Limits, Continuity and Differentiability — JEE Main Questions

38 JEE Main practice questions on Limits, Continuity and Differentiability, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. If f(x) = sec^{-1} x, then f'(x) for |x| > 1 is:

  1. 1/(√(1-x^2))
  2. 1/(x√(x^2-1))
  3. 1/(1+x^2)
  4. 1/(|x|√(x^2-1))

Answer: 1/(|x|√(x^2-1))

The derivative of sec^{-1} x is 1/(|x|√(x^2-1)) for |x| > 1. This is derived using implicit differentiation: let y = sec^{-1} x, then sec y = x. Differentiating gives sec y tan y dy/dx = 1, so dy/dx = 1/(sec y tan y) = 1/(|x|√(x^2-1)). The absolute value accounts for the sign of tan y.

2. Let f(x) = [x] be the greatest integer function. Which of the following is true about f on the interval [1,2]?

  1. f is continuous on [1,2]
  2. f is left-continuous at x=1 and right-continuous at x=2
  3. f is right-continuous at x=1 and left-continuous at x=2
  4. f is continuous only at x=1.5

Answer: f is right-continuous at x=1 and left-continuous at x=2

For f(x)=[x] on [1,2], at x=1, right-hand limit = 1 = f(1), so right-continuous. At x=2, left-hand limit = 1 but f(2)=2, so not left-continuous. At interior points, f is constant, hence continuous. Thus f is right-continuous at 1 and left-continuous at 2.

3. For which function does Rolle's theorem fail on the given interval?

  1. f(x) = x^2 - 4x + 3 on [1, 3]
  2. f(x) = sin x on [0, π]
  3. f(x) = x^3 - 3x on [-√3, √3]
  4. f(x) = |x| on [-1, 1]

Answer: f(x) = |x| on [-1, 1]

Rolle's theorem requires continuity on [a,b] and differentiability on (a,b). f(x)=|x| is continuous on [-1,1] but not differentiable at x=0, so the theorem fails. The other functions satisfy all conditions and have a c with f'(c)=0.

4. Let f(x) = x² + 1 for x ≤ 1 and f(x) = ax + b for 1 < x ≤ 3. If f is differentiable on (0,3), then the value of c in (0,3) satisfying LMVT is

  1. 1
  2. 5/6
  3. 3/2
  4. 2/3

Answer: 5/6

Continuity at x=1 gives a+b=2. Differentiability gives a=2, b=0. Average slope over [0,3] = (f(3)-f(0))/3 = (6-1)/3 = 5/3. For c in (0,1), f'(c)=2c = 5/3 ⇒ c=5/6 ∈ (0,1). For c in (1,3), f'(c)=2 ≠ 5/3. Hence c=5/6.

5. The value of lim_{x→∞} (1 + 2/x)^{3x} is

  1. e^2
  2. e^5
  3. e^3
  4. e^6

Answer: e^6

This is a 1^∞ form. Using the standard result: lim_{x→∞} (1 + a/x)^{bx} = e^{ab}. Here a=2, b=3, so limit = e^{6}. Alternatively, take natural log: bx ln(1 + a/x) → bx * (a/x) = ab, so limit = e^{ab} = e^6.

6. If x = a cos t, y = a sin t, then d^2y/dx^2 at t = π/2 is:

  1. -1/a
  2. 1/a
  3. -cosec^3 t / a
  4. -cosec^2 t / a

Answer: -1/a

First find dy/dx = (dy/dt)/(dx/dt) = (a cos t)/(-a sin t) = -cot t. Then d^2y/dx^2 = d/dt(dy/dx) * dt/dx = (cosec^2 t) * (1/(-a sin t)) = -cosec^3 t / a. At t = π/2, cosec(π/2)=1, so d^2y/dx^2 = -1/a.

7. For f(x) = |x-2|, what are the left-hand derivative (LHD) and right-hand derivative (RHD) at x=2?

  1. LHD = -1, RHD = 1
  2. LHD = 1, RHD = -1
  3. LHD = 0, RHD = 0
  4. LHD = -1, RHD = -1

Answer: LHD = -1, RHD = 1

For f(x)=|x-2|, at x=2, LHD = lim_{h→0^-} (|2+h-2|-0)/h = lim_{h→0^-} |h|/h = lim_{h→0^-} (-h)/h = -1. RHD = lim_{h→0^+} |h|/h = lim_{h→0^+} h/h = 1. Since LHD ≠ RHD, f is not differentiable at 2.

8. The value of lim_{x→0} (sin x - x)/x³ is

  1. 1/6
  2. 0
  3. -1/6
  4. -1/3

Answer: -1/6

Using series expansion: sin x = x - x³/6 + O(x⁵), so sin x - x = -x³/6 + O(x⁵). Dividing by x³ gives -1/6 + O(x²) → -1/6. Alternatively, L'Hôpital's rule three times yields (-cos x)/6 → -1/6.

9. If f(x) = sin x and g(x) = x^2, which of the following is true about h(x) = f(g(x))?

  1. h is continuous everywhere
  2. h is continuous only at x=0
  3. h is continuous only for x>0
  4. h is discontinuous at x=0

Answer: h is continuous everywhere

Since g(x)=x^2 is continuous everywhere and f(x)=sin x is continuous everywhere, the composition h(x)=sin(x^2) is continuous everywhere by the theorem on continuity of composite functions.

10. At how many points is the function f(x) = |x² - 4| not differentiable?

  1. 0
  2. 1
  3. 3
  4. 2

Answer: 2

f(x) = |x²-4| has zeros at x = ±2. At these points, left and right derivatives differ (LHD = -4, RHD = 4 at x=2; LHD = -4, RHD = 4 at x=-2), so not differentiable at exactly 2 points.

11. Evaluate: lim_{x→0^+} x ln x.

  1. -1
  2. 1
  3. 0
  4. ∞

Answer: 0

This is 0·(-∞) form. Rewrite as ln x / (1/x), which is ∞/∞. Apply L'Hôpital: derivative of ln x is 1/x, derivative of 1/x is -1/x^2. So limit = lim (1/x)/(-1/x^2) = lim (-x) = 0.

12. For f(x) = (x^2 - 1)/(x - 1), what is the limit as x approaches 1?

  1. 2
  2. 0
  3. 1
  4. undefined

Answer: 2

The limit is 2 because factorising numerator as (x-1)(x+1) and cancelling (x-1) gives x+1, which approaches 2 as x→1. The function is undefined at x=1 but the limit exists.

More Mathematics topics

This page shows 12 of 38 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.