Applications of Derivatives: JEE Main PYQ, 2020–2025

Answer: (B)

The rate of change of y with respect to x is dy/dx. Differentiate: d/dx(x³ + 2x) = 3x² + 2. At x = 2 this is 3(4) + 2 = 14.

Answer: (A)

The trip takes 600/v hours, so total cost C(v) = (600/v)(1600 + v²) = 600(1600/v + v). Then C′(v) = 600(−1600/v² + 1) = 0 gives v = 40. C′ changes from negative to positive here, so this is a minimum. C(40) = 600(40 + 40) = ₹48,000.

Answer: (A)

Minimise the squared distance, because it has the same critical points as the distance. For a point (x, x²) on the parabola, D² = x² + (x² − 5)². Put t = x² with t ≥ 0, so D² = t² − 9t + 25. Differentiating gives 2t − 9 = 0, so t = 9/2, which is admissible. The second derivative is 2 > 0, so this is a minimum. Then D²min = 81/4 − 81/2 + 25 = 19/4, and the minimum distance is √19/2. Check the boundary: at t = 0, D² = 25, which is larger.

Answer: (D)

For x > 4 every factor is positive. Passing x = 4 (odd power) f′ becomes negative on (2, 4); passing x = 2 (odd power) it is positive on (1, 2); at x = 1 (even power) it stays positive on (0, 1); passing x = 0 it is negative for x < 0. So f′ goes from − to + at x = 0 and at x = 4 (local minima) and from + to − at x = 2 (local maximum). The sum is 0 + 4 = 4.

Answer: (A)

Let the point of contact be (a, a² + 4). Since dy/dx = 2x, the tangent there is y − (a² + 4) = 2a(x − a). It passes through the origin, so −(a² + 4) = −2a², which gives a² = 4. The first-quadrant condition picks a = 2, and the slope is 2a = 4.

Answer: (A)

The denominator x² + x + 1 is always positive, so f is defined and differentiable for all real x. Differentiating, the numerator is (2x − 1)(x² + x + 1) − (x² − x + 1)(2x + 1) = (2x³ + x² + x − 1) − (2x³ − x² + x + 1) = 2x² − 2. So f′(x) = 2(x² − 1)/(x² + x + 1)². The sign of f′ is the sign of x² − 1: f′ > 0 for x < −1, f′ < 0 for −1 < x < 1, and f′ > 0 for x > 1. So f has a local maximum at x = −1 and a local minimum at x = 1. We get f(−1) = 3/1 = 3 and f(1) = 1/3. Since f → 1 as x → ±∞, which lies between 1/3 and 3, these local extrema are the global ones. Hence M = 3, m = 1/3, and M − m = 8/3.

Answer: (D)

The tangent has slope 2, so set dy/dx = 2x = 2, which gives x = 1. The point of contact on the parabola is (1, 1 + 3) = (1, 4). This point lies on y = 2x + c, so 4 = 2(1) + c and c = 2.

Answer: (A)

The slope of the tangent is m(x) = y′ = −3x² + 6x + 9. To maximize m, set m′(x) = −6x + 6 = 0, so x = 1. Since m″ = −6 < 0, this is a maximum. Then m(1) = −3 + 6 + 9 = 12.

Answer: (A)

For x = y², differentiate to get 1 = 2y y′, so m₁ = 1/(2y). For xy = k, y + x y′ = 0, so m₂ = −y/x. Orthogonality requires m₁m₂ = −1, so −1/(2x) = −1 and x = 1/2 at the point of intersection. Then y² = 1/2, so y = 1/√2 (taking y > 0 since k > 0). Hence k = xy = (1/2)(1/√2) = 1/(2√2), which satisfies 8k² = 1.

Answer: (A)

On a closed interval the absolute maximum occurs at a critical point or at an endpoint. Differentiate: f′(x) = 3x² − 3 = 0 gives x = ±1, and only x = 1 lies in [0, 2]. Evaluate f(0) = 2, f(1) = 1 − 3 + 2 = 0, f(2) = 8 − 6 + 2 = 4. The largest of these is 4, so the greatest value is 4, attained at an endpoint.