Previous year questions: Limits and Derivatives

How to use this page

Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. Stems were checked against the official PDF pages, and every numerical answer was re-derived. Start each question by identifying whether it is a limit, a differentiability question or a derivative computation, then choose the standard limit or the derivative rule that fits.

The JEE items here are mostly objective and limit-based, while the exemplar items are single-step evaluations. Time yourself: an objective limit should take about a minute, a differentiability count about two minutes, and a first-principles derivative about three.

One more note: several of the JEE items below are objective questions whose answer is a single number, and the exemplar items are evaluated by the standard limits. Attempt them without looking at the solution first, then compare the first step rather than only the final value.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

If the limit of sine of (x cubed minus 5x squared plus ax plus b) divided by (square root of (x minus 1) minus 1) times log of (x minus 1) as x tends to 2 is m, find a plus b plus m.

Answer: The denominator tends to zero, so the numerator must also tend to zero, which gives 2a plus b equal to 12. For a finite non-zero limit the polynomial must have a double root at 2, so its derivative at 2 is zero: 12 minus 20 plus a equal to 0, giving a equal to 8 and b equal to minus 4. The polynomial is then (x minus 2) squared times (x minus 1). The denominator is approximately (x minus 2) squared by 2, so the limit m is 2. Then a plus b plus m is 8 minus 4 plus 2 equal to 6. The official key is 6.

Q2 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The number of critical points of the function f(x) equal to modulus of sine x divided by x for x not zero, and equal to 1 at x equal to zero, in the interval from minus 2 pi to 2 pi is:

Answer: Critical points occur where the derivative is zero or does not exist. At x equal to zero the function jumps from minus 1 to 1, so it is not differentiable there. At x equal to plus or minus pi the modulus creates a corner, so the derivative does not exist there either. In the intervals where sine keeps one sign, setting the derivative to zero leads to tangent x equal to x, which has one solution in the interval from pi to 2 pi and its mirror image, giving two more points. Counting zero, plus or minus pi and the two symmetric tangent solutions gives 5 critical points. The official key is 5.

Q3 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Let f(x) equal the limit as y tends to 0 of (1 minus cosine of xy) times tangent of xy, all divided by y cubed. Find the number of solutions of f(x) equal to sine x with x real.

Answer: Using 1 minus cosine of xy approximately (xy) squared by 2 and tangent of xy approximately xy, the numerator is approximately x cubed y cubed by 2, so f(x) is x cubed by 2. The equation x cubed by 2 equal to sine x has the solution zero by symmetry, and near x equal to 1.2 the two curves cross; the function is odd, so the negative crossing is the mirror image. Hence three solutions. The official key is 3.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

The value of the limit as x tends to 0 of (x squared times sine squared x) divided by (x squared minus sine squared x) is:

Answer: Expand sine x as x minus x cubed by 6 near zero. Then sine squared x is approximately x squared minus x to the fourth by 3, so the denominator is approximately x to the fourth by 3. The numerator is approximately x to the fourth. Dividing gives the limit 3. The official key is 3.

Q5 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

For the function f(x) equal to e to the power sine of modulus x minus modulus x, consider two statements. Statement I: f is differentiable for all real x. Statement II: f is increasing in the interval from minus pi to minus pi by 2. Choose the correct option.

Answer: At zero the right branch is e to the sine x minus x with derivative e to the sine x cosine x minus 1, which is 0 at zero, and the left branch is e to the minus sine x plus x with derivative minus e to the minus sine x cosine x plus 1, also 0 at zero, so f is differentiable everywhere. On the interval from minus pi to minus pi by 2 the derivative simplifies to 1 plus e to the minus sine x times the modulus of cosine x, which is positive, so f is increasing there. Both statements are true. The official key is the first option, both statements true.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Let f(x) equal x cubed plus 8 for x less than 0 and x squared minus 4 for x at least 0, and let g(x) equal the cube root of (x minus 8) for x less than 0 and the square root of (x plus 4) for x at least 0. Find the number of points where the composite g of f is discontinuous.

Answer: The outer function g has a jump at zero, so the composite is discontinuous wherever f takes the value zero, and also where f itself jumps. The function f jumps at zero, is zero at x equal to minus 2 from the left branch and at x equal to 2 from the right branch. The three points minus 2, 0 and 2 therefore give discontinuities. The official key is 3.

Q7 (NCERT Exemplar, 3 marks)

Evaluate the limit as x tends to 0 of sine x divided by (1 plus cosine x), all divided by x.

Answer: Write sine x by (1 plus cosine x) as tangent of x by 2, using the half-angle identity. Then the expression is tangent of x by 2 divided by x, which is 1 by 2 times tangent of x by 2 over x by 2, and the standard tangent limit gives 1 by 2. Source: NCERT Exemplar Class 11 Mathematics, Chapter 13, Objective Type Questions, Example 22.

Q8 (NCERT Exemplar, 3 marks)

Evaluate the limit as x tends to pi by 2 of (1 minus sine x) divided by cosine x.

Answer: Substitute y equal to x minus pi by 2, so that y tends to zero. Then sine x becomes cosine y and cosine x becomes minus sine y, and the expression becomes (1 minus cosine y) by (minus sine y), which simplifies to minus tangent of y by 2. The limit as y tends to zero is 0. Source: NCERT Exemplar Class 11 Mathematics, Chapter 13, Objective Type Questions, Example 23.

Q9 (NCERT Exemplar, 3 marks)

Evaluate the limit of modulus x divided by x as x tends to 0.

Answer: From the right the quotient is 1 throughout and the right limit is 1. From the left the quotient is minus 1 throughout and the left limit is minus 1. Since the one-sided limits differ, the limit does not exist. Source: NCERT Exemplar Class 11 Mathematics, Chapter 13, Objective Type Questions, Example 24.

Q10 (NCERT Exemplar, 3 marks)

Evaluate the limit of x times sine of 1 by x as x tends to 0.

Answer: The sine factor lies between minus 1 and 1, so the whole expression lies between minus the absolute value of x and plus the absolute value of x. Both bounds tend to zero, so by the sandwich theorem the limit is 0. Source: NCERT Exemplar Class 11 Mathematics, Chapter 13, Objective Type Questions, Example 26.