Take timed practice tests on Application of Derivatives for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.
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1. The slope of the tangent to y = x² at x = 2 is:
Explanation: dy/dx = 2x. At x = 2: slope = 2×2 = 4.
2. f(x) = x² is increasing on:
Explanation: f'(x) = 2x > 0 when x > 0. So f is increasing on (0, ∞) and decreasing on (−∞, 0).
3. At a local maximum or minimum, f'(x) is:
Explanation: At a local extremum (maxima or minima), f'(x) = 0 (stationary point), provided f is differentiable there. (Also check points where f' is undefined.)
4. If the radius of a circle increases at 2 cm/s, the rate of increase of area when r = 5 cm is:
Explanation: A = πr². dA/dt = 2πr(dr/dt) = 2π(5)(2) = 20π cm²/s.
5. For f(x) = x³ − 3x, the local maximum value is:
Explanation: f'(x) = 3x² − 3 = 0 → x = ±1. f''(x) = 6x. At x = −1: f'' = −6
6. If f'(a) = 0 and f''(a) > 0, then x = a is:
Explanation: Second derivative test: f'(a) = 0 and f''(a) > 0 → local minimum. f''(a)
7. Using differentials, √25.1 ≈:
Explanation: Let f(x) = √x, f'(x) = 1/(2√x). Δy ≈ f'(25)×Δx = (1/10)(0.1) = 0.01. √25.1 ≈ 5 + 0.01 = 5.01.
8. Rolle's theorem requires f(a) = f(b) and:
Explanation: Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then ∃ c ∈ (a,b) such that f'(c) = 0.
9. The minimum value of f(x) = x + 1/x for x > 0 is:
Explanation: f'(x) = 1 − 1/x² = 0 → x = 1. f''(x) = 2/x³ > 0 → minimum at x = 1. f(1) = 1 + 1 = 2.
10. The maximum area of a rectangle with perimeter 20 m is:
Explanation: Let sides be x and y: 2(x+y) = 20 → x+y = 10. Area A = xy. By AM-GM: xy ≤ ((x+y)/2)² = 25. Maximum area = 25 m² when x = y = 5 (square).
11. f(x) = 2x³ − 9x² + 12x − 5 is decreasing on:
Explanation: f'(x) = 6x²−18x+12 = 6(x²−3x+2) = 6(x−1)(x−2). f'(x)
12. The equation of the normal to y = x² − 1 at the point (1, 0) is:
Explanation: y' = 2x → at (1,0): slope of tangent = 2. Slope of normal = −1/2. Normal: y − 0 = −(1/2)(x − 1) → 2y = −x + 1 → x + 2y − 1 = 0.
13. For f(x) = x^x (x > 0), the minimum occurs at:
Explanation: f(x) = x^x → ln f = x ln x → f'/f = ln x + 1 → f'(x) = x^x(ln x + 1) = 0 → ln x = −1 → x = 1/e. f'' > 0 at x = 1/e → minimum.
14. The equation of the tangent to y = x³ at the origin is:
Explanation: At origin: y' = 3x² = 3(0)² = 0. Slope = 0 → tangent is horizontal: y = 0.
15. A particle moves with displacement s = t³ − 6t² + 9t. The particle is at rest when t equals:
Explanation: Velocity v = ds/dt = 3t² − 12t + 9 = 3(t²−4t+3) = 3(t−1)(t−3). v = 0 at t = 1 and t = 3.
16. The interval in which f(x) = sin x + cos x is increasing on [0, 2π] is:
Explanation: f'(x) = cos x − sin x. f'(x) > 0 when cos x > sin x, i.e., tan x