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Class 12 Mathematics Part 1

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Previous year questions: Continuity and Differentiability

Continuity and Differentiability28 min readLesson 6 of 6

Q1 (CBSE Board 2024, 1 mark)

For the greatest integer function f(x) = [x] with 0 < x < 3, at how many points is the function not differentiable? Choose from 0, 1, 2 and 3. Answer: The greatest integer function jumps at every integer, so it is discontinuous at x = 1 and x = 2 inside the interval. A discontinuous function cannot be differentiable, so those two points are not differentiable. No other points in the open interval are integers, so the count is 2. The two integers inside the open interval are the only candidates, so counting them finishes the question. Each integer inside the open interval contributes one point, so the answer is a simple count. The count covers only integers strictly inside the interval, so the endpoints are ignored.

Q2 (CBSE Board 2024, 3 marks)

Differentiate y = (tan x) raised to the power x with respect to x. Answer: Take logarithms: log y = x log(tan x). Differentiating both sides gives y' over y equal to log(tan x) plus x times the derivative of log(tan x), which is x sec squared x over tan x. Multiplying by y gives (tan x) to the power x times the bracket log(tan x) plus x sec squared x over tan x. The bracket can also be written as log(tan x) plus 2x over sin 2x. Keeping the bracket unsimplified is acceptable because the exam scheme also accepts the single fraction form. The logarithm step is written first so the power rule applies legally. The log step converts the variable exponent into a product, which is what makes the derivative accessible.

Q3 (CBSE Board 2024, 3 marks)

If y = sin(tan inverse of e to the power x), find dy by dx at x = 0. Answer: Apply the chain rule twice. The derivative is cos of tan inverse of e to the power x times e to the power x divided by 1 plus e to the power 2x. At x = 0 the inner angle is tan inverse of 1, which is pi by 4, so its cosine is root 2 by 2. The fraction becomes 1 by 2, so the derivative equals root 2 by 4. The inner value at zero and the cosine of that value are the only two numbers needed. The two substitutions are the only arithmetic needed after the chain rule. The two substitution values are the only arithmetic after the chain rule is set up.

Q4 (CBSE Board 2025, 2 marks)

If e to the power y times (x + 1) equals 1, prove that dy by dx equals minus e to the power y. Answer: Rewrite the relation as e to the power y equals 1 over (x + 1). Differentiating both sides with respect to x gives e to the power y times dy by dx equal to minus 1 over (x + 1) squared. Since 1 over (x + 1) equals e to the power y, the right side is minus e to the power 2y. Dividing by e to the power y gives dy by dx equal to minus e to the power y, which proves the claim. The substitution e to the power y for the reciprocal is what removes x and leaves the claimed answer. Substituting the reciprocal after differentiating is what closes the proof in one line. The reciprocal relation is what turns the right side into a clean exponential expression.

Q5 (CBSE Board 2025, 2 marks)

Find dy by dt when y = cos cubed of (sec squared 2t). Answer: The function has three layers: a cube, a cosine and a secant squared. Differentiating the cube gives 3 cos squared(sec squared 2t). Differentiating the cosine gives minus sine of the inner expression. Differentiating sec squared 2t gives 4 sec squared 2t tan 2t. Multiplying the three pieces gives minus 12 cos squared(sec squared 2t) sin(sec squared 2t) sec squared 2t tan 2t as dy by dt. Each layer contributes one factor, and the product of the four factors is the complete derivative. The three chain factors multiply in order from the outermost function to the innermost. Each layer contributes exactly one factor to the final product.

Q6 (CBSE Board 2026, 1 mark)

If sin inverse x equals y, then dy by dx is which of cos inverse x, cos y, the reciprocal of the square root of one minus x squared, or sec y? Answer: Differentiate sin y equals x implicitly. This gives cos y times dy by dx equals 1, so dy by dx equals 1 over cos y. Since cos y equals the square root of one minus x squared, the derivative is the reciprocal of that square root. The correct choice is the third option. The implicit method also shows why the formula has a plus or minus choice hidden in the branch. Writing the implicit relation first makes the branch restriction explicit. The implicit route also explains why the answer depends only on a square root denominator.

Q7 (CBSE Board 2026, 1 mark)

If 2 cos inverse x equals y, then dy by dx equals which expression? Answer: Differentiating y equal to 2 cos inverse x gives dy by dx equal to 2 times minus 1 over the square root of one minus x squared, which is minus 2 over the square root of one minus x squared. Since sin of y by 2 equals the square root of one minus x squared, this can also be written as minus 2 cosec of y by 2. The answer matches the option written in that form. The factor of 2 in front survives the differentiation and appears in the answer. The leading factor 2 is carried through the differentiation unchanged.

Q8 (CBSE Board 2026, 2 marks)

If x = e to the power sin inverse t and y = e to the power cos inverse t, find dy by dx at t = 1/2. Answer: Differentiate each coordinate with respect to t. dx by dt equals e to the power sin inverse t over the square root of one minus t squared, and dy by dt equals minus e to the power cos inverse t over the same square root. Dividing, dy by dx equals minus y over x. At t = 1/2, sin inverse t is pi by 6 and cos inverse t is pi by 3, so y over x is e to the power pi by 6. Hence the derivative is minus e to the power pi by 6. Complementary angles make the two exponents relate simply, so the ratio is a single exponential. Complementary angles make the exponents combine into a single exponential before substitution. Complementary angles make the two exponentials collapse into one ratio.

Q9 (CBSE SQP 2021-22, 1 mark)

If e to the power x plus e to the power y equals e to the power x plus y, then dy by dx equals which of e to the power y minus x, e to the power x plus y, minus e to the power y minus x, or 2 e to the power x minus y? Answer: Differentiate the relation implicitly: e to the power x plus e to the power y times y' equals e to the power x plus y times (1 plus y'). Collecting the y' terms gives y' times (e to the power y minus e to the power x plus y) equal to e to the power x plus y minus e to the power x. Simplifying with the original relation gives y' equal to minus e to the power y minus x. The correct option is the third one. The collected coefficient of y prime is non-zero on the domain, so the division is valid. The original relation removes the extra exponential after collection. The original relation removes the remaining exponential after the y prime terms are collected.

Q10 (CBSE SQP 2021-22, 1 mark)

If x = a sec theta and y = b tan theta, find the second derivative of y with respect to x at theta = pi by 6. Answer: First, dy by dx equals (b sec squared theta) divided by (a sec theta tan theta), which simplifies to (b by a) cosec theta. Differentiate this with respect to theta and divide by dx by d theta: the result is minus (b by a squared) cot cubed theta. At theta equal to pi by 6, cot theta is root 3, so the second derivative is minus 3 root 3 b divided by a squared. Converting to cotangent before substituting keeps the value exact and shows the a squared in the denominator. Simplifying to a single trigonometric power before substituting keeps the final value exact. Simplifying to cotangent before substitution keeps the exact radical form.