The function f(x) = x³ – 3x² + 4x is increasing on:
MCQ
✅ Solution
f'(x) = 3x² – 6x + 4 = 3(x–1)² + 1 > 0 for ALL x ∈ ℝ. Hence f is strictly increasing on ℝ. (Example 8, NCERT)
4
For f(x) = x² – 4x + 6, which interval is the function DECREASING?
MCQ
✅ Solution
f'(x) = 2x – 4. Setting f'(x) = 0 gives x = 2. In (–∞, 2), f'(x) = 2x – 4 < 0, so f is decreasing there. In (2, ∞), f'(x) > 0, so f is increasing. (Example 10, NCERT)
5
The interval in which y = x²e⁻ˣ is increasing is:
MCQ
✅ Solution
dy/dx = e⁻ˣ(2x – x²) = xe⁻ˣ(2 – x). Since e⁻ˣ > 0, sign depends on x(2–x) > 0 ⟹ 0 < x < 2. So y is increasing in (0, 2). (Exercise 6.2 Q19, NCERT)
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6
A point where f'(c) = 0 or f is not differentiable is called:
MCQ
✅ Solution
By definition in NCERT: A point c in the domain of f at which either f'(c) = 0 or f is not differentiable is called a critical point of f. Note: every point of local max/min is a critical point, but not vice versa.
7
By Second Derivative Test, x = c is a point of local MINIMA if:
MCQ
✅ Solution
Theorem 4 (NCERT): x = c is a point of local minima if f'(c) = 0 and f''(c) > 0. The function curves upward at c (like a valley). Recall: local maxima requires f''(c) < 0 (curves downward like a hill).
8
Volume of a cube is increasing at 9 cm³/s. The rate of increase of surface area when edge is 10 cm is:
MCQ
✅ Solution
V = x³ ⟹ dV/dt = 3x²(dx/dt) = 9 ⟹ dx/dt = 3/x². S = 6x² ⟹ dS/dt = 12x·(dx/dt) = 12x·(3/x²) = 36/x. At x = 10: dS/dt = 36/10 = 3.6 cm²/s. (Example 2, NCERT)
9
f(x) = 4x³ – 6x² – 72x + 30 is DECREASING in the interval:
MCQ
✅ Solution
f'(x) = 12x² – 12x – 72 = 12(x–3)(x+2). f'(x) = 0 at x = –2 and x = 3. In (–2, 3): sign is (–)(+) < 0, so f is decreasing in (–2, 3). (Example 11, NCERT)
10
The absolute maximum value of f(x) = 2x³ – 15x² + 36x + 1 on [1, 5] occurs at:
MCQ
✅ Solution
f(1)=24, f(2)=29, f(3)=28, f(5)=56. Maximum value is 56 at x=5. Minimum is 24 at x=1. (Example 27, NCERT)
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11
The point on curve x² = 2y nearest to the point (0, 5) is:
MCQ
✅ Solution
Let point be (x, x²/2). Distance² = x² + (x²/2 – 5)². Minimizing: at x = ±2, y = 2. Nearest point = (2, 2). (Exercise 6.3 Q27, NCERT)
12
f(x) = x³ – 3x + 3 has a local MINIMA at:
MCQ
✅ Solution
f'(x) = 3x² – 3 = 3(x–1)(x+1). Critical points: x = ±1. At x = 1: f' changes from negative to positive → local minima. At x = –1: f' changes from positive to negative → local maxima. (Example 17, NCERT)
13
A stone dropped into a lake creates circular waves at 4 cm/s. When radius = 10 cm, area is increasing at:
MCQ
✅ Solution
dA/dt = 2πr·(dr/dt). Given dr/dt = 4 cm/s, r = 10 cm. dA/dt = 2π(10)(4) = 80π cm²/s. (Example 3, NCERT)
14
For f(x) = sin x, the function is DECREASING in:
MCQ
✅ Solution
f'(x) = cos x. In (π/2, π), cos x < 0, so f is decreasing. In (0, π/2) and (π, 3π/2), cos x > 0 so f is increasing. (Exercise 6.2 Q3, NCERT)
15
Two numbers sum to 15 and the sum of their squares is minimum. Each number equals:
For f(x) = sin x + cos x in [0, 2π], f'(x) = 0 gives x = π/4 and x = 5π/4. The number of critical points in this interval is:
Numerical
✅ Solution
f'(x) = cos x – sin x = 0 ⟹ sin x = cos x ⟹ x = π/4 and x = 5π/4 in [0, 2π]. That gives 2 critical points. (Example 13, NCERT)
24
The area of a trapezium with three equal sides of 10 cm when area is maximum equals 75√3. What is the value of the variable side x at maximum area? (Enter numerical value in cm)