4.4 MCQs-Application of Derivatives
Application of Derivatives
Rate of Change
1. If the distance s (in meters) traveled by a particle in time t (in seconds) is given by s(t) = t³ - 3t² + 4t, then the velocity at t = 2 seconds is:
4 m/s
2 m/s
0 m/s
8 m/s
Show me the answer
Answer: 1. 4 m/s
Explanation:
Velocity is the derivative of distance with respect to time: v(t) = s'(t)
s'(t) = 3t² - 6t + 4
At t = 2: v(2) = 3(2)² - 6(2) + 4 = 3(4) - 12 + 4 = 12 - 12 + 4 = 4 m/s
2. The radius of a circle is increasing at the rate of 0.7 cm/s. The rate at which the area is increasing when radius = 4 cm is:
5.6π cm²/s
11.2π cm²/s
17.6π cm²/s
8.4π cm²/s
Show me the answer
Answer: 1. 5.6π cm²/s
Explanation:
Area of circle: A = πr²
Differentiate with respect to time t: dA/dt = 2πr(dr/dt)
Given: dr/dt = 0.7 cm/s, r = 4 cm
dA/dt = 2π(4)(0.7) = 8π(0.7) = 5.6π cm²/s
Tangent and Normal
3. The slope of the tangent to the curve y = x³ - x at the point (2, 6) is:
11
12
10
13
Show me the answer
Answer: 1. 11
Explanation:
Slope of tangent = dy/dx
y = x³ - x ⇒ dy/dx = 3x² - 1
At x = 2: dy/dx = 3(2)² - 1 = 3(4) - 1 = 12 - 1 = 11
4. The equation of the tangent to the curve y = x² at the point (1, 1) is:
y = 2x - 1
y = 2x + 1
y = x - 1
y = 2x
Show me the answer
Answer: 1. y = 2x - 1
Explanation:
y = x² ⇒ dy/dx = 2x
At (1, 1): slope m = 2(1) = 2
Equation of tangent: y - y₁ = m(x - x₁)
y - 1 = 2(x - 1)
y - 1 = 2x - 2
y = 2x - 1
5. The slope of the normal to the curve y = 2x² + 3 at the point where x = 2 is:
-1/8
1/8
-8
8
Show me the answer
Answer: 1. -1/8
Explanation:
y = 2x² + 3 ⇒ dy/dx = 4x
At x = 2: slope of tangent = 4(2) = 8
Slope of normal = -1/(slope of tangent) = -1/8
Increasing and Decreasing Functions
6. The function f(x) = x³ - 3x² + 3x - 100 is:
Increasing on R
Decreasing on R
Increasing on (-∞, 1) and decreasing on (1, ∞)
Decreasing on (-∞, 1) and increasing on (1, ∞)
Show me the answer
Answer: 1. Increasing on R
Explanation:
f'(x) = 3x² - 6x + 3 = 3(x² - 2x + 1) = 3(x - 1)²
f'(x) = 3(x - 1)² ≥ 0 for all x ∈ R
f'(x) = 0 only at x = 1
Since f'(x) ≥ 0 for all x, f(x) is increasing on R
7. The interval in which the function f(x) = x² - 4x + 6 is decreasing is:
(-∞, 2)
(2, ∞)
(-∞, ∞)
None of these
Show me the answer
Answer: 1. (-∞, 2)
Explanation:
f'(x) = 2x - 4
f'(x) < 0 when 2x - 4 < 0 ⇒ x < 2
Therefore, f(x) is decreasing on (-∞, 2)
f'(x) > 0 when x > 2, so f(x) is increasing on (2, ∞)
Maxima and Minima
8. The maximum value of f(x) = sin x + cos x in [0, π/2] is:
1
√2
2
1/√2
Show me the answer
Answer: 2. √2
Explanation:
f'(x) = cos x - sin x
For critical points: f'(x) = 0 ⇒ cos x = sin x ⇒ tan x = 1 ⇒ x = π/4 in [0, π/2]
f(0) = sin 0 + cos 0 = 0 + 1 = 1
f(π/4) = sin(π/4) + cos(π/4) = 1/√2 + 1/√2 = 2/√2 = √2
f(π/2) = sin(π/2) + cos(π/2) = 1 + 0 = 1
Maximum value = √2 at x = π/4
9. The function f(x) = x + 1/x has a local minimum at:
x = 1
x = -1
x = 0
x = 2
Show me the answer
Answer: 1. x = 1
Explanation:
f(x) = x + 1/x, x ≠ 0
f'(x) = 1 - 1/x²
For critical points: f'(x) = 0 ⇒ 1 - 1/x² = 0 ⇒ x² = 1 ⇒ x = ±1
f''(x) = 2/x³
f''(1) = 2/1³ = 2 > 0 ⇒ local minimum at x = 1
f''(-1) = 2/(-1)³ = -2 < 0 ⇒ local maximum at x = -1
10. The maximum value of (1/x)ˣ is:
e
e^(1/e)
1
∞
Show me the answer
Answer: 2. e^(1/e)
Explanation:
Let y = (1/x)ˣ = x^(-x)
Taking ln: ln y = -x ln x
Differentiate: (1/y) dy/dx = -ln x - 1
For maximum/minimum: dy/dx = 0 ⇒ -ln x - 1 = 0 ⇒ ln x = -1 ⇒ x = 1/e
Second derivative test confirms it's a maximum
Maximum value: y = (1/(1/e))^(1/e) = e^(1/e)
Approximation
11. Using differentials, the approximate value of √25.3 is:
5.03
5.02
5.01
5.00
Show me the answer
Answer: 1. 5.03
Explanation:
Let y = √x
dy/dx = 1/(2√x)
Take x = 25, Δx = 0.3
√25.3 ≈ √25 + (dy/dx)Δx = 5 + [1/(2×5)]×0.3 = 5 + (1/10)×0.3 = 5 + 0.03 = 5.03
12. The approximate value of f(2.01) where f(x) = 4x² + 5x + 2 is:
28.24
28.34
28.44
28.54
Show me the answer
Answer: 2. 28.34
Explanation:
f'(x) = 8x + 5
Take x = 2, Δx = 0.01
f(2) = 4(4) + 5(2) + 2 = 16 + 10 + 2 = 28
f(2.01) ≈ f(2) + f'(2)Δx = 28 + [8(2) + 5]×0.01 = 28 + (16 + 5)×0.01 = 28 + 21×0.01 = 28 + 0.21 = 28.21
Exact calculation: f(2.01) = 4(2.01)² + 5(2.01) + 2 = 4(4.0401) + 10.05 + 2 = 16.1604 + 10.05 + 2 = 28.2104 ≈ 28.21
Mean Value Theorems
13. For the function f(x) = x(x - 1)(x - 2) in [0, 1/2], the value of c in Rolle's theorem is:
1 - 1/√3
1 + 1/√3
1/√3
1 - √3
Show me the answer
Answer: 1. 1 - 1/√3
Explanation:
f(x) = x(x - 1)(x - 2) = x³ - 3x² + 2x
f(0) = 0, f(1/2) = (1/2)(-1/2)(-3/2) = 3/8 ≠ 0
Wait, Rolle's theorem requires f(a) = f(b)
Let's check: f(0) = 0, f(2) = 0, so interval should be [0, 2]
f'(x) = 3x² - 6x + 2
By Rolle's theorem, f'(c) = 0 for some c ∈ (0, 2)
3c² - 6c + 2 = 0 ⇒ c = [6 ± √(36 - 24)]/(6) = [6 ± √12]/6 = [6 ± 2√3]/6 = 1 ± 1/√3
Both values are in (0, 2): 1 - 1/√3 ≈ 0.422 and 1 + 1/√3 ≈ 1.577
14. For f(x) = x² in [1, 3], the value of c in Lagrange's Mean Value Theorem is:
1
2
3
4
Show me the answer
Answer: 2. 2
Explanation:
Lagrange's MVT: f'(c) = [f(3) - f(1)]/(3 - 1)
f(x) = x² ⇒ f'(x) = 2x
f(3) = 9, f(1) = 1
f'(c) = (9 - 1)/(3 - 1) = 8/2 = 4
2c = 4 ⇒ c = 2 ∈ (1, 3)
Monotonicity
15. The function f(x) = 2x³ - 9x² + 12x - 5 is increasing in the interval:
(-∞, 1) ∪ (2, ∞)
(1, 2)
(-∞, 1)
(2, ∞)
Show me the answer
Answer: 1. (-∞, 1) ∪ (2, ∞)
Explanation:
f'(x) = 6x² - 18x + 12 = 6(x² - 3x + 2) = 6(x - 1)(x - 2)
f'(x) > 0 when (x - 1)(x - 2) > 0
This occurs when x < 1 or x > 2
Therefore, f(x) is increasing on (-∞, 1) ∪ (2, ∞)
Tangents and Normals (Advanced)
16. The angle between the curves y = x² and y = x³ at the point of intersection (1, 1) is:
tan⁻¹(1/7)
tan⁻¹(2/7)
tan⁻¹(3/7)
tan⁻¹(4/7)
Show me the answer
Answer: 1. tan⁻¹(1/7)
Explanation:
For y = x²: dy/dx = 2x ⇒ m₁ = 2(1) = 2
For y = x³: dy/dx = 3x² ⇒ m₂ = 3(1)² = 3
Angle between curves: tan θ = |(m₁ - m₂)/(1 + m₁m₂)|
tan θ = |(2 - 3)/(1 + 2×3)| = |(-1)/(1 + 6)| = 1/7
θ = tan⁻¹(1/7)
17. The equation of the normal to the curve x² = 4y which passes through (1, 2) is:
x + y = 3
x - y = 3
x + y = 1
x - y = 1
Show me the answer
Answer: 1. x + y = 3
Explanation:
x² = 4y ⇒ y = x²/4
dy/dx = x/2
Let point on curve be (h, k) where k = h²/4
Slope of tangent at (h, k) = h/2
Slope of normal = -2/h
Equation of normal: y - k = (-2/h)(x - h)
Passes through (1, 2): 2 - h²/4 = (-2/h)(1 - h)
Multiply by 4h: 8h - h³ = -8(1 - h) = -8 + 8h
Simplify: 8h - h³ = -8 + 8h ⇒ -h³ = -8 ⇒ h³ = 8 ⇒ h = 2
k = h²/4 = 4/4 = 1
Equation: y - 1 = (-2/2)(x - 2) = -1(x - 2)
y - 1 = -x + 2 ⇒ x + y = 3
Optimization Problems
18. The maximum area of a rectangle inscribed in a circle of radius R is:
R²
2R²
πR²
R²/2
Show me the answer
Answer: 2. 2R²
Explanation:
Let rectangle have sides 2x and 2y
Diagonal = 2R (diameter of circle)
(2x)² + (2y)² = (2R)² ⇒ 4x² + 4y² = 4R² ⇒ x² + y² = R²
Area A = (2x)(2y) = 4xy
Using y = √(R² - x²): A = 4x√(R² - x²)
Maximize A² = 16x²(R² - x²) = 16(R²x² - x⁴)
d(A²)/dx = 16(2R²x - 4x³) = 32x(R² - 2x²) = 0
x = 0 or x = R/√2
Maximum at x = R/√2, y = R/√2
Maximum area = 4(R/√2)(R/√2) = 4(R²/2) = 2R²
19. The height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is:
R/√3
2R/√3
R√3
2R√3
Show me the answer
Answer: 2. 2R/√3
Explanation:
Let cylinder have radius r and height h
From geometry: r² + (h/2)² = R² ⇒ r² = R² - h²/4
Volume V = πr²h = π(R² - h²/4)h = π(R²h - h³/4)
dV/dh = π(R² - 3h²/4) = 0 ⇒ 3h²/4 = R² ⇒ h² = 4R²/3 ⇒ h = 2R/√3
d²V/dh² = π(-3h/2) < 0 for h > 0, so it's maximum
Rolle's and LMVT Applications
20. If f(x) is differentiable and f(1) = 4, f(2) = 6, then there exists at least one c ∈ (1, 2) such that f'(c) equals:
1
2
3
4
Show me the answer
Answer: 2. 2
Explanation:
By Lagrange's Mean Value Theorem, there exists c ∈ (1, 2) such that:
f'(c) = [f(2) - f(1)]/(2 - 1) = (6 - 4)/1 = 2
Second Derivative Test
21. For the function f(x) = x³ - 3x, which of the following is true?
Local maximum at x = 1, local minimum at x = -1
Local minimum at x = 1, local maximum at x = -1
No local extremum
Both are points of inflection
Show me the answer
Answer: 2. Local minimum at x = 1, local maximum at x = -1
Explanation:
f'(x) = 3x² - 3 = 3(x² - 1)
Critical points: f'(x) = 0 ⇒ x = ±1
f''(x) = 6x
f''(-1) = -6 < 0 ⇒ local maximum at x = -1
f''(1) = 6 > 0 ⇒ local minimum at x = 1
Approximation Errors
22. If the error in measuring the radius of a sphere is 2%, then the error in calculating its volume is:
2%
4%
6%
8%
Show me the answer
Answer: 3. 6%
Explanation:
Volume V = (4/3)πr³
dV/dr = 4πr²
Relative error: ΔV/V ≈ (dV/dr)(Δr)/V = (4πr²Δr)/((4/3)πr³) = 3(Δr/r)
Percentage error in V ≈ 3 × (percentage error in r) = 3 × 2% = 6%
Curve Sketching
23. The function f(x) = x⁴ - 4x³ has:
A local minimum at x = 3
A local maximum at x = 0
A point of inflection at x = 2
All of the above
Show me the answer
Answer: 4. All of the above
Explanation:
f'(x) = 4x³ - 12x² = 4x²(x - 3)
Critical points: x = 0, 3
f''(x) = 12x² - 24x = 12x(x - 2)
f''(0) = 0 (inconclusive), f''(3) = 12(3)(1) = 36 > 0 ⇒ local minimum at x = 3
Check sign change of f' around x = 0: f' changes from negative to negative ⇒ no extremum at x = 0
f''(x) = 0 at x = 0, 2
f'' changes sign at x = 2 ⇒ point of inflection at x = 2
f'' doesn't change sign at x = 0 ⇒ not inflection point
Rate of Change Applications
24. A ladder 5 m long is leaning against a wall. The bottom is pulled away at 2 m/s. When the bottom is 3 m from the wall, the top is moving down at the rate of:
1.5 m/s
2 m/s
2.5 m/s
3 m/s
Show me the answer
Answer: 1. 1.5 m/s
Explanation:
Let x = distance from wall to bottom, y = height of top
x² + y² = 25
Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0
Given: dx/dt = 2 m/s, x = 3 m
y = √(25 - 9) = √16 = 4 m
2(3)(2) + 2(4)(dy/dt) = 0
12 + 8(dy/dt) = 0 ⇒ dy/dt = -12/8 = -1.5 m/s
Negative means moving down at 1.5 m/s
Absolute Maxima and Minima
25. The absolute maximum value of f(x) = 4x - x² + 5 on [-2, 3] is:
5
7
9
11