Which of these describes f ( x ) = ∣ x ∣ f(x) = \lvert x \rvert f ( x ) = ∣ x ∣ at x = 0 x = 0 x = 0 ?
Options for question 83 A Option A Continuous and differentiable Correct answer Your answerB Option B Neither continuous nor differentiable Correct answer Your answerC Option C Differentiable but not continuous Correct answer Your answerD Option D Continuous but not differentiable Correct answer Your answer
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Show the explanation Hide the explanation Answer D · Continuous but not differentiable
∣ x ∣ → 0 = f ( 0 ) \lvert x \rvert \to 0 = f(0) ∣ x ∣ → 0 = f ( 0 ) from both sides: continuous.
The left derivative is − 1 -1 − 1 and the right one 1: not differentiable.
Worth noting
A corner joins the pieces without a break but with two different slopes.
Read the two statements and choose the correct option.
I. Every function differentiable at x = a x = a x = a is continuous at x = a x = a x = a .
II. Every function continuous at x = a x = a x = a is differentiable at x = a x = a x = a .
Options for question 84 A Option A Only I Correct answer Your answerB Option B Only II Correct answer Your answerC Option C Both I and II Correct answer Your answerD Option D Neither I nor II Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Only I
I is true: differentiability forces the limit to equal the value.
II is false: ∣ x ∣ \lvert x \rvert ∣ x ∣ is continuous at 0 but not differentiable there.
Worth noting
The implication runs one way only.
Which of these describes f ( x ) = [ x ] f(x) = [x] f ( x ) = [ x ] at x = 1 x = 1 x = 1 ?
Options for question 85 A Option A Continuous but not differentiable Correct answer Your answerB Option B Continuous and differentiable Correct answer Your answerC Option C Differentiable but not continuous Correct answer Your answerD Option D Neither continuous nor differentiable Correct answer Your answer
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Show the explanation Hide the explanation Answer D · Neither continuous nor differentiable
[ x ] [x] [ x ] jumps from 0 to 1 at x = 1 x = 1 x = 1 : not continuous.
A function that is not continuous cannot be differentiable.
Worth noting
Differentiable but not continuous is impossible: differentiability forces continuity.
Let f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ .
What is f ( 2 ) f(2) f ( 2 ) for this sum of two moduli?
Options for question 86 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 4 4 4 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 2 2 2
∣ 2 − 1 ∣ + ∣ 2 − 3 ∣ = 1 + 1 \lvert 2 - 1 \rvert + \lvert 2 - 3 \rvert = 1 + 1 ∣ 2 − 1 ∣ + ∣ 2 − 3 ∣ = 1 + 1 .
f ( 2 ) = 2 f(2) = 2 f ( 2 ) = 2 .
Worth noting
0 drops the modulus on 2 − 3 = − 1 2 - 3 = -1 2 − 3 = − 1 .
Let f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ .
At how many points is this sum of two moduli not continuous?
Options for question 87 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer A · 0 0 0
Each modulus is continuous everywhere, and so is their sum.
0 points.
Worth noting
2 counts the corners, which spoil differentiability, not continuity.
At which value of x x x is f ( x ) = ∣ 2 x − 5 ∣ f(x) = \lvert 2x - 5 \rvert f ( x ) = ∣ 2 x − 5 ∣ not differentiable?
Options for question 88 A Option A − 5 2 - \frac{5}{2} − 2 5 Correct answer Your answerB Option B 2 2 2 Correct answer Your answerC Option C 5 2 \frac{5}{2} 2 5 Correct answer Your answerD Option D 5 5 5 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 5 2 \frac{5}{2} 2 5
A modulus has its corner where its inside is 0.
2 x − 5 = 0 2x - 5 = 0 2 x − 5 = 0 at x = 5 2 x = \frac{5}{2} x = 2 5 ; there the slope changes from − 2 -2 − 2 to 2.
Worth noting
5 reads off the constant in 2 x − 5 2x - 5 2 x − 5 without dividing by the 2 in front of x x x .
Let f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ f(x) = x\lvert x \rvert + \lvert x - 1 \rvert f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ .
What is f ( − 2 ) f(-2) f ( − 2 ) for this function?
Options for question 89 A Option A − 7 -7 − 7 Correct answer Your answerB Option B − 1 -1 − 1 Correct answer Your answerC Option C 1 1 1 Correct answer Your answerD Option D 7 7 7 Correct answer Your answer
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Show the explanation Hide the explanation Answer B · − 1 -1 − 1
x ∣ x ∣ = ( − 2 ) ( 2 ) = − 4 x\lvert x \rvert = (-2)(2) = -4 x ∣ x ∣ = ( − 2 ) ( 2 ) = − 4 and ∣ − 2 − 1 ∣ = 3 \lvert -2 - 1 \rvert = 3 ∣ − 2 − 1 ∣ = 3 .
f ( − 2 ) = − 4 + 3 = − 1 f(-2) = -4 + 3 = -1 f ( − 2 ) = − 4 + 3 = − 1 .
Worth noting
7 takes x ∣ x ∣ x\lvert x \rvert x ∣ x ∣ as x 2 = 4 x^{2} = 4 x 2 = 4 ; for negative x x x it is − x 2 -x^{2} − x 2 .
Which of these describes f ( x ) = ∣ x + 2 ∣ f(x) = \lvert x + 2 \rvert f ( x ) = ∣ x + 2 ∣ at x = 0 x = 0 x = 0 ?
Options for question 90 A Option A Continuous but not differentiable Correct answer Your answerB Option B Neither continuous nor differentiable Correct answer Your answerC Option C Differentiable but not continuous Correct answer Your answerD Option D Continuous and differentiable Correct answer Your answer
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Show the explanation Hide the explanation Answer D · Continuous and differentiable
Near 0, x + 2 > 0 x + 2 > 0 x + 2 > 0 , so f ( x ) = x + 2 f(x) = x + 2 f ( x ) = x + 2 .
A linear function there: continuous and differentiable, with f ′ ( 0 ) = 1 f'(0) = 1 f ′ ( 0 ) = 1 .
Worth noting
The corner of ∣ x + 2 ∣ \lvert x + 2 \rvert ∣ x + 2 ∣ is at x = − 2 x = -2 x = − 2 , where the inside is 0, not at the origin.
At how many points is f ( x ) = ∣ x ∣ + ∣ x − 1 ∣ f(x) = \lvert x \rvert + \lvert x - 1 \rvert f ( x ) = ∣ x ∣ + ∣ x − 1 ∣ not differentiable?
Options for question 91 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 2 2 2
Each modulus has a corner where its inside is 0: at x = 0 x = 0 x = 0 and x = 1 x = 1 x = 1 .
The two corners are at different points and do not cancel: 2 points.
Worth noting
1 merges the corners; they are at different places.
Assertion (A): f ( x ) = ∣ x ∣ f(x) = \lvert x \rvert f ( x ) = ∣ x ∣ is continuous at x = 0 x = 0 x = 0 .
Reason (R): f ( x ) = ∣ x ∣ f(x) = \lvert x \rvert f ( x ) = ∣ x ∣ is differentiable at x = 0 x = 0 x = 0 .
Options for question 92 A Option A Both A and R are true, and R is the correct explanation of A. Correct answer Your answerB Option B Both A and R are true, but R is not the correct explanation of A. Correct answer Your answerC Option C A is true, but R is false. Correct answer Your answerD Option D A is false, but R is true. Correct answer Your answer
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Show the explanation Hide the explanation Answer C · A is true, but R is false.
A is true: both one-sided limits are 0, equal to f ( 0 ) f(0) f ( 0 ) .
R is false: the one-sided derivatives are − 1 -1 − 1 and 1.
Which of these describes f ( x ) = x ∣ x ∣ f(x) = x\lvert x \rvert f ( x ) = x ∣ x ∣ at x = 0 x = 0 x = 0 ?
Options for question 93 A Option A Continuous and differentiable Correct answer Your answerB Option B Continuous but not differentiable Correct answer Your answerC Option C Neither continuous nor differentiable Correct answer Your answerD Option D Differentiable but not continuous Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Continuous and differentiable
f ( x ) = − x 2 f(x) = -x^{2} f ( x ) = − x 2 for x < 0 x < 0 x < 0 and x 2 x^{2} x 2 for x ≥ 0 x \ge 0 x ≥ 0 ; both pieces give 0 at 0.
Both one-sided derivatives are 0: differentiable.
Worth noting
Multiplying ∣ x ∣ \lvert x \rvert ∣ x ∣ by x x x flattens the corner: both slopes become 0.
At how many points is f ( x ) = ∣ x 2 − 1 ∣ f(x) = \lvert x^{2} - 1 \rvert f ( x ) = ∣ x 2 − 1 ∣ not differentiable?
Options for question 94 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 2 2 2
x 2 − 1 = 0 x^{2} - 1 = 0 x 2 − 1 = 0 at x = ± 1 x = \pm 1 x = ± 1 , where it changes sign.
Each gives a corner: 2 points.
Worth noting
1 counts only x = 1 x = 1 x = 1 ; x = − 1 x = -1 x = − 1 is a root too.
Which of these describes f ( x ) = { x 2 , x ≤ 1 2 x − 1 , x > 1 f(x) = \begin{cases} x^{2}, & x \le 1 \\ 2x - 1, & x > 1 \end{cases} f ( x ) = { x 2 , 2 x − 1 , x ≤ 1 x > 1 at x = 1 x = 1 x = 1 ?
Options for question 95 A Option A Continuous but not differentiable Correct answer Your answerB Option B Neither continuous nor differentiable Correct answer Your answerC Option C Differentiable but not continuous Correct answer Your answerD Option D Continuous and differentiable Correct answer Your answer
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Show the explanation Hide the explanation Answer D · Continuous and differentiable
Both pieces give 1 at x = 1 x = 1 x = 1 : continuous.
The slopes are 2 x = 2 2x = 2 2 x = 2 and 2: equal, so differentiable.
Worth noting
A change of formula is smooth when both the values and the slopes match.
Which of these describes f ( x ) = { x 2 , x ≤ 1 x , x > 1 f(x) = \begin{cases} x^{2}, & x \le 1 \\ x, & x > 1 \end{cases} f ( x ) = { x 2 , x , x ≤ 1 x > 1 at x = 1 x = 1 x = 1 ?
Options for question 96 A Option A Continuous but not differentiable Correct answer Your answerB Option B Continuous and differentiable Correct answer Your answerC Option C Neither continuous nor differentiable Correct answer Your answerD Option D Differentiable but not continuous Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Continuous but not differentiable
Both pieces give 1 at x = 1 x = 1 x = 1 : continuous.
The slopes are 2 on the left and 1 on the right: not differentiable.
Worth noting
Matching values is not enough; the slopes must match too.
Which of these describes f ( x ) = ∣ x − 3 ∣ f(x) = \lvert x - 3 \rvert f ( x ) = ∣ x − 3 ∣ at x = 3 x = 3 x = 3 ?
Options for question 97 A Option A Continuous but not differentiable Correct answer Your answerB Option B Continuous and differentiable Correct answer Your answerC Option C Neither continuous nor differentiable Correct answer Your answerD Option D Differentiable but not continuous Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Continuous but not differentiable
∣ x − 3 ∣ → 0 = f ( 3 ) \lvert x - 3 \rvert \to 0 = f(3) ∣ x − 3 ∣ → 0 = f ( 3 ) from both sides: continuous.
Slopes − 1 -1 − 1 and 1: a corner, not differentiable.
Worth noting
∣ x − 3 ∣ \lvert x - 3 \rvert ∣ x − 3 ∣ is ∣ x ∣ \lvert x \rvert ∣ x ∣ moved 3 to the right, corner and all.
What are the left-hand and right-hand derivatives of f ( x ) = ∣ x − 1 ∣ + x f(x) = \lvert x - 1 \rvert + x f ( x ) = ∣ x − 1 ∣ + x at x = 1 x = 1 x = 1 ?
Options for question 98 A Option A − 1 -1 − 1 and 1 1 1 Correct answer Your answerB Option B 1 1 1 and 1 1 1 Correct answer Your answerC Option C 0 0 0 and 2 2 2 Correct answer Your answerD Option D 0 0 0 and 1 1 1 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 0 0 0 and 2 2 2
Left of 1: f = 1 − x + x = 1 f = 1 - x + x = 1 f = 1 − x + x = 1 , slope 0.
Right of 1: f = x − 1 + x = 2 x − 1 f = x - 1 + x = 2x - 1 f = x − 1 + x = 2 x − 1 , slope 2.
Worth noting
− 1 -1 − 1 and 1 are the slopes of ∣ x − 1 ∣ \lvert x - 1 \rvert ∣ x − 1 ∣ alone; the + x +x + x adds 1 to each.
Assertion (A): f ( x ) = [ x ] f(x) = [x] f ( x ) = [ x ] is not differentiable at x = 2 x = 2 x = 2 .
Reason (R): f ( x ) = [ x ] f(x) = [x] f ( x ) = [ x ] is not continuous at x = 2 x = 2 x = 2 .
Options for question 99 A Option A Both A and R are true, and R is the correct explanation of A. Correct answer Your answerB Option B Both A and R are true, but R is not the correct explanation of A. Correct answer Your answerC Option C A is true, but R is false. Correct answer Your answerD Option D A is false, but R is true. Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Both A and R are true, and R is the correct explanation of A.
R is true: [ x ] [x] [ x ] jumps from 1 to 2 at x = 2 x = 2 x = 2 .
A is true, and R is the reason: a function that is not continuous cannot be differentiable.
At how many points is f ( x ) = ∣ x ( x − 2 ) ∣ f(x) = \lvert x(x - 2) \rvert f ( x ) = ∣ x ( x − 2 )∣ not differentiable?
Options for question 100 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 2 2 2
x ( x − 2 ) x(x - 2) x ( x − 2 ) changes sign at x = 0 x = 0 x = 0 and x = 2 x = 2 x = 2 .
The modulus makes a corner at each: 2 points.
Worth noting
1 counts only one of the two roots; each root of x ( x − 2 ) x(x - 2) x ( x − 2 ) gives a corner.
Let f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ .
At how many points is this sum of two moduli not differentiable?
Options for question 101 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 2 2 2
Corners at x = 1 x = 1 x = 1 and x = 3 x = 3 x = 3 , where each inside is 0.
2 points.
Worth noting
1 counts only the midpoint, where nothing special happens.
Which of these describes f ( x ) = ∣ x 2 − 9 ∣ f(x) = \lvert x^{2} - 9 \rvert f ( x ) = ∣ x 2 − 9 ∣ at x = 3 x = 3 x = 3 ?
Options for question 102 A Option A Continuous and differentiable Correct answer Your answerB Option B Neither continuous nor differentiable Correct answer Your answerC Option C Differentiable but not continuous Correct answer Your answerD Option D Continuous but not differentiable Correct answer Your answer
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Show the explanation Hide the explanation Answer D · Continuous but not differentiable
Both sides tend to 0 = f ( 3 ) f(3) f ( 3 ) : continuous.
Slopes − 2 x = − 6 -2x = -6 − 2 x = − 6 and 2 x = 6 2x = 6 2 x = 6 : a corner.
Worth noting
The modulus folds the curve up where x 2 − 9 x^{2} - 9 x 2 − 9 changes sign, leaving a corner.
Which of these describes f ( x ) = e ∣ x ∣ f(x) = e^{\lvert x \rvert} f ( x ) = e ∣ x ∣ at x = 0 x = 0 x = 0 ?
Options for question 103 A Option A Continuous and differentiable Correct answer Your answerB Option B Continuous but not differentiable Correct answer Your answerC Option C Neither continuous nor differentiable Correct answer Your answerD Option D Differentiable but not continuous Correct answer Your answer
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Show the explanation Hide the explanation Answer B · Continuous but not differentiable
Both sides tend to e 0 = 1 = f ( 0 ) e^{0} = 1 = f(0) e 0 = 1 = f ( 0 ) : continuous.
Slopes − 1 -1 − 1 on the left and 1 on the right: not differentiable.
Worth noting
∣ x ∣ \lvert x \rvert ∣ x ∣ inside a smooth function keeps its corner.
Assertion (A): No value of k k k makes f ( x ) = ∣ x − 2 ∣ + k x f(x) = \lvert x - 2 \rvert + kx f ( x ) = ∣ x − 2 ∣ + k x differentiable at x = 2 x = 2 x = 2 .
Reason (R): Adding k x kx k x adds the same k k k to both one-sided derivatives at x = 2 x = 2 x = 2 .
Options for question 104 A Option A Both A and R are true, and R is the correct explanation of A. Correct answer Your answerB Option B Both A and R are true, but R is not the correct explanation of A. Correct answer Your answerC Option C A is true, but R is false. Correct answer Your answerD Option D A is false, but R is true. Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Both A and R are true, and R is the correct explanation of A.
R is true: the one-sided derivatives at 2 are k − 1 k - 1 k − 1 and k + 1 k + 1 k + 1 .
They differ by 2 whatever k k k is, so they are never equal: A is true, and R is why.
Worth noting
Adding a line tilts both slopes of the corner equally and cannot close the gap between them.
What is f ′ ( 1 ) f'(1) f ′ ( 1 ) for f ( x ) = { x 2 + 3 x , x ≤ 1 5 x − 1 , x > 1 f(x) = \begin{cases} x^{2} + 3x, & x \le 1 \\ 5x - 1, & x > 1 \end{cases} f ( x ) = { x 2 + 3 x , 5 x − 1 , x ≤ 1 x > 1 ?
Options for question 105 A Option A 2 2 2 Correct answer Your answerB Option B 3 3 3 Correct answer Your answerC Option C 4 4 4 Correct answer Your answerD Option D 5 5 5 Correct answer Your answer
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Show the explanation Hide the explanation Answer D · 5 5 5
Values at 1: 1 + 3 = 4 1 + 3 = 4 1 + 3 = 4 and 5 − 1 = 4 5 - 1 = 4 5 − 1 = 4 , so f f f is continuous there.
Slopes: 2 x + 3 = 5 2x + 3 = 5 2 x + 3 = 5 on the left and 5 on the right; they agree, so f ′ ( 1 ) = 5 f'(1) = 5 f ′ ( 1 ) = 5 .
Worth noting
4 is f ( 1 ) f(1) f ( 1 ) , the value at the join, not the slope.
Which of these functions are differentiable at x = 0 x = 0 x = 0 ?
I. f ( x ) = ∣ x ∣ sin x f(x) = \lvert x \rvert \sin x f ( x ) = ∣ x ∣ sin x
II. f ( x ) = ∣ sin x ∣ f(x) = \lvert \sin x \rvert f ( x ) = ∣ sin x ∣
III. f ( x ) = cos ∣ x ∣ f(x) = \cos \lvert x \rvert f ( x ) = cos ∣ x ∣
IV. f ( x ) = x + ∣ x ∣ f(x) = x + \lvert x \rvert f ( x ) = x + ∣ x ∣
Options for question 106 A Option A III only Correct answer Your answerB Option B I, II and III only Correct answer Your answerC Option C I and III only Correct answer Your answerD Option D I, III and IV only Correct answer Your answer
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Show the explanation Hide the explanation Answer C · I and III only
I is − x sin x -x \sin x − x sin x on the left and x sin x x \sin x x sin x on the right, both with slope 0 at 0. III is cos x \cos x cos x , since cosine is even. Both are differentiable.
II has slopes − cos 0 = − 1 -\cos 0 = -1 − cos 0 = − 1 and cos 0 = 1 \cos 0 = 1 cos 0 = 1 , a corner. IV is 0 on the left and 2 x 2x 2 x on the right, with slopes 0 and 2.
Worth noting
A factor that is 0 at the corner, like sin x \sin x sin x in I, or an even outer function, like cos \cos cos in III, smooths ∣ x ∣ \lvert x \rvert ∣ x ∣ out.
Let f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ f(x) = x\lvert x \rvert + \lvert x - 1 \rvert f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ .
What is f ′ ( 0 ) f'(0) f ′ ( 0 ) for this function?
Options for question 107 A Option A − 2 -2 − 2 Correct answer Your answerB Option B − 1 -1 − 1 Correct answer Your answerC Option C 0 0 0 Correct answer Your answerD Option D 1 1 1 Correct answer Your answer
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Show the explanation Hide the explanation Answer B · − 1 -1 − 1
Near 0, x − 1 < 0 x - 1 < 0 x − 1 < 0 , so ∣ x − 1 ∣ = 1 − x \lvert x - 1 \rvert = 1 - x ∣ x − 1 ∣ = 1 − x , with slope − 1 -1 − 1 .
x ∣ x ∣ x\lvert x \rvert x ∣ x ∣ has slope 2 ∣ x ∣ = 0 2\lvert x \rvert = 0 2 ∣ x ∣ = 0 at 0, so f ′ ( 0 ) = 0 − 1 = − 1 f'(0) = 0 - 1 = -1 f ′ ( 0 ) = 0 − 1 = − 1 .
Worth noting
1 takes the slope of ∣ x − 1 ∣ \lvert x - 1 \rvert ∣ x − 1 ∣ as + 1 +1 + 1 ; to the left of 1 it falls, with slope − 1 -1 − 1 .
Match each function in List I with its left-hand and right-hand derivatives at x = 0 x = 0 x = 0 in List II.
List I List II A. ∣ x ∣ \lvert x \rvert ∣ x ∣ I. − 2 -2 − 2 and 2 2 2 B. x + ∣ x ∣ x + \lvert x \rvert x + ∣ x ∣ II. − 3 -3 − 3 and − 1 -1 − 1 C. ∣ x ∣ − 2 x \lvert x \rvert - 2x ∣ x ∣ − 2 x III. 0 0 0 and 2 2 2 D. ∣ sin 2 x ∣ \lvert \sin 2x \rvert ∣ sin 2 x ∣ IV. − 1 -1 − 1 and 1 1 1
Options for question 108 A Option A A-I, B-III, C-II, D-IV Correct answer Your answerB Option B A-IV, B-III, C-II, D-I Correct answer Your answerC Option C A-IV, B-II, C-III, D-I Correct answer Your answerD Option D A-I, B-II, C-III, D-IV Correct answer Your answer
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Show the explanation Hide the explanation Answer B · A-IV, B-III, C-II, D-I
Write each on the two sides of 0: ∣ x ∣ \lvert x \rvert ∣ x ∣ is − x -x − x then x x x ; x + ∣ x ∣ x + \lvert x \rvert x + ∣ x ∣ is 0 then 2 x 2x 2 x ; ∣ x ∣ − 2 x \lvert x \rvert - 2x ∣ x ∣ − 2 x is − 3 x -3x − 3 x then − x -x − x ; ∣ sin 2 x ∣ \lvert \sin 2x \rvert ∣ sin 2 x ∣ is − sin 2 x -\sin 2x − sin 2 x then sin 2 x \sin 2x sin 2 x .
Their slopes at 0, left then right, are − 1 , 1 -1, 1 − 1 , 1 ; 0 , 2 0, 2 0 , 2 ; − 3 , − 1 -3, -1 − 3 , − 1 ; − 2 , 2 -2, 2 − 2 , 2 .
Worth noting
∣ sin 2 x ∣ \lvert \sin 2x \rvert ∣ sin 2 x ∣ is not ∣ x ∣ \lvert x \rvert ∣ x ∣ again: the 2 inside doubles both slopes.
At how many points in the open interval ( 0 , 3 ) (0, 3) ( 0 , 3 ) is f ( x ) = [ x ] + ∣ x − 2 ∣ f(x) = [x] + \lvert x - 2 \rvert f ( x ) = [ x ] + ∣ x − 2 ∣ not differentiable?
Options for question 109 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · 2 2 2
[ x ] [x] [ x ] jumps at 1 and 2, so f f f is not even continuous there, and hence not differentiable.
∣ x − 2 ∣ \lvert x - 2 \rvert ∣ x − 2 ∣ has its corner at 2, a point already counted. So 2 points.
Worth noting
3 counts x = 2 x = 2 x = 2 twice, once for the jump and once for the corner; it is one point.
Let f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ f(x) = x\lvert x \rvert + \lvert x - 1 \rvert f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ .
At how many points is this function not differentiable?
Options for question 110 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer B · 1 1 1
x ∣ x ∣ x\lvert x \rvert x ∣ x ∣ is − x 2 -x^{2} − x 2 for x < 0 x < 0 x < 0 and x 2 x^{2} x 2 for x ≥ 0 x \ge 0 x ≥ 0 : slope 0 on both sides of 0, so no corner there.
∣ x − 1 ∣ \lvert x - 1 \rvert ∣ x − 1 ∣ has a corner at 1, and the smooth x ∣ x ∣ x\lvert x \rvert x ∣ x ∣ cannot remove it: 1 point.
Worth noting
2 counts x = 0 x = 0 x = 0 too; x ∣ x ∣ x\lvert x \rvert x ∣ x ∣ is smooth there.
What are the left-hand and right-hand derivatives of f ( x ) = ∣ x 2 − 4 ∣ f(x) = \lvert x^{2} - 4 \rvert f ( x ) = ∣ x 2 − 4 ∣ at x = 2 x = 2 x = 2 ?
Options for question 111 A Option A 4 4 4 and 4 4 4 Correct answer Your answerB Option B − 2 -2 − 2 and 2 2 2 Correct answer Your answerC Option C 0 0 0 and 4 4 4 Correct answer Your answerD Option D − 4 -4 − 4 and 4 4 4 Correct answer Your answer
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Show the explanation Hide the explanation Answer D · − 4 -4 − 4 and 4 4 4
Just left of 2, x 2 − 4 < 0 x^{2} - 4 < 0 x 2 − 4 < 0 , so f = 4 − x 2 f = 4 - x^{2} f = 4 − x 2 with slope − 2 x = − 4 -2x = -4 − 2 x = − 4 .
Just right of 2, f = x 2 − 4 f = x^{2} - 4 f = x 2 − 4 with slope 2 x = 4 2x = 4 2 x = 4 .
Worth noting
− 2 -2 − 2 and 2 drop the factor x = 2 x = 2 x = 2 in the derivative 2 x 2x 2 x .
For which a a a and b b b is f ( x ) = { a x 2 + 1 , x ≤ 1 x + b , x > 1 f(x) = \begin{cases} ax^{2} + 1, & x \le 1 \\ x + b, & x > 1 \end{cases} f ( x ) = { a x 2 + 1 , x + b , x ≤ 1 x > 1 differentiable at x = 1 x = 1 x = 1 ?
Options for question 112 A Option A a = 1 2 , b = 1 2 a = \frac{1}{2},\ b = \frac{1}{2} a = 2 1 , b = 2 1 Correct answer Your answerB Option B a = 1 , b = 1 a = 1,\ b = 1 a = 1 , b = 1 Correct answer Your answerC Option C a = 1 2 , b = 3 2 a = \frac{1}{2},\ b = \frac{3}{2} a = 2 1 , b = 2 3 Correct answer Your answerD Option D a = 2 , b = 2 a = 2,\ b = 2 a = 2 , b = 2 Correct answer Your answer
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Show the explanation Hide the explanation Answer A · a = 1 2 , b = 1 2 a = \frac{1}{2},\ b = \frac{1}{2} a = 2 1 , b = 2 1
Values must match at x = 1 x = 1 x = 1 : a + 1 = b + 1 a + 1 = b + 1 a + 1 = b + 1 .
Slopes must match: 2 a = 1 2 a = 1 2 a = 1 , so a = 1 2 a = \frac{1}{2} a = 2 1 , b = 1 2 b = \frac{1}{2} b = 2 1 .
Worth noting
a = 1 , b = 1 a = 1,\ b = 1 a = 1 , b = 1 joins the values but leaves slopes 2 and 1.
Match each function in List I with its behaviour at x = 0 x = 0 x = 0 in List II.
List I List II A. x 2 x^{2} x 2 I. Neither continuous nor differentiable B. ∣ x ∣ \lvert x \rvert ∣ x ∣ II. Differentiable, with derivative 0 C. ∣ x ∣ x \frac{\lvert x \rvert}{x} x ∣ x ∣ with value 0 at 0 III. Differentiable, though built from ∣ x ∣ \lvert x \rvert ∣ x ∣ D. ∣ x ∣ 3 \lvert x \rvert^{3} ∣ x ∣ 3 IV. Continuous but not differentiable
Options for question 113 A Option A A-IV, B-II, C-I, D-III Correct answer Your answerB Option B A-II, B-IV, C-I, D-III Correct answer Your answerC Option C A-II, B-IV, C-III, D-I Correct answer Your answerD Option D A-IV, B-II, C-III, D-I Correct answer Your answer
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Show the explanation Hide the explanation Answer B · A-II, B-IV, C-I, D-III
x 2 x^{2} x 2 has slope 0 at 0; ∣ x ∣ \lvert x \rvert ∣ x ∣ has a corner; ∣ x ∣ x \frac{\lvert x \rvert}{x} x ∣ x ∣ jumps from − 1 -1 − 1 to 1.
∣ x ∣ 3 \lvert x \rvert^{3} ∣ x ∣ 3 has one-sided slopes ± 3 x 2 = 0 \pm 3x^{2} = 0 ± 3 x 2 = 0 : differentiable.
At how many points is f ( x ) = ∣ x − 1 ∣ + ∣ x − 2 ∣ + ∣ x − 3 ∣ f(x) = \lvert x - 1 \rvert + \lvert x - 2 \rvert + \lvert x - 3 \rvert f ( x ) = ∣ x − 1 ∣ + ∣ x − 2 ∣ + ∣ x − 3 ∣ not differentiable?
Options for question 114 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer D · 3 3 3
The corners are at 1, 2 and 3.
At each, the slope jumps by 2, so all 3 are points of non-differentiability.
Worth noting
1 guesses that the middle corner is the only one; each modulus brings its own.
For which a a a and b b b is f ( x ) = { x 3 , x ≤ 2 a x + b , x > 2 f(x) = \begin{cases} x^{3}, & x \le 2 \\ ax + b, & x > 2 \end{cases} f ( x ) = { x 3 , a x + b , x ≤ 2 x > 2 differentiable at x = 2 x = 2 x = 2 ?
Options for question 115 A Option A a = 4 , b = 0 a = 4,\ b = 0 a = 4 , b = 0 Correct answer Your answerB Option B a = 12 , b = 8 a = 12,\ b = 8 a = 12 , b = 8 Correct answer Your answerC Option C a = 12 , b = − 16 a = 12,\ b = -16 a = 12 , b = − 16 Correct answer Your answerD Option D a = 6 , b = − 4 a = 6,\ b = -4 a = 6 , b = − 4 Correct answer Your answer
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Show the explanation Hide the explanation Answer C · a = 12 , b = − 16 a = 12,\ b = -16 a = 12 , b = − 16
Values must match at x = 2 x = 2 x = 2 : 8 = 2 a + b 8 = 2 a + b 8 = 2 a + b .
Slopes must match: 12 = a 12 = a 12 = a , so a = 12 a = 12 a = 12 , b = − 16 b = -16 b = − 16 .
Worth noting
a = 4 , b = 0 a = 4,\ b = 0 a = 4 , b = 0 only joins the values at 2; the slope of x 3 x^{3} x 3 there is 3 x 2 = 12 3x^{2} = 12 3 x 2 = 12 .
At how many points is f ( x ) = ∣ sin x ∣ f(x) = \lvert \sin x \rvert f ( x ) = ∣ sin x ∣ on ( 0 , 2 π ) (0, 2\pi) ( 0 , 2 π ) not differentiable?
Options for question 116 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer B · 1 1 1
sin x = 0 \sin x = 0 sin x = 0 inside ( 0 , 2 π ) (0, 2\pi) ( 0 , 2 π ) only at x = π x = \pi x = π .
There ∣ sin x ∣ \lvert \sin x \rvert ∣ sin x ∣ has a corner: 1 point.
Worth noting
2 or 3 count 0 and 2 π 2\pi 2 π , which the open interval leaves out.
At how many points is f ( x ) = x ∣ x ∣ f(x) = x\lvert x \rvert f ( x ) = x ∣ x ∣ not differentiable?
Options for question 117 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer A · 0 0 0
f ( x ) = − x 2 f(x) = -x^{2} f ( x ) = − x 2 for x < 0 x < 0 x < 0 and x 2 x^{2} x 2 for x ≥ 0 x \ge 0 x ≥ 0 .
Both slopes are 0 at 0, so it is differentiable everywhere: 0 points.
Worth noting
1 treats it like ∣ x ∣ \lvert x \rvert ∣ x ∣ ; the extra factor x x x removes the corner.
Let f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert f ( x ) = ∣ x − 1 ∣ + ∣ x − 3 ∣ .
What is f ′ ( 2 ) f'(2) f ′ ( 2 ) for this sum of two moduli?
Options for question 118 A Option A − 2 -2 − 2 Correct answer Your answerB Option B 0 0 0 Correct answer Your answerC Option C 1 1 1 Correct answer Your answerD Option D 2 2 2 Correct answer Your answer
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Show the explanation Hide the explanation Answer B · 0 0 0
For 1 < x < 3 1 < x < 3 1 < x < 3 : f ( x ) = ( x − 1 ) + ( 3 − x ) = 2 f(x) = (x - 1) + (3 - x) = 2 f ( x ) = ( x − 1 ) + ( 3 − x ) = 2 , a constant.
So f ′ ( 2 ) = 0 f'(2) = 0 f ′ ( 2 ) = 0 .
Worth noting
2 adds the slopes + 1 +1 + 1 and + 1 +1 + 1 ; between 1 and 3 the second modulus has slope − 1 -1 − 1 .
Which of these describes f ( x ) = ∣ sin x ∣ f(x) = \lvert \sin x \rvert f ( x ) = ∣ sin x ∣ at x = π x = \pi x = π ?
Options for question 119 A Option A Continuous and differentiable Correct answer Your answerB Option B Neither continuous nor differentiable Correct answer Your answerC Option C Differentiable but not continuous Correct answer Your answerD Option D Continuous but not differentiable Correct answer Your answer
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Show the explanation Hide the explanation Answer D · Continuous but not differentiable
Both sides tend to 0 = f ( π ) f(\pi) f ( π ) : continuous.
Slopes cos π = − 1 \cos \pi = -1 cos π = − 1 on the left and − cos π = 1 -\cos \pi = 1 − cos π = 1 on the right: a corner.
Worth noting
Where sin x \sin x sin x crosses 0, the modulus turns its slope round.
For which a a a and b b b is f ( x ) = { a x + b , x < 2 x 2 − 1 , x ≥ 2 f(x) = \begin{cases} ax + b, & x < 2 \\ x^{2} - 1, & x \ge 2 \end{cases} f ( x ) = { a x + b , x 2 − 1 , x < 2 x ≥ 2 differentiable at x = 2 x = 2 x = 2 ?
Options for question 120 A Option A a = 4 , b = − 5 a = 4,\ b = -5 a = 4 , b = − 5 Correct answer Your answerB Option B a = 4 , b = 3 a = 4,\ b = 3 a = 4 , b = 3 Correct answer Your answerC Option C a = 2 , b = − 1 a = 2,\ b = -1 a = 2 , b = − 1 Correct answer Your answerD Option D a = − 5 , b = 4 a = -5,\ b = 4 a = − 5 , b = 4 Correct answer Your answer
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Show the explanation Hide the explanation Answer A · a = 4 , b = − 5 a = 4,\ b = -5 a = 4 , b = − 5
Values match: 2 a + b = 3 2a + b = 3 2 a + b = 3 . Slopes match: a = 4 a = 4 a = 4 .
So b = 3 − 8 = − 5 b = 3 - 8 = -5 b = 3 − 8 = − 5 .
Worth noting
a = 2 , b = − 1 a = 2,\ b = -1 a = 2 , b = − 1 joins the values but takes the slope as 2 instead of 2 x = 4 2x = 4 2 x = 4 .
Read the two statements and choose the correct option.
I. f ( x ) = x 2 sin 1 x f(x) = x^{2} \sin \frac{1}{x} f ( x ) = x 2 sin x 1 for x ≠ 0 x \ne 0 x = 0 , with f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 , is differentiable at x = 0 x = 0 x = 0 .
II. g ( x ) = x sin 1 x g(x) = x \sin \frac{1}{x} g ( x ) = x sin x 1 for x ≠ 0 x \ne 0 x = 0 , with g ( 0 ) = 0 g(0) = 0 g ( 0 ) = 0 , is differentiable at x = 0 x = 0 x = 0 .
Options for question 121 A Option A Only I Correct answer Your answerB Option B Only II Correct answer Your answerC Option C Both I and II Correct answer Your answerD Option D Neither I nor II Correct answer Your answer
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Show the explanation Hide the explanation Answer A · Only I
I: f ( h ) − f ( 0 ) h = h sin 1 h → 0 \frac{f(h) - f(0)}{h} = h \sin \frac{1}{h} \to 0 h f ( h ) − f ( 0 ) = h sin h 1 → 0 , so f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0 . True.
II: g ( h ) − g ( 0 ) h = sin 1 h \frac{g(h) - g(0)}{h} = \sin \frac{1}{h} h g ( h ) − g ( 0 ) = sin h 1 , which keeps swinging between − 1 -1 − 1 and 1 and has no limit. False, though g g g is continuous at 0.
Worth noting
Continuity at 0 does not give differentiability: the difference quotient of g g g never settles.
What are the left-hand and right-hand derivatives of f ( x ) = [ x ] sin π x f(x) = [x] \sin \pi x f ( x ) = [ x ] sin π x at x = 1 x = 1 x = 1 ?
Options for question 122 A Option A 0 0 0 and − π -\pi − π Correct answer Your answerB Option B − π -\pi − π and − π -\pi − π Correct answer Your answerC Option C π \pi π and − π -\pi − π Correct answer Your answerD Option D 0 0 0 and π \pi π Correct answer Your answer
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Show the explanation Hide the explanation Answer A · 0 0 0 and − π -\pi − π
Just left of 1, [ x ] = 0 [x] = 0 [ x ] = 0 , so f ( x ) = 0 f(x) = 0 f ( x ) = 0 there: the left-hand derivative is 0.
At 1 and just right of it, [ x ] = 1 [x] = 1 [ x ] = 1 , so f ( x ) = sin π x f(x) = \sin \pi x f ( x ) = sin π x , with slope π cos π = − π \pi \cos \pi = -\pi π cos π = − π at 1.
Worth noting
− π -\pi − π and − π -\pi − π treats [ x ] [x] [ x ] as 1 on both sides; to the left of 1 it is 0, which flattens f f f .
Let f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ f(x) = x\lvert x \rvert + \lvert x - 1 \rvert f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ .
What is the right-hand derivative of this function at x = 1 x = 1 x = 1 ?
Options for question 123 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer D · 3 3 3
For x > 1 x > 1 x > 1 , x ∣ x ∣ = x 2 x\lvert x \rvert = x^{2} x ∣ x ∣ = x 2 and ∣ x − 1 ∣ = x − 1 \lvert x - 1 \rvert = x - 1 ∣ x − 1 ∣ = x − 1 , so f ( x ) = x 2 + x − 1 f(x) = x^{2} + x - 1 f ( x ) = x 2 + x − 1 .
Its slope at 1 is 2 ( 1 ) + 1 = 3 2(1) + 1 = 3 2 ( 1 ) + 1 = 3 .
Worth noting
1 is the left-hand derivative, the slope of x 2 + 1 − x x^{2} + 1 - x x 2 + 1 − x ; the corner at 1 makes the two sides differ.
For which a a a and b b b is f ( x ) = { a x 2 + b , x < 1 1 x , x ≥ 1 f(x) = \begin{cases} ax^{2} + b, & x < 1 \\ \frac{1}{x}, & x \ge 1 \end{cases} f ( x ) = { a x 2 + b , x 1 , x < 1 x ≥ 1 differentiable at x = 1 x = 1 x = 1 ?
Options for question 124 A Option A a = 1 2 , b = 1 2 a = \frac{1}{2},\ b = \frac{1}{2} a = 2 1 , b = 2 1 Correct answer Your answerB Option B a = − 1 , b = 2 a = -1,\ b = 2 a = − 1 , b = 2 Correct answer Your answerC Option C a = − 1 2 , b = 1 2 a = -\frac{1}{2},\ b = \frac{1}{2} a = − 2 1 , b = 2 1 Correct answer Your answerD Option D a = − 1 2 , b = 3 2 a = -\frac{1}{2},\ b = \frac{3}{2} a = − 2 1 , b = 2 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer D · a = − 1 2 , b = 3 2 a = -\frac{1}{2},\ b = \frac{3}{2} a = − 2 1 , b = 2 3
Values must match at x = 1 x = 1 x = 1 : a + b = 1 a + b = 1 a + b = 1 . Slopes must match: 2 a = − 1 x 2 = − 1 2a = -\frac{1}{x^{2}} = -1 2 a = − x 2 1 = − 1 at x = 1 x = 1 x = 1 .
So a = − 1 2 a = -\frac{1}{2} a = − 2 1 and b = 1 − a = 3 2 b = 1 - a = \frac{3}{2} b = 1 − a = 2 3 .
Worth noting
a = 1 2 , b = 1 2 a = \frac{1}{2},\ b = \frac{1}{2} a = 2 1 , b = 2 1 takes the slope of 1 x \frac{1}{x} x 1 at 1 as + 1 +1 + 1 ; it is − 1 x 2 = − 1 -\frac{1}{x^{2}} = -1 − x 2 1 = − 1 .
At how many points is f ( x ) = ∣ ( x − 1 ) 2 ( x − 3 ) ∣ f(x) = \lvert (x - 1)^{2}(x - 3) \rvert f ( x ) = ∣( x − 1 ) 2 ( x − 3 )∣ not differentiable?
Options for question 125 A Option A 0 0 0 Correct answer Your answerB Option B 1 1 1 Correct answer Your answerC Option C 2 2 2 Correct answer Your answerD Option D 3 3 3 Correct answer Your answer
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Show the explanation Hide the explanation Answer B · 1 1 1
u = ( x − 1 ) 2 ( x − 3 ) u = (x - 1)^{2}(x - 3) u = ( x − 1 ) 2 ( x − 3 ) is 0 at x = 1 x = 1 x = 1 and x = 3 x = 3 x = 3 . At 3 it changes sign, so ∣ u ∣ \lvert u \rvert ∣ u ∣ has a corner there, with slopes − 4 -4 − 4 and 4.
At 1 the double root keeps u ≤ 0 u \le 0 u ≤ 0 on both sides, so f = − u f = -u f = − u near 1, a smooth curve with slope 0. So 1 point.
Worth noting
2 counts the double root 1 as well; there u u u touches 0 without changing sign, so the modulus does not fold it.