Limits, Continuity and Differentiability: JEE Advanced PYQ, 2007–2026

Answer: (D)

Differentiating: 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + y)/(x + 2y). At (1, 2) this is −4/5.

Answer: (A)

At a non-integer point [x] is locally constant, so f is clearly differentiable there. At an integer n, f(n) = n·0 = 0. For x near n, f(x)/(x − n) = [x]·sin²(πx)/(x − n) = [x]·π²(x − n)·(sin(π(x − n))/(π(x − n)))². Since [x] stays bounded near n, this tends to 0 as x → n from either side. So f'(n) = 0 exists, and f is differentiable on all of R.

Answer: (A)

A function is twice differentiable at 0 only if f, f' and f'' all agree from both sides. Continuity: c = e⁰ = 1. First derivative: f'(0⁻) = 2e⁰ = 2 and f'(0⁺) = b, so b = 2. Second derivative: f''(0⁻) = 4e⁰ = 4 and f''(0⁺) = 2a, so a = 2. Therefore f(2) = 2·4 + 2·2 + 1 = 13.

Answer: (C)

The inner expression repeats, so y = √(x + y), i.e. y² = x + y. At x = 2: y² − y − 2 = 0 gives y = 2 (positive root). Differentiate implicitly: 2y·y′ = 1 + y′, so y′(2y − 1) = 1 and y′ = 1/(2y − 1) = 1/3.

Answer: (A)

The radical repeats itself, so y = √(ln x + y), which gives y² = ln x + y. Differentiate implicitly: 2y y′ = 1/x + y′, so y′ = 1/(x(2y − 1)). At x = e: y² = 1 + y, so y² − y − 1 = 0. Since y > 0, y = (1 + √5)/2 and 2y − 1 = √5. Therefore y′ = 1/(e√5).

Answer: (A)

Differentiate with the product rule. y' = e^{2x}(2 sin x + cos x). Then y'' = e^{2x}[2(2 sin x + cos x) + (2 cos x − sin x)] = e^{2x}(3 sin x + 4 cos x). Substitute into y'' + a y' + b y = 0 and divide by e^{2x}: (3 + 2a + b) sin x + (4 + a) cos x = 0 for all x. So a = −4, and then 3 − 8 + b = 0 gives b = 5. Hence a + b = 1.

Answer: (A)

Because rationals and irrationals are both dense, f is continuous at a point a only if a² = 2a − 1, that is (a − 1)² = 0, so only at a = 1. Differentiability needs continuity, so 1 is the only candidate. Here f(1) = 1. For rational x, [f(x) − f(1)]/(x − 1) = (x² − 1)/(x − 1) = x + 1 → 2. For irrational x, [f(x) − f(1)]/(x − 1) = (2x − 2)/(x − 1) = 2. Both families of quotients have limit 2, so f′(1) = 2 exists. Hence f is differentiable at exactly one point.

Answer: (A)

Twice differentiability at 1 requires the left and right values of f, f′ and f″ to agree at x = 1. Left side: f(1) = 1, f′(1) = 3, f″(1) = 6. Right side: f(1) = a + b + c, f′(1) = 2a + b, f″(1) = 2a. So 2a = 6 gives a = 3. Then 2a + b = 3 gives b = −3. Then a + b + c = 1 gives c = 1. Hence abc = 3 × (−3) × 1 = −9.

Answer: (A)

Use the definition at 0 at each stage. f′(0) = lim h²|h|/h = lim h|h| = 0, and for x ≠ 0, f′(x) = 3x|x|. Then f″(0) = lim 3h|h|/h = lim 3|h| = 0, and f″(x) = 6|x| for x ≠ 0. Finally f‴(0) = lim 6|h|/h, which is +6 from the right and −6 from the left, so it does not exist. The smallest such n is 3.

Answer: (A)

Write f(a + 2h) − f(a − 3h) = [f(a + 2h) − f(a)] + [f(a) − f(a − 3h)]. Then [f(a + 2h) − f(a)]/h = 2·[f(a + 2h) − f(a)]/(2h) → 2f′(a). Also [f(a) − f(a − 3h)]/h = 3·[f(a − 3h) − f(a)]/(−3h) → 3f′(a). The total is 5f′(a) = 20.