Applications of Derivatives: JEE Advanced PYQ, 2007–2026

Answer: (A)

The slope of the tangent at a point is the derivative there. dy/dx = 2x − 3, so at x = 3 the slope is 3. The tangent through (3, 2) is y − 2 = 3(x − 3), i.e. 3x − y − 7 = 0. The normal would use −1/3, and forgetting the −3 term gives slope 6.

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: (B)

dy/dt = 2 and dx/dt = 2t, so dy/dx = 1/t. At t = 2 the slope is 1/2.

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)

dy/dx = 1/(2√x), which equals 1/4 at x = 4. The normal is perpendicular to the tangent, so its slope is −4. The normal through (4, 2) is y − 2 = −4(x − 4), i.e. 4x + y − 18 = 0.

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)

Differentiate: f′(x) = cos x (1 + cos x) − sin²x = 2cos²x + cos x − 1 = (2cos x − 1)(cos x + 1). On (0, π), cos x + 1 > 0, so the sign of f′ is the sign of 2cos x − 1. This is positive for x < π/3 and negative for x > π/3, so f has a maximum at x = π/3. The end values are f(0) = f(π) = 0. Hence the greatest value is f(π/3) = (√3/2)(1 + 1/2) = 3√3/4.