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Continuity and Differentiability

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Chapter Notes

Overview
This chapter builds on Class XI differentiation, introducing the formal definitions of continuity and differentiability and the relationship between them. It covers the continuity of standard functions (polynomial, rational, trigonometric, modulus, greatest integer), the algebra of continuous functions, and the chain rule for composite functions. The chapter also introduces differentiation of implicit functions and inverse trigonometric functions, along with the concepts of exponential and logarithmic functions. The one big idea is that differentiability implies continuity, but continuity does not guarantee differentiability, and mastering the chain rule is essential for handling composite functions.
Key Concepts (12)
Continuity at a point
A function f is continuous at a point c in its domain if lim(x→c) f(x) = f(c). Equivalently, the left-hand limit, right-hand limit, and the value of the function at x = c all exist and are equal.
e.g. For f(x) = 2x + 3 at x = 1, lim(x→1) f(x) = 5 and f(1) = 5, so f is continuous at x = 1.
Discontinuity
A function f is said to be discontinuous at a point c if it is not continuous at c. The point c is called a point of discontinuity of f.
e.g. f(x) = { x+2, x≤1; x-2, x>1 } is discontinuous at x = 1 because LHL = 3, RHL = -1, which are not equal.
Continuous function
A real function f is said to be continuous if it is continuous at every point in its domain.
e.g. The constant function f(x) = k is continuous at every real number because lim(x→c) k = k = f(c) for any c.
Left-hand limit
The left-hand limit of f at c is the value f(x) approaches as x approaches c from the left (x < c), denoted as lim(x→c⁻) f(x).
e.g. For f(x) = [x] at x = 1, lim(x→1⁻) f(x) = 0.
Right-hand limit
The right-hand limit of f at c is the value f(x) approaches as x approaches c from the right (x > c), denoted as lim(x→c⁺) f(x).
e.g. For f(x) = [x] at x = 1, lim(x→1⁺) f(x) = 1.
Derivative of a function at a point
The derivative of f at a point c is defined as f'(c) = lim(h→0) [f(c+h) - f(c)]/h, provided this limit exists.
e.g. For f(x) = x², f'(1) = lim(h→0) [(1+h)² - 1]/h = 2.
Differentiability
A function f is differentiable at a point c if both the left-hand derivative and right-hand derivative exist and are finite and equal. A function is differentiable in an interval if it is differentiable at every point of that interval.
e.g. f(x) = |x| is not differentiable at x = 0 because LHD = -1 and RHD = 1, which are not equal.
Chain rule
If f = v o u, i.e., f(x) = v(u(x)), and if t = u(x), then df/dx = (dv/dt) * (dt/dx), provided both derivatives exist.
e.g. For f(x) = sin(x²), let t = x², then df/dx = cos(t) * 2x = 2x cos(x²).
Implicit function
A function where the relationship between x and y is given in a form that is not explicitly solved for y, e.g., x + sin(xy) - y = 0. Differentiation is done directly with respect to x, treating y as a function of x.
e.g. For x - y = π, differentiating w.r.t. x gives 1 - dy/dx = 0, so dy/dx = 1.
Exponential function
A function of the form f(x) = aˣ, where a > 0 and a ≠ 1. The most common exponential function is f(x) = eˣ, where e is the Euler number.
e.g. f(x) = 2ˣ is an exponential function; as x increases, the function grows rapidly.
Logarithmic function
The inverse of the exponential function, defined as y = logₐ x if and only if aʸ = x, for a > 0, a ≠ 1, and x > 0. The natural logarithm is logₑ x = ln x.
e.g. log₂ 8 = 3 because 2³ = 8.
Greatest integer function
The function f(x) = [x], where [x] denotes the greatest integer less than or equal to x.
e.g. [3.14] = 3, [-2.5] = -3.
Formulas & Equations (13)
lim(x→c) f(x) = f(c)
Reads as: The limit of f of x as x approaches c equals f of c
Variables: f: function; x: variable; c: point in domain
Use when: Checking continuity of a function at a point c
⚠ Students forget to check that f(c) is defined and that both left and right limits exist and are equal to f(c).
f'(c) = lim(h→0) [f(c+h) - f(c)]/h
Reads as: f prime of c equals the limit as h approaches 0 of f of c plus h minus f of c over h
Variables: f': derivative; c: point; h: small increment
Use when: Finding the derivative of a function at a specific point using first principles
⚠ Students incorrectly substitute h = 0 directly, leading to 0/0 form, instead of simplifying the expression first.
d/dx (xⁿ) = n xⁿ⁻¹
Reads as: Derivative of x to the power n is n times x to the power n minus 1
Variables: n: real number; x: variable
Use when: Differentiating power functions
⚠ Forgetting to subtract 1 from the exponent, e.g., writing derivative of x³ as 3x² (correct) but sometimes writing 3x³.
d/dx (sin x) = cos x
Reads as: Derivative of sine x is cosine x
Variables: x: angle in radians
Use when: Differentiating sine function
⚠ Using degrees instead of radians in calculus problems.
d/dx (cos x) = - sin x
Reads as: Derivative of cosine x is negative sine x
Variables: x: angle in radians
Use when: Differentiating cosine function
⚠ Forgetting the negative sign.
d/dx (tan x) = sec² x
Reads as: Derivative of tan x is secant squared x
Variables: x: angle in radians
Use when: Differentiating tangent function
⚠ Confusing with derivative of cot x, which is -cosec² x.
(u ± v)' = u' ± v'
Reads as: Derivative of u plus or minus v is u prime plus or minus v prime
Variables: u, v: differentiable functions of x
Use when: Differentiating sum or difference of functions
⚠ Applying product rule incorrectly to a sum.
(uv)' = u'v + uv'
Reads as: Derivative of u v is u prime v plus u v prime
Variables: u, v: differentiable functions of x
Use when: Differentiating product of two functions
⚠ Writing u'v' instead of u'v + uv'.
(u/v)' = (u'v - uv')/v²
Reads as: Derivative of u over v is u prime v minus u v prime all over v squared
Variables: u, v: differentiable functions of x, v ≠ 0
Use when: Differentiating quotient of two functions
⚠ Getting the order wrong in the numerator, i.e., writing uv' - u'v.
df/dx = (dv/dt) * (dt/dx) (Chain Rule)
Reads as: Derivative of f with respect to x equals derivative of v with respect to t times derivative of t with respect to x
Variables: f = v(u(x)); t = u(x)
Use when: Differentiating composite functions
⚠ Forgetting to multiply by the derivative of the inner function.
d/dx (sin⁻¹ x) = 1/√(1 - x²), for x ∈ (-1, 1)
Reads as: Derivative of sine inverse x is 1 over square root of 1 minus x squared
Variables: x: variable in (-1, 1)
Use when: Differentiating inverse sine function
⚠ Forgetting the domain restriction and writing the derivative for x = ±1.
d/dx (cos⁻¹ x) = -1/√(1 - x²), for x ∈ (-1, 1)
Reads as: Derivative of cosine inverse x is negative 1 over square root of 1 minus x squared
Variables: x: variable in (-1, 1)
Use when: Differentiating inverse cosine function
⚠ Forgetting the negative sign.
d/dx (tan⁻¹ x) = 1/(1 + x²), for x ∈ R
Reads as: Derivative of tan inverse x is 1 over 1 plus x squared
Variables: x: any real number
Use when: Differentiating inverse tangent function
⚠ Confusing with derivative of cot⁻¹ x, which is -1/(1 + x²).
How It Works
Checking continuity of a function at a point
  1. Step 1: Verify that the function is defined at the point c, i.e., f(c) exists.
  2. Step 2: Compute the left-hand limit, lim(x→c⁻) f(x).
  3. Step 3: Compute the right-hand limit, lim(x→c⁺) f(x).
  4. Step 4: Check if both limits exist, are finite, and equal to each other.
  5. Step 5: Check if this common limit equals f(c). If yes, the function is continuous at c; otherwise, it is discontinuous.
💡 Students often forget to check that f(c) is defined, especially in piecewise functions where the value at the point is given separately.
Differentiating a composite function using the chain rule
  1. Step 1: Identify the outer function (v) and the inner function (u) such that f(x) = v(u(x)).
  2. Step 2: Let t = u(x).
  3. Step 3: Differentiate the outer function with respect to t: dv/dt.
  4. Step 4: Differentiate the inner function with respect to x: dt/dx.
  5. Step 5: Multiply the two derivatives: df/dx = (dv/dt) * (dt/dx).
  6. Step 6: Substitute t back in terms of x to express the final answer.
💡 The most common mistake is forgetting to multiply by the derivative of the inner function (dt/dx). For example, differentiating sin(x²) as cos(x²) instead of 2x cos(x²).
Differentiating an implicit function
  1. Step 1: Differentiate both sides of the equation with respect to x, treating y as a function of x.
  2. Step 2: Apply the chain rule whenever differentiating a term involving y, i.e., d/dx (y) = dy/dx, d/dx (y²) = 2y dy/dx, etc.
  3. Step 3: Collect all terms containing dy/dx on one side of the equation.
  4. Step 4: Factor out dy/dx and solve for it.
  5. Step 5: Simplify the expression if possible.
💡 Students often forget to apply the chain rule to y terms, e.g., differentiating y² as 2y instead of 2y dy/dx.
Deriving the derivative of sin⁻¹ x
  1. Step 1: Let y = sin⁻¹ x. Then x = sin y.
  2. Step 2: Differentiate both sides w.r.t. x: 1 = cos y * (dy/dx).
  3. Step 3: Solve for dy/dx: dy/dx = 1/cos y.
  4. Step 4: Express cos y in terms of x: cos y = √(1 - sin² y) = √(1 - x²). (Choose positive root because y ∈ [-π/2, π/2] where cos y ≥ 0.)
  5. Step 5: Therefore, dy/dx = 1/√(1 - x²) for x ∈ (-1, 1).
💡 Students often forget the domain restriction x ∈ (-1, 1) and the justification for taking the positive square root.
Exam Traps — Don't Make These Mistakes
⚠ Confusing continuity with differentiability
✗ If a function is continuous at a point, it must be differentiable at that point.
→ Continuity is a necessary condition for differentiability, but not sufficient. The classic counterexample is f(x) = |x|, which is continuous at x = 0 but not differentiable there because the left and right derivatives are different.
✓ Every differentiable function is continuous, but the converse is not true. A function can be continuous at a point without being differentiable there.
⚠ Forgetting the chain rule in composite functions
✗ Derivative of sin(x²) is cos(x²).
→ The student differentiated only the outer function (sin) and forgot to multiply by the derivative of the inner function (x²), which is 2x.
✓ Derivative of sin(x²) is cos(x²) * 2x = 2x cos(x²).
⚠ Incorrect application of quotient rule order
✗ Derivative of u/v is (uv' - u'v)/v².
→ The order in the numerator is reversed. The correct formula is (u'v - uv')/v². Getting the order wrong leads to a sign error.
✓ Derivative of u/v is (u'v - uv')/v². Remember the mnemonic: 'low d-high minus high d-low over low squared'.
⚠ Ignoring domain restrictions for inverse trigonometric derivatives
✗ d/dx (sin⁻¹ x) = 1/√(1 - x²) for all x.
→ The derivative of sin⁻¹ x is defined only for x ∈ (-1, 1). At x = ±1, the denominator becomes zero and the derivative does not exist.
✓ d/dx (sin⁻¹ x) = 1/√(1 - x²) for x ∈ (-1, 1).
⚠ Misapplying the product rule to a constant times a function
✗ Derivative of 5x² is (0)(x²) + (5)(2x) = 10x, which is correct but the method is overkill.
→ While the answer is correct, students often apply the product rule unnecessarily and then make errors. The simpler approach is to use the constant multiple rule: d/dx (k f(x)) = k f'(x).
✓ For a constant times a function, use the constant multiple rule: d/dx (5x²) = 5 * d/dx (x²) = 5 * 2x = 10x.
Compare & Contrast
Relationship between the two concepts
Continuity
Differentiability
Differentiability implies continuity (every differentiable function is continuous), but continuity does not imply differentiability (e.g., f(x) = |x| is continuous at x = 0 but not differentiable there).
Way of expressing the relationship between x and y
Explicit function
Implicit function
In an explicit function, y is expressed directly in terms of x (e.g., y = x²). In an implicit function, the relationship is given without solving for y (e.g., x² + y² = 1), and differentiation is done directly on the equation.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Very Short Answer (2 marks), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total approximately 11-12 marks
Most asked: Continuity of piecewise functions at a point (finding values of constants for continuity) · Differentiability of modulus and greatest integer functions · Chain rule for composite functions · Derivatives of inverse trigonometric functions · Implicit differentiation
Don't skip: Finding values of constants (a, b, k) to make a piecewise function continuous at a point (5 marks); Differentiating composite functions using chain rule and simplifying (3-5 marks); Implicit differentiation problems involving trigonometric or inverse trigonometric functions (3-5 marks)
One-Line Revision (read aloud before the exam)
  1. A function f is continuous at x = c if lim(x→c) f(x) = f(c).
  2. Every polynomial function is continuous on R.
  3. Every rational function is continuous on its domain (where denominator ≠ 0).
  4. The greatest integer function f(x) = [x] is discontinuous at every integer.
  5. Differentiability implies continuity, but continuity does not imply differentiability.
  6. f(x) = |x| is continuous at x = 0 but not differentiable at x = 0.
  7. The chain rule: if f = v(u(x)), then df/dx = (dv/dt) * (dt/dx), where t = u(x).
  8. For implicit functions, differentiate both sides w.r.t. x, treating y as a function of x, and solve for dy/dx.
  9. d/dx (sin⁻¹ x) = 1/√(1 - x²) for x ∈ (-1, 1).
  10. d/dx (cos⁻¹ x) = -1/√(1 - x²) for x ∈ (-1, 1).
  11. d/dx (tan⁻¹ x) = 1/(1 + x²) for all real x.
  12. The product rule: (uv)' = u'v + uv'.
  13. The quotient rule: (u/v)' = (u'v - uv')/v².
  14. A function is differentiable at a point if its left-hand derivative and right-hand derivative exist and are equal.
  15. Exponential functions grow faster than polynomial functions for large x.

MCQs — 10 Questions with Answers

Q1. Which of the following functions is continuous at x = 0?

  • A. f(x) = 1 if x ≤ 0, and 2 if x > 0
  • B. f(x) = |x| ✓
  • C. f(x) = 1/x for x ≠ 0, and 0 at x = 0
  • D. f(x) = x + 3 if x ≠ 0, and 5 if x = 0

Answer: B — For f(x) = |x|, we have f(0) = 0, lim(x→0−) |x| = 0, and lim(x→0+) |x| = 0, so all three values match.

Q2. A function f(x) = x² + 2x is continuous at x = 1 because:

  • A. f(1) is defined
  • B. The limit lim(x→1) (x² + 2x) exists
  • C. f(1) = 3 and lim(x→1) (x² + 2x) = 3 ✓
  • D. The function is a polynomial

Answer: C — Continuity requires that the function value equals the limit at that point; here f(1) = 1 + 2 = 3 and the limit also equals 3.

Q3. The function f(x) = 5 (constant function) is continuous at:

  • A. Only at x = 0
  • B. Only at positive values of x
  • C. At every real number c ✓
  • D. Only at x = 5

Answer: C — For any point c, f(c) = 5 and lim(x→c) 5 = 5, satisfying the continuity condition at every real number.

Q4. For the piecewise function f(x) = {x if x ≤ 1, 3 if x > 1}, at which point is there a jump discontinuity?

  • A. x = 0
  • B. x = 1 ✓
  • C. x = 3
  • D. The function is continuous everywhere

Answer: B — At x = 1, lim(x→1−) f(x) = 1 but lim(x→1+) f(x) = 3, and these left and right limits are unequal, creating a jump.

Q5. If f(x) = (x³ - 1)/(x - 1) for x ≠ 1 and f(1) = 3, then f is:

  • A. Continuous at x = 1 because f(1) = 3
  • B. Discontinuous at x = 1 because the formula is undefined there
  • C. Continuous at x = 1 because lim(x→1) [(x³ - 1)/(x - 1)] = 3 = f(1) ✓
  • D. Discontinuous at x = 1 because lim(x→1) [(x³ - 1)/(x - 1)] = 2

Answer: C — Simplifying: (x³ - 1)/(x - 1) = x² + x + 1, so lim(x→1) (x² + x + 1) = 3 = f(1), making it continuous.

Q6. Which statement about continuity is INCORRECT?

  • A. A function continuous on [a,b] must be continuous at every interior point
  • B. If f(c) is undefined, then f is continuous at c ✓
  • C. All polynomial functions are continuous on ℝ
  • D. If lim(x→c) f(x) exists and equals f(c), then f is continuous at c

Answer: B — If f(c) is undefined, the function cannot be continuous at c because continuity requires f(c) to be defined and equal the limit.

Q7. Consider f(x) = |x - 2|. At x = 2, the function is continuous. Which of the following justifies this?

  • A. f(2) = 0 and lim(x→2−) |x - 2| = lim(x→2+) |x - 2| = 0
  • B. The absolute value function is always continuous
  • C. The left and right limits are equal to f(2) = 0
  • D. All of the above ✓

Answer: D — All three reasons are true: f(2) = 0, both one-sided limits equal 0, and |x - 2| is a standard continuous function type.

Q8. The function f(x) = (x² - 4)/(x - 2) for x ≠ 2 has a removable discontinuity at x = 2 because:

  • A. f(2) is undefined
  • B. lim(x→2) [(x² - 4)/(x - 2)] = 4 exists but f(2) is undefined ✓
  • C. The left and right limits are unequal at x = 2
  • D. The denominator becomes zero

Answer: B — The limit lim(x→2) (x + 2) = 4 exists even though f(2) is undefined, making this a removable discontinuity that could be 'fixed' by redefining f(2) = 4.

Q9. Assertion (A): The function f(x) = x³ + x² - 1 is continuous at x = 0. Reason (R): All polynomial functions are continuous everywhere on ℝ.

  • A. Both (A) and (R) are true, and (R) is the correct reason for (A) ✓
  • B. Both (A) and (R) are true, but (R) is not the correct reason for (A)
  • C. (A) is true but (R) is false
  • D. (A) is false but (R) is true

Answer: A — Both statements are true: f(0) = -1 and lim(x→0) (x³ + x² - 1) = -1, confirming continuity; and (R) correctly explains why all polynomials are continuous.

Q10. A function is defined as f(x) = {2x + 1 if x < 1, k if x = 1, x² if x > 1}. For what value of k is f continuous at x = 1?

  • A. k = 1
  • B. k = 2
  • C. k = 3 ✓
  • D. k = 4

Answer: C — For continuity at x = 1: lim(x→1−) (2x + 1) = 3 and lim(x→1+) x² = 1, but we need left limit = right limit = f(1); actually lim(x→1−) (2x + 1) = 3, so k = 3.

Flashcards

Definition of continuity of f at point c

Function f is continuous at c if lim(x→c) f(x) = f(c), meaning the left limit, right limit, and function value all exist and are equal.

When is a function called continuous on its domain?

A function is continuous if it is continuous at every point in its domain.

What does it mean graphically when a function is continuous?

The graph can be drawn without lifting the pen from the paper, showing no jumps, breaks, or holes at any point.

Example: Is f(x) = 2x + 3 continuous at x = 1?

Yes, because f(1) = 5 and lim(x→1) (2x + 3) = 5, so the limit equals the function value.

Example: Is f(x) = |x| continuous at x = 0?

Yes, because f(0) = 0, lim(x→0−) |x| = 0, and lim(x→0+) |x| = 0, so all three are equal.

What is a point of discontinuity?

A point c where the function is not continuous, meaning lim(x→c) f(x) ≠ f(c) or the limit does not exist.

Piecewise function: f(x) = 1 if x ≤ 0, and 2 if x > 0. Is it continuous at x = 0?

No, because lim(x→0−) f(x) = 1 but lim(x→0+) f(x) = 2, so left and right limits don't match.

For a function defined on [a,b], what condition ensures continuity at endpoint a?

The right-hand limit must exist and equal f(a), written as lim(x→a+) f(x) = f(a).

Is a constant function f(x) = k continuous everywhere?

Yes, because for any point c, f(c) = k and lim(x→c) k = k, satisfying the continuity condition.

Is the identity function f(x) = x continuous on all real numbers?

Yes, because for any point c, f(c) = c and lim(x→c) x = c, so continuity holds at every real number.

Important Board Questions

Define continuity of a function at a point c. Give one example of a function that is continuous at x = 0 and one that is discontinuous at x = 0. [2 marks]

State the three conditions: f(c) defined, limit exists, and limit = f(c). For continuous example use f(x) = |x| or x²; for discontinuous use a piecewise function with unequal one-sided limits at x = 0.

Examine the continuity of the function f(x) = {(x² - 9)/(x - 3) if x ≠ 3, and 6 if x = 3} at x = 3. Show all working steps and state whether it is continuous or discontinuous, and why. [5 marks]

Simplify (x² - 9)/(x - 3) to find lim(x→3) f(x); compare this limit with f(3) = 6; continuity requires all three values to match. Check: (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3 for x ≠ 3.

Prove that the function f(x) = |x| is continuous at every real number. Explain your proof with reference to the definition of continuity and handle the cases x < 0, x = 0, and x > 0 separately. [6 marks]

Rewrite |x| as a piecewise function: -x if x < 0, x if x ≥ 0. For each region and the boundary point x = 0, show lim(x→c) |x| = |c| = f(c) using polynomial continuity for x ≠ 0 and matching one-sided limits at x = 0. Conclude general continuity.

Assertion-Reason Questions for Continuity and Differentiability

Q1. Assertion (A):

The function f(x) = x³ + x² - 1 is continuous at x = 2.

Reason (R):

For any polynomial function, lim(x→c) f(x) = f(c) for all real numbers c.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. All polynomial functions are continuous everywhere because they are sums and products of continuous functions (identity and constant functions). The reason correctly explains why the assertion is true.

Q2. Assertion (A):

A function that is continuous on [a, b] must be continuous at both x = a and x = b.

Reason (R):

Continuity at the endpoints requires that lim(x→a⁻) f(x) = f(a) and lim(x→b⁺) f(x) = f(b).

Answer: C — A is true but R is false — The assertion is true: a function continuous on [a, b] is continuous at both endpoints. However, the reason is false because the left-hand limit at a and the right-hand limit at b do not exist (they are outside the domain). Instead, continuity at a means lim(x→a⁺) f(x) = f(a), and at b means lim(x→b⁻) f(x) = f(b).

Q3. Assertion (A):

The function f(x) = |x| is not continuous at x = 0.

Reason (R):

The left-hand and right-hand limits of |x| at x = 0 are equal, but the function has a sharp corner.

Answer: D — A is false but R is true — The assertion is false: |x| is continuous at x = 0 because lim(x→0⁻) |x| = 0, lim(x→0⁺) |x| = 0, and f(0) = 0, all equal. The reason is also problematic: while the function does have a sharp corner (making it non-differentiable), continuity depends on the limit equaling the function value, not on smoothness. Having a corner does not imply discontinuity.

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Case-Based Questions (CBQ) for Continuity and Differentiability

Case Study:

An engineering team in India is designing a traffic light control system that adjusts the light duration based on the density of vehicles detected by sensors. The system uses a function f(t) that represents the light duration (in seconds) based on vehicle density t (in vehicles per minute). The function is defined as: f(t) = 5t if t ≤ 100, and f(t) = 500 + 2(t - 100) if t > 100. The engineers want to ensure the system operates smoothly without sudden jumps in light duration that could confuse drivers.

Q1. Is the function f(t) continuous at t = 100? Verify using the definition of continuity.

  • A. No, because f(100) from the left equals 500 but from the right equals 500, creating a jump discontinuity
  • B. Yes, because lim(t→100⁻) f(t) = 500, lim(t→100⁺) f(t) = 500, and f(100) = 500 ✓
  • C. No, because the two expressions for f(t) are different
  • D. Yes, but only if we redefine the function at t = 100

Answer: B

Q2. What is the light duration at a vehicle density of 150 vehicles per minute?

  • A. 500 seconds
  • B. 600 seconds ✓
  • C. 750 seconds
  • D. 400 seconds

Answer: B

Q3. If continuity is essential for the traffic system to function safely, which statement is true?

  • A. The function is continuous everywhere, so the system is safe
  • B. The function is discontinuous at t = 100, so drivers may experience sudden changes
  • C. The function is continuous at all density values, making smooth transitions for drivers ✓
  • D. Continuity is not relevant to traffic light operations

Answer: C

Q4. At what vehicle density does the light duration first reach 600 seconds?

  • A. At t = 100 vehicles/min
  • B. At t = 120 vehicles/min
  • C. At t = 150 vehicles/min ✓
  • D. Never, the maximum duration is 500 seconds

Answer: C

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Previous Year Questions (PYQ) for Continuity and Differentiability

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 2 marks

Check whether function f(x) defined as f(x) = {(x³ + 2(x-3), 3 ≤ x; 6, x = 3; (6x - x³)/3, x < 3} is continuous at x = 3 or not?

CBSE 2026 · Set 65/1/1 · 5 marks

If x = cos t, y = cos mt, prove that (1 – x²) d²y/dx² – x dy/dx + m²y = 0.

CBSE 2026 · Set 65/1/2 · 2 marks

Check whether function f(x) defined as f(x) = {x/(3-x), x < 3; (x-6)/(x-3), x ≥ 3} is continuous at x = 3 or not?

CBSE 2026 · Set 65/1/2 · 5 marks

If x = 3 sin t – sin 3t, y = 3 cos t – cos 3t, find dy/dx and prove that d²y/dx² = –cosec³(2t) · cosec t/3

CBSE 2026 · Set 65/1/3 · 2 marks

Check whether function f(x) defined as f(x) = {x²(3 - 2x) for 0 ≤ x ≤ 3; 6x - x for x > 3} is continuous at x = 3 or not?

CBSE 2026 · Set 65/1/3 · 5 marks

If x = a(sin t – t cos t) and y = b(cos t + t sin t), then find dy/dx and d²y/dx².

CBSE 2026 · Set 65/2/1 · 1 mark

If e⁻ˣ + e⁻ʸ = 2, then dy/dx is

  • eˣ⁻ʸ
  • eʸ⁻ˣ
  • -eˣ⁻ʸ
  • -eʸ⁻ˣ

CBSE 2026 · Set 65/2/1 · 2 marks

If x = a sin³ t, y = b cos³ t, then find dy/dx at t = π/4.

CBSE 2026 · Set 65/2/1 · 1 mark

Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes. The equation of one such track is given as follows: f(x) = { (x⁴ - 4x²)/24, 0 ≤ x ≤ 4; x - 3, x > 4 }

Find f'(x) for 0 < x < 3.

CBSE 2026 · Set 65/2/2 · 1 mark

If e^(x + y) = 3x, then dy/dx is

  • y – 3e^(–x)
  • y – e^(–x)
  • (e^y – 1)/(e^x)
  • (e^y – 3)/(e^x)

CBSE 2026 · Set 65/2/2 · 2 marks

If x = sin t – cos t, y = sin t cos t, find dy/dx at t = π/4.

CBSE 2026 · Set 65(B) R Mathematics · 1 mark

The value of k for which f(x) = {sin 7x/5x, x ≠ 0; k, x = 0} is continuous at x = 0 is :

  • 5/7
  • 7/5
  • 0
  • 1/5

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Class 12 Maths Formula Sheet

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Class 12 Mathematics Previous Year Questions, Chapter-wise

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