44 JEE Main practice questions on Differential Calculus and Applications, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.
Answer: 7/(4√2)
Distance from point (x, x²+1) on parabola to line x − y − 1 = 0 is |x − (x²+1) − 1|/√2 = |−x² + x − 2|/√2. The quadratic −x² + x − 2 is always negative (discriminant < 0), so distance = (x² − x + 2)/√2. Minimise f(x) = x² − x + 2: f'(x) = 2x − 1 = 0 ⇒ x = 1/2. Minimum f = (1/4) − (1/2) + 2 = 7/4. Minimum distance = (7/4)/√2 = 7/(4√2).
Answer: 32πR^3/81
Let distance from sphere centre to cone base be d. Then height h = R + d, base radius r = √(R^2 - d^2). Volume V = (1/3)π(R^2 - d^2)(R + d) = (1/3)π(R - d)(R + d)^2. Differentiate: V'(d) = (1/3)π(R + d)(R - 3d). Set V' = 0 gives d = R/3. Then h = 4R/3, r^2 = 8R^2/9, V_max = (1/3)π(8R^2/9)(4R/3) = 32πR^3/81.
Answer: 4R/3
Let the base be at distance d from the sphere's centre. Then height h = R + d and base radius r satisfies r² = R² − d². Volume V = (1/3)πr²h = (1/3)π(R² − d²)(R + d). Differentiate with respect to d and set to zero: dV/dd = (1/3)π(R + d)(R − 3d) = 0 ⇒ d = R/3. Hence h = R + R/3 = 4R/3.
Answer: 50/(4+π) m²
Let radius = r, rectangle height = h. Perimeter: 2r + 2h + πr = 10. Area: 2rh + ½πr². Solve h = (10 - 2r - πr)/2. Substitute: A(r) = r(10 - 2r - πr) + ½πr² = 10r - (2+π/2)r². dA/dr = 10 - (4+π)r = 0 ⇒ r = 10/(4+π). Then h = r. Max area = 10r - (2+π/2)r² = 50/(4+π) m².
Answer: 3
The quadratic x^2 - 4x + 3 = (x-1)(x-3) has zeros at x=1 and x=3. On [0,4], f(x) = x^2 - 4x + 3 for x in [0,1]∪[3,4] and f(x) = -(x^2 - 4x + 3) for x in [1,3]. Evaluate at endpoints 0,4 and critical points 1,2,3: f(0)=3, f(1)=0, f(2)=1, f(3)=0, f(4)=3. Maximum is 3.
Answer: 4
Let x be the landing distance downstream. Rowing distance = √(9+x²), walking distance = 5-x. Total time T(x)=√(9+x²)/4+(5-x)/5. Set T'(x)=x/(4√(9+x²))-1/5=0 → 5x=4√(9+x²) → 25x²=16(9+x²) → 9x²=144 → x=4. Second derivative positive confirms minimum.
Answer: 200√2
Let side parallel to wall be x, other side y = 200/x. Cost C = 5(x + 2y) = 5(x + 400/x). Differentiate: dC/dx = 5(1 - 400/x^2) = 0 gives x = 20. Then y = 10. Minimum cost = 5(20 + 20) = Rs 200. Second derivative positive confirms minimum.
Answer: strictly increasing on (a,b)
By the first-derivative monotonicity test, if f'(x) > 0 on an open interval (a,b), then f is strictly increasing on that interval. This follows from the Mean Value Theorem: for any x1<x2 in (a,b), f(x2)-f(x1)=f'(c)(x2-x1)>0.
Answer: ln(1+x) ≤ x
Define f(x)=x-ln(1+x). f(0)=0, f'(x)=1-1/(1+x)=x/(1+x)>0 for x>0, so f is strictly increasing. Hence f(x)>0 for x>0, i.e., x>ln(1+x). Thus ln(1+x) < x, which implies ln(1+x) ≤ x (since strict inequality implies non-strict).
Answer: -1/e
Differentiate: f'(x) = e^x + x e^x = (1 + x)e^x. Set f'(x) = 0 ⇒ x = -1 (since e^x > 0). Second derivative: f''(x) = (2 + x)e^x. At x = -1, f''(-1) = e^{-1} > 0, so local minimum. Minimum value f(-1) = -1 * e^{-1} = -1/e.
Answer: 3
f'(x)=3x^2-12=3(x-2)(x+2). Critical points at x=-2 and x=2. f(-2)=17>0, f(2)=-15<0. As x→-∞, f→-∞; as x→∞, f→∞. By Intermediate Value Theorem, there is one root in each of (-∞,-2), (-2,2), (2,∞). So exactly 3 real roots.
Answer: Critical points in (a,b) and endpoints a, b
The algorithm for absolute extrema on a closed interval requires evaluating the function at all critical points in the open interval and at the endpoints. The largest value among these is the absolute maximum.
This page shows 12 of 44 questions on this topic. The full set, with progress tracking and five agent perspectives per question, is in the JupiteX app — browse the exam catalogue or browse the Learn library.