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📐 Mathematics  ·  Class 12  ·  JEE

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.

Take the timed Continuity and Differentiability chapterwise test →

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|: Continuous but NOT Differentiable at 0x=0 (sharp "kink")LHD = -1RHD = +1LHD ≠ RHD at the kink → not differentiable, even though the curve has no gap (continuous)

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.

  • A e<sup>3x</sup>(3cos(2x) - 2sin(2x))
  • B e<sup>3x</sup>(3cos(2x) + 2sin(2x))
  • C 3e<sup>3x</sup> cos(2x)
  • D e<sup>3x</sup>(2cos(2x) - 3sin(2x))
Show answer & explanation

Answer: A. e<sup>3x</sup>(3cos(2x) - 2sin(2x))

Why: Product rule: d/dx[e<sup>3x</sup>]cos(2x) + e<sup>3x</sup> d/dx[cos(2x)] = 3e<sup>3x</sup>cos(2x) + e<sup>3x</sup>(-2sin(2x)) = e<sup>3x</sup>(3cos(2x) - 2sin(2x)).

Q28.

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
  • 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
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:

  • A continuous everywhere
  • B plainly discontinuous
  • C entirely constant
  • D strictly periodic
Show answer & explanation

Answer: A. continuous everywhere

Why: Differentiability everywhere forces continuity everywhere.

Q59.

The derivative of the quotient u/v is:

  • A (u′v − uv′)/v²
  • B (u′v + uv′)/v²
  • C u′/v′
  • D just uv′
Show answer & explanation

Answer: A. (u′v − uv′)/v²

Why: The quotient rule gives (u/v)′ = (u′v − uv′)/v².

Q60.

For y = e<sup>2x</sup>, the derivative dy/dx equals:

  • A 2e<sup>2x</sup>
  • B e<sup>2x</sup>
  • C 2x·e<sup>2x</sup>
  • D e<sup>2x</sup>/2
Show answer & explanation

Answer: A. 2e<sup>2x</sup>

Why: By the chain rule, dy/dx = e<sup>2x</sup>·2 = 2e<sup>2x</sup>.

Q61.

The function f(x) = |x| is continuous everywhere but not differentiable at:

  • A x = 1
  • B x = 0
  • C x = −1
  • D nowhere
Show answer & explanation

Answer: B. x = 0

Why: There is a corner at x = 0 where the left and right derivatives differ.

Q62.

The value of k that makes f(x) = (sinx)/x for x ≠ 0 and f(0) = k continuous at 0 is:

  • A 0
  • B 1
  • C π
  • D undefined
Show answer & explanation

Answer: B. 1

Why: Since (sinx)/x → 1 as x → 0, we need k = 1.

Q63.

If f(x) = x²·sin(1/x) for x ≠ 0 and f(0) = 0, then f'(0) equals:

  • A 0
  • B 1
  • C does not exist
  • D
Show answer & explanation

Answer: A. 0

Why: f'(0) = lim (x sin(1/x)) as x → 0, which is 0 by the squeeze theorem.

Q64.

The number of points of discontinuity of the greatest integer function [x] on the open interval (0, 3) is:

  • A 1
  • B 2
  • C 3
  • D 0
Show answer & explanation

Answer: B. 2

Why: It jumps at the integers 1 and 2 inside the interval, giving 2 points.

Q65.

If y = x<sup>sinx</sup>, then dy/dx equals:

  • A x<sup>sinx</sup>(cosx·lnx + sinx/x)
  • B x<sup>sinx</sup>·cosx
  • C sinx·x<sup>sinx − 1</sup>
  • D x<sup>sinx</sup>·lnx
Show answer & explanation

Answer: A. x<sup>sinx</sup>(cosx·lnx + sinx/x)

Why: Logarithmic differentiation gives dy/dx = x<sup>sinx</sup>(cosx·lnx + sinx/x).

Q66.

The function f(x) = e<sup>−|x|</sup> is continuous everywhere and non-differentiable at:

  • A nowhere
  • B x = 0
  • C x = 1
  • D everywhere
Show answer & explanation

Answer: B. x = 0

Why: The corner in |x| at 0 makes f non-differentiable only at x = 0.

Q67.

If f is differentiable with f(x + y) = f(x)f(y), f(0) = 1 and f'(0) = 2, then f'(x) equals:

  • A f(x)
  • B 2f(x)
  • C 2x
  • D 2
Show answer & explanation

Answer: B. 2f(x)

Why: Such an f is e<sup>2x</sup>, and f'(x) = 2e<sup>2x</sup> = 2f(x).

Q68.

By Rolle's theorem, the value c in (−2, 2) for f(x) = x² − 4 is:

  • A −1
  • B 0
  • C 1
  • D 2
Show answer & explanation

Answer: B. 0

Why: f'(x) = 2x = 0 gives c = 0.