Limits, Continuity and Differentiability: JEE Main PYQ, 2020–2025

Answer: (D)

|g(x)| can fail to be differentiable only where g(x) = 0 and g changes sign with g'(x) ≠ 0. Here g(x) = sin x is zero at x = nπ, and cos nπ ≠ 0. Of the given points only x = π is such a zero. At π, the left derivative is −cos π·(−1)... explicitly, for x slightly less than π, f = sin x with slope cos π = −1, and for x slightly more than π, f = −sin x with slope −cos π = 1. The one-sided derivatives are −1 and 1, so f is not differentiable at π. At π/2, π/4 and 3π/2, sin x is nonzero, so f equals ±sin x locally and is smooth.

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)

max{x, x²} switches between the two expressions only where x = x², so x(x − 1) = 0 and x = 0 or x = 1. For x < 0, x² > x, so f = x² with slope 2x. For 0 < x < 1, x > x², so f = x with slope 1. For x > 1, f = x² again. At x = 0 the slopes are 0 (left) and 1 (right). At x = 1 the slopes are 1 (left) and 2 (right). Both differ, so f is not differentiable at 0 or 1, giving 2 points.

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: (C)

dy/dx = [1 + x/√(x² + 9)]/(x + √(x² + 9)) = [(√(x² + 9) + x)/√(x² + 9)]/(x + √(x² + 9)) = 1/√(x² + 9). At x = 4 this is 1/5.

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)

Suppose f + g is continuous at a. Then g = (f + g) − f is continuous at a, as the difference of two continuous functions. This contradicts the given condition, so f + g must be discontinuous at a.