Continuity and Differentiability - Practice Questions with Answers
68 free MCQs on Continuity and Differentiability with worked answers and explanations. When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.
Below are 68 practice questions on Continuity and Differentiability, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Continuity and Differentiability notes.
f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 - since these one-sided derivatives disagree, the function is not differentiable at that point.
Easy - 20 questions
Q1.
A function f is continuous at x = a if:
A f(a) is defined, without checking the limit
B lim(x->a) f(x) exists, without comparing it to f(a)
C lim(x->a) f(x) = f(a)
D f is differentiable at a, a stronger but different condition
Show answer & explanation
Answer: C. lim(x->a) f(x) = f(a)
Why: Continuity at a point requires the limit to exist and equal the function value there: lim(x->a) f(x) = f(a).
Q2.
Every differentiable function is:
A Discontinuous
B Continuous
C Periodic
D Unbounded
Show answer & explanation
Answer: B. Continuous
Why: Differentiability implies continuity at that point, though the converse is not true.
Q3.
The function f(x) = |x| at x = 0 is:
A Differentiable but not continuous
B Continuous and differentiable
C Continuous but not differentiable
D Neither continuous nor differentiable
Show answer & explanation
Answer: C. Continuous but not differentiable
Why: f(x) = |x| is continuous everywhere, but at x = 0 the left-hand derivative is -1 and the right-hand derivative is +1, so it is not differentiable there.
Q4.
d/dx (sin-1 x) =
A 1/sqrt(1-x<sup>2</sup>)
B -1/sqrt(1-x<sup>2</sup>)
C 1/(1+x<sup>2</sup>)
D 1/sqrt(x<sup>2</sup>-1)
Show answer & explanation
Answer: A. 1/sqrt(1-x<sup>2</sup>)
Why: The derivative of sin inverse x is 1/sqrt(1-x<sup>2</sup>), valid for -1 < x < 1.
Q5.
d/dx (cos-1 x) =
A 1/sqrt(1-x<sup>2</sup>)
B -1/sqrt(1-x<sup>2</sup>)
C -1/(1+x<sup>2</sup>)
D 1/(1+x<sup>2</sup>)
Show answer & explanation
Answer: B. -1/sqrt(1-x<sup>2</sup>)
Why: The derivative of cos inverse x is -1/sqrt(1-x<sup>2</sup>).
Q6.
d/dx (tan-1 x) =
A 1/(1+x<sup>2</sup>)
B -1/(1+x<sup>2</sup>)
C 1/sqrt(1-x<sup>2</sup>)
D 1/(1-x<sup>2</sup>)
Show answer & explanation
Answer: A. 1/(1+x<sup>2</sup>)
Why: The derivative of tan inverse x is 1/(1+x<sup>2</sup>) for all real x.
Q7.
A function with LHL not equal to RHL at a point has what kind of discontinuity?
A Removable
B Jump discontinuity
C No discontinuity
D Infinite discontinuity
Show answer & explanation
Answer: B. Jump discontinuity
Why: When the left-hand limit and right-hand limit exist but are unequal, the function has a jump (first kind) discontinuity.
Q8.
If y = e<sup>x</sup>, then dy/dx =
A x e<sup>x-1</sup>
B e<sup>x</sup>
C x e<sup>x</sup>
D e
Show answer & explanation
Answer: B. e<sup>x</sup>
Why: The exponential function e<sup>x</sup> is its own derivative.
Q9.
Rolle's Theorem requires f(a) and f(b) to satisfy:
A f(a) > f(b)
B f(a) = f(b)
C f(a) < f(b)
D f(a) times f(b) = 0
Show answer & explanation
Answer: B. f(a) = f(b)
Why: Rolle's Theorem requires f continuous on [a,b], differentiable on (a,b), and f(a) = f(b); then some c in (a,b) has f'(c) = 0.
Q10.
If x = at<sup>2</sup> and y = 2at (parametric form), then dy/dx is found using:
A dy/dx = dx/dt divided by dy/dt
B dy/dx = (dy/dt) divided by (dx/dt)
C dy/dx = dy/dt times dx/dt
D dy/dx = dt/dy
Show answer & explanation
Answer: B. dy/dx = (dy/dt) divided by (dx/dt)
Why: For parametric curves, dy/dx = (dy/dt) / (dx/dt), provided dx/dt is not zero.
Q11.
Which of these is an example of a removable discontinuity?
A f(x) = 1/x at x = 0, an infinite discontinuity that grows without bound
B f(x) = [x] at integer points, a jump discontinuity in the step function
C f(x) = (x<sup>2</sup>-1)/(x-1) at x = 1, undefined there but limit exists
D f(x) = x<sup>2</sup> everywhere, a polynomial that is already smooth and continuous
Show answer & explanation
Answer: C. f(x) = (x<sup>2</sup>-1)/(x-1) at x = 1, undefined there but limit exists
Why: (x<sup>2</sup>-1)/(x-1) simplifies to x+1 for x not equal to 1, so the limit as x approaches 1 exists (equals 2) even though f(1) is undefined; redefining f(1)=2 removes the gap.
Q12.
If f and g are continuous at x = a, then f/g is continuous at a provided:
A f(a) = 0
B g(a) = 0
C g(a) is not equal to 0
D f(a) is not equal to 0
Show answer & explanation
Answer: C. g(a) is not equal to 0
Why: The quotient of two continuous functions is continuous wherever the denominator is nonzero.
Q13.
A function f is continuous at x = a if the limit of f(x) as x → a equals:
A f(a)
B 0
C infinity
D 1
Show answer & explanation
Answer: A. f(a)
Why: Continuity requires the limit at a to exist and equal the function value f(a).
Q14.
A function whose graph can be drawn without lifting the pen is:
A continuous
B discontinuous
C undefined
D constant only
Show answer & explanation
Answer: A. continuous
Why: A continuous function has an unbroken graph.
Q15.
If a function is differentiable at a point, then at that point it is also:
A continuous
B discontinuous
C undefined
D zero
Show answer & explanation
Answer: A. continuous
Why: Differentiability implies continuity, though not the reverse.
Q16.
The derivative measures the ___ of a function at a point:
A instantaneous rate of change
B the total enclosed area
C the average value only
D the solid volume value
Show answer & explanation
Answer: A. instantaneous rate of change
Why: The derivative is the instantaneous rate of change (the slope of the tangent).
Q17.
The function f(x) = |x| fails to be differentiable at x =:
A 0
B 1
C −1
D 2
Show answer & explanation
Answer: A. 0
Why: At x = 0 the graph of |x| has a sharp corner, so no unique tangent exists.
Q18.
A jump or break in a graph indicates a point of:
A discontinuity
B continuity
C differentiability
D symmetry
Show answer & explanation
Answer: A. discontinuity
Why: A jump means the function is discontinuous there.
Q19.
The chain rule is used to differentiate a ___ function:
A composite
B constant
C linear-only
D polynomial-only
Show answer & explanation
Answer: A. composite
Why: The chain rule handles composite functions of the form f(g(x)).
Q20.
The derivative of any constant is:
A 0
B 1
C the constant itself
D undefined
Show answer & explanation
Answer: A. 0
Why: A constant does not change, so its rate of change is zero.
Medium - 20 questions
Q21.
If f(x) = x sin(1/x) for x not 0 and f(0) = 0, is f continuous at x = 0?
A No, the limit does not exist since sin(1/x) oscillates without settling
B Yes, since x sin(1/x) is squeezed between -|x| and |x| which go to 0
C No, because sin(1/x) oscillates infinitely often near x = 0
D Cannot be determined without evaluating the one-sided limits separately
Show answer & explanation
Answer: B. Yes, since x sin(1/x) is squeezed between -|x| and |x| which go to 0
Why: Since |x sin(1/x)| <= |x|, by the squeeze theorem the limit as x approaches 0 is 0, which equals f(0), so f is continuous at 0.
Q22.
Differentiate y = sin(x<sup>2</sup> + 1) using the chain rule.
A cos(x<sup>2</sup>+1)
B 2x cos(x<sup>2</sup>+1)
C 2x sin(x<sup>2</sup>+1)
D cos(2x)
Show answer & explanation
Answer: B. 2x cos(x<sup>2</sup>+1)
Why: Let u = x<sup>2</sup>+1. dy/dx = cos(u) times du/dx = cos(x<sup>2</sup>+1) times 2x = 2x cos(x<sup>2</sup>+1).
Q23.
If x<sup>2</sup> + y<sup>2</sup> = 25, find dy/dx using implicit differentiation.
A dy/dx = x/y
B dy/dx = -x/y
C dy/dx = y/x
D dy/dx = -y/x
Show answer & explanation
Answer: B. dy/dx = -x/y
Why: Differentiating both sides: 2x + 2y(dy/dx) = 0, so dy/dx = -x/y.
Q24.
Differentiate y = x<sup>x</sup> using logarithmic differentiation.
A dy/dx = x<sup>x</sup>
B dy/dx = x<sup>x</sup> (1 + ln x)
C dy/dx = x times x<sup>x-1</sup>
D dy/dx = x<sup>x</sup> ln x
Show answer & explanation
Answer: B. dy/dx = x<sup>x</sup> (1 + ln x)
Why: Take ln: ln y = x ln x. Differentiate: (1/y)(dy/dx) = ln x + 1. So dy/dx = y(1 + ln x) = x<sup>x</sup>(1 + ln x).
Q25.
If y = tan-1(2x/(1-x<sup>2</sup>)), which substitution simplifies the derivative using a standard identity?
A x = sin(theta), reducing the expression to a sine double-angle form
B x = tan(theta), since 2x/(1-x<sup>2</sup>) is tan(2theta) when x = tan(theta)
C x = cos(theta), reducing the expression to a cosine double-angle form
D x = sec(theta), used for expressions involving square roots of x²-1
Show answer & explanation
Answer: B. x = tan(theta), since 2x/(1-x<sup>2</sup>) is tan(2theta) when x = tan(theta)
Why: Using x = tan(theta), 2x/(1-x<sup>2</sup>) = tan(2theta), so y = 2theta = 2tan-1(x), giving dy/dx = 2/(1+x<sup>2</sup>) directly.
Q26.
If x = a cos(t) and y = a sin(t), find dy/dx.
A tan(t)
B -tan(t)
C -cot(t)
D cot(t)
Show answer & explanation
Answer: C. -cot(t)
Why: dx/dt = -a sin(t), dy/dt = a cos(t). dy/dx = (a cos t)/(-a sin t) = -cot(t).
Q27.
Differentiate y = e<sup>3x</sup> cos(2x) using the product rule.
For f(x) = |x - 2|, at which point is f not differentiable?
A x = 0
B x = 1
C x = 2
D f is differentiable everywhere
Show answer & explanation
Answer: C. x = 2
Why: Similar to |x| shifted, |x-2| has a sharp corner at x = 2 where the left and right derivatives (-1 and +1) disagree.
Q29.
By Rolle's Theorem applied to f(x) = x<sup>2</sup> - 4x + 3 on [1,3], at which point is f'(c) = 0?
A c = 1
B c = 2
C c = 3
D c = 1.5
Show answer & explanation
Answer: B. c = 2
Why: f(1) = 0 = f(3), satisfying the hypothesis. f'(x) = 2x - 4 = 0 gives x = 2, which lies in (1,3).
Q30.
Differentiate y = sin-1(x) + cos-1(x).
A 0
B 1
C 2/sqrt(1-x<sup>2</sup>)
D pi/2
Show answer & explanation
Answer: A. 0
Why: Since sin-1(x) + cos-1(x) = pi/2 (a constant) for all x in [-1,1], its derivative is 0.
Q31.
Find dy/dx if y = (sin x)<sup>x.</sup>
A y[x cot x + ln(sin x)]
B y[x cot x]
C y[ln(sin x)]
D x(sin x)<sup>x-1</sup> cos x
Show answer & explanation
Answer: A. y[x cot x + ln(sin x)]
Why: Take ln: ln y = x ln(sin x). Differentiating: (1/y)(dy/dx) = ln(sin x) + x cot x. So dy/dx = y[x cot x + ln(sin x)].
Q32.
Differentiate y = ln(x<sup>2</sup> + 1) with respect to x.
A 2x/(x<sup>2</sup>+1)
B 1/(x<sup>2</sup>+1)
C 2/(x<sup>2</sup>+1)
D x/(x<sup>2</sup>+1)
Show answer & explanation
Answer: A. 2x/(x<sup>2</sup>+1)
Why: By chain rule, dy/dx = 1/(x<sup>2</sup>+1) times 2x = 2x/(x<sup>2</sup>+1).
Q33.
Differentiate y = tan-1(x<sup>2</sup>).
A 2x/(1+x<sup>4</sup>)
B 1/(1+x<sup>4</sup>)
C 2x/(1+x<sup>2</sup>)
D x<sup>2</sup>/(1+x<sup>4</sup>)
Show answer & explanation
Answer: A. 2x/(1+x<sup>4</sup>)
Why: By chain rule with u = x<sup>2</sup>: dy/dx = 1/(1+u<sup>2</sup>) times du/dx = 1/(1+x<sup>4</sup>) times 2x = 2x/(1+x<sup>4</sup>).
Q34.
By the chain rule, the derivative of y = f(g(x)) is:
A f′(g(x))·g′(x)
B the product f′(x)·g′(x)
C the value f(g′(x))
D the value g′(f(x))
Show answer & explanation
Answer: A. f′(g(x))·g′(x)
Why: Differentiate the outer function then multiply by the derivative of the inner function.
Q35.
The derivative of eˣ with respect to x is:
A eˣ
B x·eˣ⁻¹
C just 1
D the number e
Show answer & explanation
Answer: A. eˣ
Why: The exponential function eˣ is its own derivative.
Q36.
The derivative of ln x with respect to x is:
A 1/x
B x
C ln x
D eˣ
Show answer & explanation
Answer: A. 1/x
Why: d(ln x)/dx = 1/x.
Q37.
A function that is continuous but not differentiable has a:
A sharp corner
B smooth curve
C flat line
D single point
Show answer & explanation
Answer: A. sharp corner
Why: A corner (as in |x| at 0) makes a function continuous yet not differentiable there.
Q38.
The derivative of cos x with respect to x is:
A −sin x
B sin x
C cos x
D −cos x
Show answer & explanation
Answer: A. −sin x
Why: d(cos x)/dx = −sin x.
Q39.
Logarithmic differentiation is used when the variable appears in both the base and the:
A exponent
B denominator
C coefficient
D constant term
Show answer & explanation
Answer: A. exponent
Why: For expressions like xˣ, taking logarithms first makes differentiation possible.
Q40.
The derivative of a product u·v is given by:
A u′v + uv′
B u′v′
C just uv
D u′v − uv′
Show answer & explanation
Answer: A. u′v + uv′
Why: The product rule gives (uv)′ = u′v + uv′.
Hard - 28 questions
Q41.
For what value of k is f(x) = {kx+1 if x<=5, 3x-5 if x>5} continuous at x = 5?
A k = 1
B k = 9/5
C k = 5
D k = 3
Show answer & explanation
Answer: B. k = 9/5
Why: Continuity requires left limit = right limit = f(5). Left: lim(x→5⁻) = 5k+1. Right: lim(x→5⁺) = 3(5)−5 = 10. Setting 5k+1 = 10 gives 5k = 9, so k = 9/5.
Q42.
If y = tan-1[(sqrt(1+x<sup>2</sup>) - 1)/x], find dy/dx.
A 1/(2(1+x<sup>2</sup>))
B 1/(1+x<sup>2</sup>)
C 2/(1+x<sup>2</sup>)
D -1/(2(1+x<sup>2</sup>))
Show answer & explanation
Answer: A. 1/(2(1+x<sup>2</sup>))
Why: Substituting x = tan(theta), the expression simplifies to tan-1[tan(theta/2)] = theta/2 = (1/2)tan-1(x), so dy/dx = (1/2) times 1/(1+x<sup>2</sup>) = 1/(2(1+x<sup>2</sup>)).
Q43.
If y = x<sup>sin x</sup>, find dy/dx.
A x<sup>sin x</sup> [cos x ln x + (sin x)/x]
B x<sup>sin x</sup> cos x ln x, missing the second additive term
C x<sup>sin x</sup> (sin x)/x, missing the logarithmic term
D sin x times x<sup>sin x - 1</sup>, a plain power-rule shortcut
Show answer & explanation
Answer: A. x<sup>sin x</sup> [cos x ln x + (sin x)/x]
Why: Take ln: ln y = sin x times ln x. Differentiate: (1/y)(dy/dx) = cos x ln x + sin x times (1/x). So dy/dx = y[cos x ln x + (sin x)/x] = x<sup>sin x</sup>[cos x ln x + (sin x)/x].
Q44.
If x = a(theta - sin theta) and y = a(1 - cos theta), find dy/dx at theta = pi/2.
A 1
B 0
C -1
D 2
Show answer & explanation
Answer: A. 1
Why: dx/d(theta) = a(1-cos theta), dy/d(theta) = a sin(theta). dy/dx = sin(theta)/(1-cos theta). At theta=pi/2: sin=1, cos=0, so dy/dx = 1/(1-0) = 1.
Q45.
For f(x) = x|x|, which statement is correct about differentiability at x = 0?
A f is not continuous at x=0
B f is continuous but not differentiable at x=0
C f is differentiable at x=0 with f'(0)=0
D f is differentiable at x=0 with f'(0)=1
Show answer & explanation
Answer: C. f is differentiable at x=0 with f'(0)=0
Why: f(x) = x<sup>2</sup> for x>=0 and f(x) = -x<sup>2</sup> for x<0. Both one-sided derivatives at 0 equal 0, so f is differentiable at x=0 with f'(0)=0.
Q46.
If y = sin-1(2x sqrt(1-x<sup>2</sup>)) for -1/sqrt(2) <= x <= 1/sqrt(2), find dy/dx.
A 2/sqrt(1-x<sup>2</sup>)
B -2/sqrt(1-x<sup>2</sup>)
C 1/sqrt(1-x<sup>2</sup>)
D 2/(1-x<sup>2</sup>)
Show answer & explanation
Answer: A. 2/sqrt(1-x<sup>2</sup>)
Why: Substituting x = sin(theta), 2x sqrt(1-x<sup>2</sup>) = sin(2theta), so y = sin-1[sin(2theta)] = 2theta = 2sin-1(x) in this range, giving dy/dx = 2/sqrt(1-x<sup>2</sup>).
Q47.
Find d<sup>2y</sup>/dx<sup>2</sup> if y = e<sup>x</sup> sin x.
A 2e<sup>x</sup> cos x
B 2e<sup>x</sup>(cos x - sin x)
C 2e<sup>x</sup> sin x
D e<sup>x</sup>(cos x - sin x)
Show answer & explanation
Answer: A. 2e<sup>x</sup> cos x
Why: Product rule: dy/dx = eˣ sin x + eˣ cos x = eˣ(sin x + cos x). Differentiate again: d²y/dx² = eˣ(sin x + cos x) + eˣ(cos x − sin x) = eˣ(2cos x) = 2eˣ cos x.
Q48.
If f(x) = |x|<sup>3</sup>, is f twice differentiable at x = 0?
A No, f is not even continuous at the origin because the cube of |x| is mistakenly assumed to blow up there
B No, f is not differentiable at 0 because of the sharp absolute-value kink visible in the graph near zero
C Yes, f is differentiable but f'' may not exist at 0 in the naive cubic sense, yet checking shows f''(0)=0 exists
D f is not defined at x=0 since the cube of an absolute value is wrongly treated as an undefined operation
Show answer & explanation
Answer: C. Yes, f is differentiable but f'' may not exist at 0 in the naive cubic sense, yet checking shows f''(0)=0 exists
Why: f(x) = x<sup>3</sup> for x>=0 and -x<sup>3</sup> for x<0, so f'(x) = 3x<sup>2</sup> for x>=0 and -3x<sup>2</sup> for x<0; f'(0)=0 from both sides, and f''(x) = 6|x|, giving f''(0) = 0, so f is twice differentiable at 0.
Q49.
Verify Rolle's theorem applicability for f(x) = sin x on [0, pi]. What is f'(c) = 0 satisfied at?
A c = 0
B c = pi/2
C c = pi/4
D c = pi
Show answer & explanation
Answer: B. c = pi/2
Why: Rolle's theorem conditions: sin x is continuous on [0,π], differentiable on (0,π), and f(0) = sin 0 = 0 = sin π = f(π). So theorem applies. f'(x) = cos x = 0 at c = π/2 ∈ (0,π).
Q50.
If y = log(x + sqrt(x<sup>2</sup>+1)), find dy/dx.
A 1/sqrt(x<sup>2</sup>+1)
B 1/(x+sqrt(x<sup>2</sup>+1))
C x/sqrt(x<sup>2</sup>+1)
D sqrt(x<sup>2</sup>+1)/x
Show answer & explanation
Answer: A. 1/sqrt(x<sup>2</sup>+1)
Why: Let u = x + sqrt(x<sup>2</sup>+1). du/dx = 1 + x/sqrt(x<sup>2</sup>+1) = (sqrt(x<sup>2</sup>+1)+x)/sqrt(x<sup>2</sup>+1) = u/sqrt(x<sup>2</sup>+1). dy/dx = (1/u)(du/dx) = 1/sqrt(x<sup>2</sup>+1).
Q51.
If x<sup>y</sup> = y<sup>x</sup>, find dy/dx using implicit and logarithmic differentiation.
A dy/dx = y(y - x ln y)/(x(x - y ln x))
B dy/dx = y/x, obtained by treating the exponents as if they cancelled directly
C dy/dx = x/y, obtained by inverting the y/x ratio without full log differentiation
D dy/dx = ln(y/x), mistakenly equating the derivative with the log of the ratio
Show answer & explanation
Answer: A. dy/dx = y(y - x ln y)/(x(x - y ln x))
Why: Taking ln: y ln x = x ln y. Differentiating implicitly: (dy/dx)ln x + y/x = ln y + x(1/y)(dy/dx). Solving for dy/dx gives dy/dx = y(y - x ln y) / (x(x - y ln x)).
Q52.
For y = xˣ, the derivative dy/dx equals:
A xˣ(1 + ln x)
B simply x·xˣ⁻¹
C simply just xˣ
D simply ln x
Show answer & explanation
Answer: A. xˣ(1 + ln x)
Why: Using logarithmic differentiation, dy/dx = xˣ(1 + ln x).
Q53.
The function f(x) = |x| is continuous at x = 0 but not:
A differentiable
B well defined
C real-valued
D clearly bounded
Show answer & explanation
Answer: A. differentiable
Why: It is continuous everywhere but has no derivative at the corner x = 0.
Q54.
By Rolle’s theorem, if f(a) = f(b) then there is a point c where f′(c) equals:
A 0
B 1
C f(a)
D b − a
Show answer & explanation
Answer: A. 0
Why: Rolle’s theorem guarantees a stationary point f′(c) = 0 between a and b.
Q55.
The Mean Value Theorem states that f′(c) = [f(b) − f(a)] divided by:
A (b − a)
B (a − b)
C (b + a)
D ab
Show answer & explanation
Answer: A. (b − a)
Why: The MVT equates the derivative at some c to the average rate of change over [a, b].
Q56.
If y = sin(x²), then dy/dx equals:
A 2x·cos(x²)
B cos(x²)
C 2x·cos x
D cos(2x)
Show answer & explanation
Answer: A. 2x·cos(x²)
Why: By the chain rule, dy/dx = cos(x²)·(2x).
Q57.
The derivative of tan x with respect to x is:
A sec²x
B −cosec²x
C sec x tan x
D cot x
Show answer & explanation
Answer: A. sec²x
Why: d(tan x)/dx = sec²x.
Q58.
A function differentiable at every point is necessarily: