# 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:
1. 4 m/s
2. 2 m/s
3. 0 m/s
4. 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:
1. 5.6π cm²/s
2. 11.2π cm²/s
3. 17.6π cm²/s
4. 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:
1. 11
2. 12
3. 10
4. 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:
1. y = 2x - 1
2. y = 2x + 1
3. y = x - 1
4. 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. -1/8
2. 1/8
3. -8
4. 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:
1. Increasing on R
2. Decreasing on R
3. Increasing on (-∞, 1) and decreasing on (1, ∞)
4. 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:
1. (-∞, 2)
2. (2, ∞)
3. (-∞, ∞)
4. 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. 1
2. √2
3. 2
4. 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:
1. x = 1
2. x = -1
3. x = 0
4. 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:
1. e
2. e^(1/e)
3. 1
4. ∞
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:
1. 5.03
2. 5.02
3. 5.01
4. 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:
1. 28.24
2. 28.34
3. 28.44
4. 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 - 1/√3
2. 1 + 1/√3
3. 1/√3
4. 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. 1
2. 2
3. 3
4. 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. (-∞, 1) ∪ (2, ∞)
2. (1, 2)
3. (-∞, 1)
4. (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:
1. tan⁻¹(1/7)
2. tan⁻¹(2/7)
3. tan⁻¹(3/7)
4. 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:
1. x + y = 3
2. x - y = 3
3. x + y = 1
4. 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:
1. R²
2. 2R²
3. πR²
4. 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:
1. R/√3
2. 2R/√3
3. R√3
4. 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. 1
2. 2
3. 3
4. 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?
1. Local maximum at x = 1, local minimum at x = -1
2. Local minimum at x = 1, local maximum at x = -1
3. No local extremum
4. 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:
1. 2%
2. 4%
3. 6%
4. 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:
1. A local minimum at x = 3
2. A local maximum at x = 0
3. A point of inflection at x = 2
4. 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. 1.5 m/s
2. 2 m/s
3. 2.5 m/s
4. 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:
1. 5
2. 7
3. 9
4. 11
Show me the answer
**Answer:** 3. 9
**Explanation:**
* f'(x) = 4 - 2x
* Critical point: f'(x) = 0 ⇒ 4 - 2x = 0 ⇒ x = 2
* Evaluate:
* f(-2) = 4(-2) - (-2)² + 5 = -8 - 4 + 5 = -7
* f(2) = 4(2) - (2)² + 5 = 8 - 4 + 5 = 9
* f(3) = 4(3) - (3)² + 5 = 12 - 9 + 5 = 8
* Absolute maximum = 9 at x = 2