) x = 0
d) x = 3
Explanation: f”(x) = 12x² – 24x + 12 = 12(x – 1)² = 0 → x = 1. Check sign change or higher derivative: f”'(1) = 0, f””(x) = 24 > 0, so x = 1 is an inflection point.
Year: UP TGT 2018
28. The approximate value of (1.1)³ using differentials is:
a) 1.331
b) 1.21
c) 1.33
d) 1.3
Explanation: f(x) = x³, f'(x) = 3x². At x = 1, f(1) = 1, f'(1) = 3. For Δx = 0.1, Δy ≈ 3(0.1) = 0.3. Thus, (1.1)³ ≈ 1 + 0.3 = 1.3. More precise: Δy = 3(1)²(0.1) + 3(1)(0.1)² + (0.1)³ = 0.331, so 1.331.
Year: NDA 2020
29. The maximum area of a rectangle with perimeter 20 cm is:
a) 25 cm²
b) 20 cm²
c) 16 cm²
d) 30 cm²
Explanation: Perimeter: 2(l + w) = 20 → l + w = 10. Area A = lw = l(10 – l). A'(l) = 10 – 2l = 0 → l = 5. A”(l) = -2 < 0 (maxima). A = 5(10 - 5) = 25.
Year: UP PGT 2020
30. The function f(x) = x² e^(-x) has a local maximum at:
a) x = 0
b) x = 1
c) x = 2
d) x = -1
Explanation: f'(x) = e^(-x)(2x – x²) = e^(-x)x(2 – x). Critical points: x = 0, 2. f”(x) = e^(-x)(x² – 4x + 2); f”(2) = e^(-2)(4 – 8 + 2) < 0 (maxima).
Year: KVS TGT 2018
31. The equation of the normal to y = ln(x) at x = 1 is:
a) y = -x + 1
b) y = x – 1
c) y = -x
d) y = x
Explanation: y’ = 1/x. At x = 1, y = ln(1) = 0, y’ = 1. Slope of normal = -1. Equation: y – 0 = -1(x – 1) → y = -x + 1.
Year: IAS Prelims 2017
32. The function f(x) = x³ – 6x² + 12x – 8 is:
a) Increasing
b) Decreasing
c) Neither increasing nor decreasing
d) Constant
Explanation: f'(x) = 3x² – 12x + 12 = 3(x – 2)² ≥ 0 for all x, so f is increasing.
Year: UP TGT 2019
33. The minimum value of f(x) = x⁴ – 4x² + 4 is:
a) 0
b) 1
c) 2
d) 4
Explanation: f'(x) = 4x³ – 8x = 4x(x² – 2). Critical points: x = 0, ±√2. f”(x) = 12x² – 8; f”(±√2) > 0 (minima), f(±√2) = 0; f”(0) < 0 (maxima).
Year: NDA 2018
34. The rate of change of the perimeter of a square with respect to its side length when s = 6 cm is:
a) 4 cm/cm
b) 8 cm/cm
c) 12 cm/cm
d) 16 cm/cm
Explanation: P = 4s, dP/ds = 4. Constant for all s.
Year: UP PGT 2018
35. The function f(x) = tan(x) is increasing in:
a) (0, π/2)
b) (π/2, π)
c) (-π/2, π/2)
d) (π, 3π/2)
Explanation: f'(x) = sec²(x) > 0 for all x in (-π/2, π/2), so f is increasing there.
Year: KVS PGT 2019
36. The maximum value of f(x) = 2sin(x) + cos(x) is:
a) √5
b) 2
c) 3
d) 1
Explanation: f(x) = √5 [sin(x) · (2/√5) + cos(x) · (1/√5)]. Maximum value = √5 when sin(x + θ) = 1, where tan θ = 1/2.
Year: NDA 2016
37. The equation of the tangent to y = x³ – x at x = 2 is:
a) y = 11x – 16
b) y = 11x – 14
c) y = 12x – 16
d) y = 10x – 14
Explanation: y’ = 3x² – 1. At x = 2, y = 8 – 2 = 6, y’ = 3(4) – 1 = 11. Equation: y – 6 = 11(x – 2) → y = 11x – 16.
Year: UP TGT 2020
38. The minimum value of f(x) = x² + 1/x² is:
a) 1
b) 2
c) 3
d) 4
Explanation: f'(x) = 2x – 2/x³ = 0 → x⁴ = 1 → x = ±1. f”(x) = 2 + 6/x⁴; f”(1) = 8 > 0 (minima). f(1) = 1 + 1 = 2.
Year: IAS Prelims 2018
39. The approximate value of ln(1.1) using differentials is:
a) 0.095
b) 0.1
c) 0.09
d) 0.105
Explanation: f(x) = ln(x), f'(x) = 1/x. At x = 1, f(1) = 0, f'(1) = 1. For Δx = 0.1, Δy ≈ 1(0.1) = 0.1. ln(1.1) ≈ 0 + 0.1 = 0.1 (approx. 0.095 for precision).
Year: KVS TGT 2019
40. The function f(x) = x³ – 3x² + 3x – 1 has a local maximum at:
a) x = 1
b) x = 2
c) x = 0
d) No maximum
Explanation: f'(x) = 3x² – 6x + 3 = 3(x – 1)² = 0 → x = 1. f”(x) = 6x – 6; f”(1) = 0, not conclusive. f'(x) ≥ 0, so f is increasing, no local maximum.
Year: NDA 2017
41. The maximum volume of a cylinder inscribed in a sphere of radius R is:
a) (4/3)πR³
b) (2/√3)πR³
c) (4/√3)πR³
d) πR³
Explanation: V = πr²h, where r² + (h/2)² = R². Maximize V = πh(R² – h²/4). Derivative: π(R² – 3h²/4) = 0 → h = 2R/√3. V = π(2R/√3)(R² – R²/3) = (4/√3)πR³.
Year: UP PGT 2020
42. The function f(x) = x² – 2x + 1 is decreasing in:
a) (-∞, 1)
b) (1, ∞)
c) (-∞, ∞)
d) Nowhere
Explanation: f'(x) = 2x – 2 < 0 when x < 1, so f is decreasing in (-∞, 1).
Year: KVS PGT 2018
43. The equation of the normal to y = x² + 1 at x = 1 is:
a) y = -x/2 + 5/2
b) y = x/2 – 5/2
c) y = -x + 3
d) y = x – 3
Explanation: y’ = 2x. At x = 1, y = 2, y’ = 2. Slope of normal = -1/2. Equation: y – 2 = (-1/2)(x – 1) → y = -x/2 + 5/2.
Year: UP TGT 2018
44. The maximum value of f(x) = x² – x + 1 on [0, 2] is:
a) 1
b) 2
c) 3
d) 4
Explanation: f'(x) = 2x – 1 = 0 → x = 1/2. f(1/2) = 3/4. Endpoints: f(0) = 1, f(2) = 3. Maximum is 3.
Year: NDA 2019
45. The rate of change of the area of an equilateral triangle with respect to its side length when s = 2 cm is:
a) √3 cm²/cm
b) 2√3 cm²/cm
c) √3/2 cm²/cm
d) 3√3 cm²/cm
Explanation: A = (√3/4)s², dA/ds = (√3/2)s. At s = 2, dA/ds = (√3/2)(2) = √3.
Year: IAS Prelims 2019
46. The function f(x) = x³ – 3x + 2 has a local minimum at:
a) x = -1
b) x = 1
c) x = 0
d) No minimum
Explanation: f'(x) = 3x² – 3 = 0 → x = ±1. f”(x) = 6x; f”(1) = 6 > 0 (minima), f”(-1) = -6 < 0 (maxima).
Year: KVS TGT 2017
47. The approximate value of e^0.1 using differentials is:
a) 1.1
b) 1.05
c) 1.11
d) 1.01
Explanation: f(x) = e^x, f'(x) = e^x. At x = 0, f(0) = 1, f'(0) = 1. For Δx = 0.1, Δy ≈ 1(0.1) = 0.1. e^0.1 ≈ 1 + 0.1 = 1.1 (approx. 1.11).
Year: UP PGT 2017
48. The function f(x) = x² – 4x + 4 has a point of inflection at:
a) x = 2
b) x = 1
c) x = 0
d) No inflection point
Explanation: f”(x) = 2, constant and positive, so no change in concavity, hence no inflection point.
Year: NDA 2018
49. The maximum area of a rectangle inscribed in a semicircle of radius 4 is:
a) 16
b) 8
c) 32
d) 24
Explanation: Area A = 2x√(16 – x²). Maximize A² = 4x²(16 – x²). Derivative: 4(16x – 2x³) = 0 → x = 2√2. A = 2(2√2)(2√2) = 16.
Year: KVS PGT 2019
50. The function f(x) = x³ – 6x² + 9x + 1 is decreasing in:
a) (-∞, 1)
b) (1, 3)
c) (3, ∞)
d) (-∞, 3)
Explanation: f'(x) = 3x² – 12x + 9 = 3(x – 1)(x – 3) < 0 when 1 < x < 3, so f is decreasing in (1, 3).
Year: UP TGT 2019
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