📝 Differentiability implies continuity (11 MCQs)

📖 From Calculus • 3. The Derivation • 11 questions available

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What is Differentiability implies continuity?

Definition:
If a function is differentiable at a point, it must also be continuous at that point, because the existence of a finite derivative requires the function values to approach the same limit from both sides without jumps or breaks.

Example:
$f(x) = x^2$ is differentiable everywhere, so it is continuous everywhere; however, a step function is continuous nowhere differentiable at the jump.

Reason:
This theorem provides a quick test for non-differentiability, as discontinuous functions cannot have derivatives, saving time when analyzing piecewise functions and identifying problematic points in calculus problems.

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📝 All Differentiability implies continuity MCQs

Q1. Which of the following statements correctly expresses the contrapositive of the theorem: 'If a function f is differentiable at x₀, then f is continuous at x₀'?

A.If f is continuous at x₀, then f is differentiable at x₀.

B.If f is not continuous at x₀, then f is not differentiable at x₀. ✅

C.If f is differentiable at x₀, then f is not continuous at x₀.

D.If f is not differentiable at x₀, then f is continuous at x₀.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The contrapositive of 'P ⇒ Q' is '¬Q ⇒ ¬P'. Here P is 'f is differentiable at x₀' and Q is 'f is continuous at x₀'. Therefore the correct contrapositive states that if f fails to be continuous at x₀, then it cannot be differentiable there, which matches option B.

Q2. Given f(x)=|x|, which statement correctly describes its differentiability and continuity at x=0?

A.f is continuous and differentiable at 0.

B.f is continuous but not differentiable at 0. ✅

C.f is differentiable but not continuous at 0.

D.f is neither continuous nor differentiable at 0.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The absolute value function is continuous everywhere because the left‑hand and right‑hand limits equal the function value at each point. At x = 0, however, the left‑hand derivative equals –1 while the right‑hand derivative equals +1, so the derivative does not exist. Hence the function is continuous but not differentiable at 0, which is option B.

Q3. Consider the piecewise function f(x)=\begin{cases}2x+1,&x\le 1\\3x-2,&x>1\end{cases}. Which statement is true about differentiability at x=1?

A.f is differentiable at 1 because the left and right derivatives are equal.

B.f is not continuous at 1, hence not differentiable. ✅

C.f is continuous but not differentiable because the slopes differ.

D.f is differentiable and its derivative at 1 is 2.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For x ≤ 1 the formula gives f(1)=3, while the limit from the right approaches 1, so the function has a jump discontinuity at x = 1. Since differentiability implies continuity, the lack of continuity guarantees non‑differentiability. Option B correctly captures this reasoning.

Q4. If a function fails to have a limit as x→x₀, which of the following must be true regarding its differentiability at x₀?

A.It may still be differentiable if the derivative is defined via one‑sided limits.

B.It is guaranteed to be differentiable because limits are not needed.

C.It cannot be differentiable because differentiability implies the existence of the limit of the difference quotient. ✅

D.Differentiability is independent of the existence of the limit of the function.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Differentiability at a point requires the limit defining the derivative to exist, which in turn uses the function values approaching the point. If the function itself lacks a limit as x→x₀, the difference quotient cannot settle to a finite number, so the function cannot be differentiable there. Option C states this necessity.

Q5. Which of the following pairs of functions illustrates that continuity does not imply differentiability?

A.f(x)=x^2 and g(x)=\sin x

B.f(x)=\sqrt{x}\;(x\ge0) and g(x)=x^3

C.f(x)=e^x and g(x)=\tan x

D.f(x)=|x| and g(x)=x^3 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The pair that demonstrates continuity without differentiability is the absolute value function together with a smooth function. |x| is continuous everywhere but fails to have a derivative at 0 due to the corner, while x³ is both continuous and differentiable. This contrast is captured by option D.

Q6. The classic Weierstrass function W(x)=\sum_{n=0}^{\infty} a^n\cos(b^n\pi x) with 0<a<1 and b odd integer, is known to be continuous everywhere. Which conclusion follows from the differentiability–continuity theorem?

A.Since W is continuous, it must be differentiable at some points.

B.Continuity guarantees differentiability at every rational point.

C.Because W is continuous, it may still fail to be differentiable at any point. ✅

D.W cannot be continuous unless it is differentiable.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The theorem only tells us that differentiability forces continuity; it does not assert the converse. The Weierstrass function is continuous everywhere yet nowhere differentiable, which is consistent with the theorem. Option C accurately reflects that continuity does not guarantee differentiability at any point.

Q7. If a function f has a vertical tangent at x₀, what can be inferred about its differentiability and continuity at x₀?

A.It is differentiable and continuous at x₀.

B.It is continuous but not differentiable at x₀. ✅

C.It is neither continuous nor differentiable at x₀.

D.It may be differentiable in the extended sense but is definitely discontinuous.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A vertical tangent means the slope approaches infinite magnitude, so the derivative does not exist as a finite number. Nevertheless, the function values approach a single limit, preserving continuity. Therefore the function is continuous at the point but not differentiable, which is described by option B.

Q8. Using the definition of derivative, which expression correctly represents f'(x₀) for a function f that is known to be continuous at x₀?

A.f'(x₀)=\displaystyle\lim_{h\to0}\frac{f(x₀+h)-f(x₀)}{h}

B.f'(x₀)=\displaystyle\lim_{x\to x₀}\frac{f(x)-f(x₀)}{x-x₀}

C.Both A and B are equivalent definitions. ✅

D.Neither A nor B defines the derivative.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Both expressions represent the same limit when we substitute h = x – x₀, so they are equivalent formulations of the derivative. The product of limits rule allows us to rewrite the limit in either form, making option C the correct choice.

Q9. Suppose a function f is continuous on [a,b] and differentiable on (a,b) except at a single interior point c where it has a cusp. Which statement is consistent with the differentiability–continuity relationship?

A.f must be discontinuous at c.

B.f is continuous at c but not differentiable there. ✅

C.f is differentiable at c because continuity holds on the whole interval.

D.f cannot have a cusp if it is continuous on the interval.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A cusp creates a sharp point where the left‑hand and right‑hand derivatives differ, preventing differentiability. However, the function values from both sides still match, ensuring continuity at the cusp. Thus the function is continuous at c but not differentiable there, as stated in option B.

Q10. Given that a function g is not differentiable at x₀, which of the following statements can be logically deduced using the contrapositive of Theorem 2.2.3?

A.g is continuous at x₀.

B.g may be continuous or discontinuous; no conclusion can be drawn. ✅

C.g is not continuous at x₀.

D.g has a vertical tangent at x₀.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The contrapositive of 'If f is differentiable then f is continuous' is 'If f is not continuous then f is not differentiable.' From the premise that f is not differentiable we cannot infer non‑continuity; the function might still be continuous. Hence no definitive statement about continuity can be drawn, matching option B.

Q11. Let h(x)=\begin{cases}\frac{\sin x}{x}, & x\neq0 \\ 1, & x=0\end{cases}. Which statement correctly describes the differentiability of h at x=0?

A.h is differentiable at 0 because the limit defining the derivative exists.

B.h is not differentiable at 0 because the function is not continuous there.

C.h is continuous at 0 but not differentiable because the derivative limit does not exist. ✅

D.h is neither continuous nor differentiable at 0.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The function h(x) equals sin x / x for x≠0 and is defined to be 1 at x=0, making it continuous because $\lim_{x\to0}\sin x / x =1$ . To test differentiability, we examine $\lim_{h\to0}[h(x+h)-h(x)]/h$ ; this limit does not exist, so h is not differentiable at 0 despite being continuous. Option C correctly describes this.

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