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Continuity and Differentiability for CUET Maths

About 5 min to learn · 125 questions

Continuity and differentiability open the calculus of CUET Maths Section B1. The questions are short and exact: is a function continuous at a point, which constant makes it continuous, where is it not differentiable. Learn each rule here, from testing a function at a point and finding the constant to the algebra of continuous functions, corners and differentiability, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area
Section B1: Mathematics
On this page · 10 sections
  1. 01What continuity means
  2. 02Continuity at a point
  3. 03Choosing a constant for continuity
  4. 04Continuous functions
  5. 05Differentiability
  6. 06How CUET asks this chapter
  7. 07Exam shortcuts
  8. 08Common mistakes
  9. 09Quick revision
  10. 10Check your understanding

What continuity means

A parking lot charges ₹20 for the first two hours and then ₹15 for every further hour or part of one. Plot the charge against time and the graph jumps at 2 hours, 3 hours, 4 hours. A function is continuous at a point when its graph has no such break there: you could draw it through that point without lifting the pen. This chapter turns that picture into a test you can run on a formula.

When is a function continuous at a point?

ff is continuous at x=ax = a when three numbers agree: the left-hand limit, the right-hand limit and the value f(a)f(a).

lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = f(a)

Continuity at a point

If the two one-sided limits differ, the graph jumps. If they agree but differ from f(a)f(a), the limit exists but the function has been given the wrong value there. Either way, it is not continuous.

Example 1 · A piecewise function at its join

Is f(x)=2x+3f(x) = 2x + 3 for x<1x < 1 and f(x)=6−xf(x) = 6 - x for x≥1x \ge 1 continuous at x=1x = 1?

  1. Left-hand limit: 2(1)+3=52(1) + 3 = 5.

  2. Right-hand limit and value: 6−1=56 - 1 = 5.

Answer

All three are 5, so ff is continuous at x=1x = 1.

Choosing a constant for continuity

How do you find the constant that makes a function continuous?

Write the left-hand limit, the right-hand limit and the value at the joining point in terms of the constant, and set them equal. For a formula of the form 00\frac{0}{0} at the point, the constant must equal the limit.

Example 2 · Joining two pieces

For which kk is f(x)=kx+5f(x) = kx + 5 for x≤2x \le 2 and f(x)=x2−1f(x) = x^{2} - 1 for x>2x > 2 continuous at x=2x = 2?

  1. The pieces must meet: 2k+5=22−1=32k + 5 = 2^{2} - 1 = 3.

  2. 2k=−22k = -2.

Answer
k=−1k = -1

Example 3 · A constant from a limit

f(x)=1−cos⁡6xx2f(x) = \frac{1 - \cos 6x}{x^{2}} for x≠0x \ne 0 and f(0)=kf(0) = k. Find kk for continuity at 0.

  1. 1−cos⁡6x=2sin⁡23x1 - \cos 6x = 2\sin^{2} 3x, so f(x)=2(sin⁡3xx)2f(x) = 2\left(\frac{\sin 3x}{x}\right)^{2}.

  2. sin⁡3xx→3\frac{\sin 3x}{x} \to 3, so the limit is 2×9=182 \times 9 = 18.

Answer
k=18k = 18

Key idea

Limits to keep ready: sin⁡axbx→ab\frac{\sin ax}{bx} \to \frac{a}{b}, eax−1x→a\frac{e^{ax} - 1}{x} \to a, 1−cos⁡axx2→a22\frac{1 - \cos ax}{x^{2}} \to \frac{a^{2}}{2}, and xn−anx−a→nan−1\frac{x^{n} - a^{n}}{x - a} \to na^{n - 1}.

Continuous functions

Which functions are continuous?

Polynomials, sin⁡x\sin x, cos⁡x\cos x, exe^{x} and ∣x∣\lvert x \rvert are continuous everywhere; log⁡x\log x, tan⁡x\tan x and rational functions are continuous at every point of their domain. Sums, differences, products and compositions of continuous functions are continuous, and so is a quotient wherever the denominator is not 0. The greatest integer function [x][x] jumps at every integer.

Watch out

Continuity is checked only at points of the domain. 1x\frac{1}{x} is a continuous function: 0 is not in its domain, so it is not a break of the function.

Differentiability

What makes a function not differentiable at a point?

ff is differentiable at aa when the left-hand and right-hand derivatives agree. A break, a jump, or a corner where two different slopes meet spoils it. A function that is differentiable at a point is continuous there, but not the other way round: ∣x∣\lvert x \rvert is continuous at 0 with a corner.

Example 4 · A smooth join

Is f(x)=x2f(x) = x^{2} for x≤2x \le 2 and f(x)=4x−4f(x) = 4x - 4 for x>2x > 2 differentiable at x=2x = 2?

  1. Values: 22=42^{2} = 4 and 4(2)−4=44(2) - 4 = 4: continuous.

  2. Slopes: 2x=42x = 4 on the left and 4 on the right: equal.

Answer

Differentiable at x=2x = 2, with derivative 4.

Example 5 · Counting corners

At how many points is f(x)=∣x+2∣+∣x−5∣f(x) = \lvert x + 2 \rvert + \lvert x - 5 \rvert not differentiable?

  1. Each modulus has a corner where its inside is 0: x=−2x = -2 and x=5x = 5.

  2. The corners are at different points, so neither cancels the other.

Answer

2 points

How CUET asks this chapter

  • At a point: whether a piecewise or 00\frac{0}{0} function is continuous, or the value of a one-sided limit.
  • Constants: the kk that makes a function continuous, or the aa and bb that make it continuous or differentiable.
  • Continuous functions: which functions are continuous everywhere, the algebra of continuous functions, points of discontinuity of [x][x].
  • Differentiability: corners of ∣x∣\lvert x \rvert-type functions, one-sided derivatives, and the link with continuity.

Exam shortcuts

  • Continuity at aa: left limit = right limit = f(a)f(a).
  • Differentiability at a join: the values match and the slopes match.
  • ∣x−a∣\lvert x - a \rvert has one corner, at aa; x∣x∣x\lvert x \rvert and ∣x∣3\lvert x \rvert^{3} have none.
  • Differentiable ⇒ continuous; continuous does not imply differentiable.

Common mistakes

  • Treating a change of formula as a break even when the pieces meet.
  • Forgetting that a limit can exist while the value is different.
  • Counting points outside the domain as discontinuities.
  • Checking only the values at a join, not the slopes, for differentiability.

Quick revision

Continuity

  • LHL = RHL = f(a)f(a)
  • Constant: set the three equal

Continuous functions

  • Sums, products, compositions stay continuous
  • [x][x] jumps at integers

Differentiability

  • Left and right derivatives agree
  • Corner: continuous, not differentiable

Next in Section B1, the rules that differentiate composite, inverse and exponential functions: chain rule, logarithmic differentiation and rate of change.

Next · 09Quick revision

Check your understanding

125 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Continuity at a pointQ1 of 125Q1

Continuity at a point

30 questions

How does f(x)={2x+1,x≤1x2+2,x>1f(x) = \begin{cases} 2x + 1, & x \le 1 \\ x^{2} + 2, & x > 1 \end{cases} behave at x=1x = 1?

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AnswerB · Continuous at the point

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 33, the right limit 33 and f(1)=3f(1) = 3.

Worth noting

Both pieces give 3 at x=1x = 1; a change of formula is not a break if the pieces meet.

For f(x)={4x−1,x<3x2,x≥3f(x) = \begin{cases} 4x - 1, & x < 3 \\ x^{2}, & x \ge 3 \end{cases}, what is the left-hand limit at x=3x = 3?

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AnswerC · 1111

  1. The left-hand limit uses the piece for xx below 3.

  2. That piece tends to 1111.

Worth noting

9 is the right-hand piece at 3; the left limit uses 4x−14x - 1.

How does f(x)={3−x,x<2x−1,x≥2f(x) = \begin{cases} 3 - x, & x < 2 \\ x - 1, & x \ge 2 \end{cases} behave at x=2x = 2?

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AnswerC · Continuous at the point

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 11, the right limit 11 and f(2)=1f(2) = 1.

Worth noting

3−2=13 - 2 = 1 and 2−1=12 - 1 = 1: the two pieces meet at x=2x = 2.

For f(x)={x2+1,x<35x−4,x≥3f(x) = \begin{cases} x^{2} + 1, & x < 3 \\ 5x - 4, & x \ge 3 \end{cases}, what is the left-hand limit at x=3x = 3?

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AnswerB · 1010

  1. Use the values of ff for xx just below 3.

  2. They tend to 1010.

Worth noting

11 is the right-hand piece at 3.

How does f(x)=∣x−2∣f(x) = \lvert x - 2 \rvert behave at x=2x = 2?

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AnswerD · Continuous at the point

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 00, the right limit 00 and f(2)=0f(2) = 0.

Worth noting

A corner is not a break: both sides tend to 0, which is f(2)f(2).

For f(x)=[x]f(x) = [x], what is the left-hand limit at x=2x = 2?

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AnswerB · 11

  1. Use the values of ff for xx just below 2.

  2. They tend to 11.

Worth noting

2 is the value at 2 and the right-hand limit; just below 2, [x]=1[x] = 1.

f(x)=x+1f(x) = x + 1 for x<1x < 1, f(1)=2f(1) = 2, and f(x)=3x−1f(x) = 3x - 1 for x>1x > 1.

What is the left-hand limit of this three-piece ff at x=1x = 1?

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AnswerC · 22

  1. For x<1x < 1, f(x)=x+1f(x) = x + 1.

  2. As x→1−x \to 1^{-}, x+1→2x + 1 \to 2.

Worth noting

1 is the value of xx, not of x+1x + 1.

f(x)=x+1f(x) = x + 1 for x<1x < 1, f(1)=2f(1) = 2, and f(x)=3x−1f(x) = 3x - 1 for x>1x > 1.

What is the right-hand limit of this three-piece ff at x=1x = 1?

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AnswerB · 22

  1. For x>1x > 1, f(x)=3x−1f(x) = 3x - 1.

  2. As x→1+x \to 1^{+}, 3x−1→23x - 1 \to 2.

Worth noting

3 drops the −1-1 of 3x−13x - 1.

How does f(x)=[x]f(x) = [x] behave at x=52x = \frac{5}{2}?

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AnswerB · Continuous at the point

  1. On (2,3)(2, 3), [x]=2[x] = 2: ff is constant near 52\frac{5}{2}.

  2. Both one-sided limits and f(52)f\left(\frac{5}{2}\right) equal 2: continuous.

Worth noting

[x][x] jumps only at integers, and 52\frac{5}{2} is not one.

How does f(x)=x2−9x−3f(x) = \frac{x^{2} - 9}{x - 3} behave at x=3x = 3?

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AnswerA · Not continuous: the function has no value there

  1. At x=3x = 3 the formula gives 00\frac{0}{0}: 3 is not in the domain, so f(3)f(3) does not exist.

  2. The limit x+3→6x + 3 \to 6 exists, but with no value at 3 there is nothing for it to equal.

Worth noting

A limit is not a value: x2−9x−3\frac{x^{2} - 9}{x - 3} tends to 6 at 3 but is not defined there.

For f(x)=x+∣x−1∣x−1f(x) = x + \frac{\lvert x - 1 \rvert}{x - 1}, what is the right-hand limit at x=1x = 1?

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AnswerC · 22

  1. For x>1x > 1, ∣x−1∣=x−1\lvert x - 1 \rvert = x - 1, so ∣x−1∣x−1=1\frac{\lvert x - 1 \rvert}{x - 1} = 1.

  2. So f(x)=x+1→2f(x) = x + 1 \to 2 as x→1+x \to 1^{+}.

Worth noting

0 uses ∣x−1∣x−1=−1\frac{\lvert x - 1 \rvert}{x - 1} = -1, which holds only to the left of 1.

Read the two statements and choose the correct option.

I. A function can have a limit at x=ax = a without being continuous there.

II. If ff is continuous at x=ax = a, then lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

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AnswerC · Both I and II

  1. I: f(x)=x2f(x) = x^{2} for x≠2x \ne 2 with f(2)=5f(2) = 5 has limit 4 at 2 but is not continuous. True.

  2. II is the definition of continuity at aa. True.

How does f(x)={x2,x≠25,x=2f(x) = \begin{cases} x^{2}, & x \ne 2 \\ 5, & x = 2 \end{cases} behave at x=2x = 2?

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AnswerD · Not continuous: the limit exists but differs from the value

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 44, the right limit 44 and f(2)=5f(2) = 5.

Worth noting

The limit 4 exists, but the function is given the value 5 there.

For f(x)={x2−1,x≤27−x,x>2f(x) = \begin{cases} x^{2} - 1, & x \le 2 \\ 7 - x, & x > 2 \end{cases}, what is the right-hand limit at x=2x = 2?

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AnswerC · 55

  1. The right-hand limit uses the piece for xx above 2.

  2. That piece tends to 55.

Worth noting

3 is f(2)f(2), from the left piece; the right-hand limit uses 7−x7 - x.

How does f(x)={∣x∣x,x≠00,x=0f(x) = \begin{cases} \frac{\lvert x \rvert}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} behave at x=0x = 0?

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AnswerA · Not continuous: the left and right limits differ

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is −1-1, the right limit 11 and f(0)=0f(0) = 0.

Worth noting

Approaching from the left gives −1-1, from the right 1: there is no single limit.

How does f(x)={x+3,x<12x−1,x≥1f(x) = \begin{cases} x + 3, & x < 1 \\ 2x - 1, & x \ge 1 \end{cases} behave at x=1x = 1?

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AnswerD · Not continuous: the left and right limits differ

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 44, the right limit 11 and f(1)=1f(1) = 1.

Worth noting

The left piece tends to 4 and the right gives 1: they do not meet.

f(x)=x+1f(x) = x + 1 for x<1x < 1, f(1)=2f(1) = 2, and f(x)=3x−1f(x) = 3x - 1 for x>1x > 1.

Which of these describes this three-piece ff at x=1x = 1?

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AnswerB · Continuous but not differentiable

  1. It is continuous at 1.

  2. The slopes are 1 on the left and 3 on the right: not differentiable.

Worth noting

Matching values make it continuous; differentiable needs matching slopes as well.

How does f(x)={x2−4x−2,x≠23,x=2f(x) = \begin{cases} \frac{x^{2} - 4}{x - 2}, & x \ne 2 \\ 3, & x = 2 \end{cases} behave at x=2x = 2?

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AnswerB · Not continuous: the limit exists but differs from the value

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 44, the right limit 44 and f(2)=3f(2) = 3.

Worth noting

The limit is x+2→4x + 2 \to 4, but the value given at 2 is 3.

How does f(x)=x−[x]f(x) = x - [x] behave at x=1x = 1?

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AnswerA · Not continuous: the left and right limits differ

  1. Just left of 1, [x]=0[x] = 0, so f(x)=x→1f(x) = x \to 1. At 1 and just right of it, [x]=1[x] = 1, so f(x)=x−1→0f(x) = x - 1 \to 0.

  2. The left limit 1 and right limit 0 differ.

Worth noting

The fractional part drops back to 0 at every integer.

f(x)=x+1f(x) = x + 1 for x<1x < 1, f(1)=2f(1) = 2, and f(x)=3x−1f(x) = 3x - 1 for x>1x > 1.

What is the right-hand derivative of this three-piece ff at x=1x = 1?

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AnswerD · 33

  1. lim⁡h→0+f(1+h)−f(1)h=lim⁡h→0+3(1+h)−1−2h\lim_{h \to 0^{+}} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^{+}} \frac{3(1 + h) - 1 - 2}{h}.

  2. =lim⁡h→0+3hh=3= \lim_{h \to 0^{+}} \frac{3h}{h} = 3.

Worth noting

1 is the left-hand derivative, the slope of x+1x + 1.

For f(x)=∣x−2∣+[x]f(x) = \lvert x - 2 \rvert + [x], what is lim⁡h→0+f(2−h)\lim_{h \to 0^{+}} f(2 - h)?

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AnswerB · 11

  1. lim⁡h→0+f(2−h)\lim_{h \to 0^{+}} f(2 - h) is the left-hand limit at 2: x=2−hx = 2 - h is just below 2.

  2. There ∣x−2∣=h→0\lvert x - 2 \rvert = h \to 0 and [x]=1[x] = 1, so the limit is 1.

Worth noting

2 takes [x]=2[x] = 2, its value at 2 and to the right; just below 2 it is 1.

Read the two statements and choose the correct option.

I. If lim⁡x→a−f(x)=f(a)\lim_{x \to a^{-}} f(x) = f(a), then ff is continuous at x=ax = a.

II. If ff is continuous at x=ax = a, then lim⁡h→0f(a+h)=f(a)\lim_{h \to 0} f(a + h) = f(a).

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AnswerB · Only II

  1. I is false: f(x)=xf(x) = x for x≤1x \le 1 and x+1x + 1 for x>1x > 1 has left limit 1=f(1)1 = f(1) but right limit 2.

  2. II is true: lim⁡h→0f(a+h)\lim_{h \to 0} f(a + h) is lim⁡x→af(x)\lim_{x \to a} f(x) written with x=a+hx = a + h, and continuity makes it f(a)f(a).

Worth noting

Continuity needs both one-sided limits to equal f(a)f(a); one side is not enough.

Which of these functions are continuous at x=1x = 1?

I. f(x)=x2+x−2x−1f(x) = \frac{x^{2} + x - 2}{x - 1} for x≠1x \ne 1, and f(1)=3f(1) = 3

II. f(x)=x−∣x−1∣f(x) = x - \lvert x - 1 \rvert

III. f(x)=[x]+∣x−1∣f(x) = [x] + \lvert x - 1 \rvert

IV. f(x)=2xf(x) = 2x for x<1x < 1, and f(x)=x2+1f(x) = x^{2} + 1 for x≥1x \ge 1

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AnswerA · I, II and IV only

  1. I: x2+x−2x−1=x+2→3=f(1)\frac{x^{2} + x - 2}{x - 1} = x + 2 \to 3 = f(1). II: a difference of continuous functions. IV: both pieces give 2 at x=1x = 1. All three are continuous.

  2. III: just left of 1, [x]+∣x−1∣→0[x] + \lvert x - 1 \rvert \to 0, while at 1 it is 1+0=11 + 0 = 1. A jump.

Worth noting

The corner of ∣x−1∣\lvert x - 1 \rvert in II does not break continuity; the jump of [x][x] in III does.

How does f(x)={x2∣x∣,x≠00,x=0f(x) = \begin{cases} \frac{x^{2}}{\lvert x \rvert}, & x \ne 0 \\ 0, & x = 0 \end{cases} behave at x=0x = 0?

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AnswerD · Continuous at the point

  1. For x≠0x \ne 0, x2∣x∣=∣x∣\frac{x^{2}}{\lvert x \rvert} = \lvert x \rvert, since x2=∣x∣2x^{2} = \lvert x \rvert^{2}.

  2. So both one-sided limits are 0, equal to f(0)f(0): continuous.

Worth noting

Unlike ∣x∣x\frac{\lvert x \rvert}{x}, this quotient has x2x^{2} on top, which is positive on both sides: no jump.

How does f(x)={x3−8x−2,x≠212,x=2f(x) = \begin{cases} \frac{x^{3} - 8}{x - 2}, & x \ne 2 \\ 12, & x = 2 \end{cases} behave at x=2x = 2?

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AnswerB · Continuous at the point

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 1212, the right limit 1212 and f(2)=12f(2) = 12.

Worth noting

At x=2x = 2 the formula is 00\frac{0}{0}, but the limit is x2+2x+4=12x^{2} + 2x + 4 = 12, the value given.

How does f(x)={sin⁡3xx,x≠01,x=0f(x) = \begin{cases} \frac{\sin 3x}{x}, & x \ne 0 \\ 1, & x = 0 \end{cases} behave at x=0x = 0?

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AnswerA · Not continuous: the limit exists but differs from the value

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 33, the right limit 33 and f(0)=1f(0) = 1.

Worth noting

sin⁡3xx→3\frac{\sin 3x}{x} \to 3, not 1: the 3 inside the sine comes out.

How does f(x)={sin⁡xx,x≠01,x=0f(x) = \begin{cases} \frac{\sin x}{x}, & x \ne 0 \\ 1, & x = 0 \end{cases} behave at x=0x = 0?

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AnswerD · Continuous at the point

  1. Continuity at aa needs the left limit, the right limit and f(a)f(a) to be equal.

  2. Here the left limit is 11, the right limit 11 and f(0)=1f(0) = 1.

Worth noting

sin⁡xx→1\frac{\sin x}{x} \to 1, and the value at 0 is chosen to be 1.

For f(x)=[x]2−[x2]f(x) = [x]^{2} - [x^{2}], what is the left-hand limit at x=2x = 2?

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AnswerB · −2-2

  1. Just left of 2: [x]=1[x] = 1, so [x]2=1[x]^{2} = 1; and x2x^{2} is just below 4, so [x2]=3[x^{2}] = 3.

  2. The left-hand limit is 1−3=−21 - 3 = -2.

Worth noting

0 is f(2)=4−4f(2) = 4 - 4; just below 2 both brackets drop, [x][x] to 1 and [x2][x^{2}] to 3.

How does f(x)={xsin⁡1x,x≠00,x=0f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} behave at x=0x = 0?

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AnswerA · Continuous at the point

  1. −∣x∣≤xsin⁡1x≤∣x∣-\lvert x \rvert \le x \sin \frac{1}{x} \le \lvert x \rvert, because sin⁡1x\sin \frac{1}{x} stays between −1-1 and 1.

  2. Both bounds tend to 0, so xsin⁡1x→0=f(0)x \sin \frac{1}{x} \to 0 = f(0) from either side: continuous.

Worth noting

sin⁡1x\sin \frac{1}{x} alone has no limit at 0, but the factor xx squeezes the product to 0.

How does f(x)={1−cos⁡2xx,x≠02,x=0f(x) = \begin{cases} \frac{\sqrt{1 - \cos 2x}}{x}, & x \ne 0 \\ \sqrt{2}, & x = 0 \end{cases} behave at x=0x = 0?

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AnswerB · Not continuous: the left and right limits differ

  1. 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^{2} x, so 1−cos⁡2x=2∣sin⁡x∣\sqrt{1 - \cos 2x} = \sqrt{2}\lvert \sin x \rvert and f(x)=2∣sin⁡x∣xf(x) = \frac{\sqrt{2}\lvert \sin x \rvert}{x}.

  2. From the right this tends to 2\sqrt{2}, from the left to −2-\sqrt{2}: the one-sided limits differ.

Worth noting

Writing 2sin⁡2x\sqrt{2\sin^{2} x} as 2sin⁡x\sqrt{2}\sin x drops the modulus; to the left of 0, sin⁡x<0\sin x < 0.

Choosing a constant for continuity

27 questions

For which value of kk is f(x)={kx+3,x≤1x2+4,x>1f(x) = \begin{cases} kx + 3, & x \le 1 \\ x^{2} + 4, & x > 1 \end{cases} continuous at x=1x = 1?

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AnswerC · 22

  1. The pieces must meet at x=1x = 1: k+3=5k + 3 = 5.

  2. So k=2k = 2.

Worth noting

1 makes the pieces equal at x=0x = 0, not at the joining point x=1x = 1.

f(x)=x2−16x−4f(x) = \frac{x^{2} - 16}{x - 4} for x≠4x \ne 4, and f(4)=kf(4) = k. For which kk is ff continuous at x=4x = 4?

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AnswerC · 88

  1. x2−16x−4=x+4\frac{x^{2} - 16}{x - 4} = x + 4 for x≠4x \ne 4.

  2. So the limit at 4 is 8, and kk must be 8.

Worth noting

4 takes only the xx of x+4x + 4.

f(x)=x3−27x−3f(x) = \frac{x^{3} - 27}{x - 3} for x≠3x \ne 3, and f(3)=kf(3) = k. For which kk is ff continuous at x=3x = 3?

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AnswerD · 2727

  1. x3−27x−3=x2+3x+9\frac{x^{3} - 27}{x - 3} = x^{2} + 3x + 9 for x≠3x \ne 3.

  2. At 3: 9+9+9=279 + 9 + 9 = 27.

Worth noting

9 keeps only the x2x^{2} term of x2+3x+9x^{2} + 3x + 9.

For which value of kk is f(x)={kx−1,x<2k+3,x≥2f(x) = \begin{cases} kx - 1, & x < 2 \\ k + 3, & x \ge 2 \end{cases} continuous at x=2x = 2?

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AnswerD · 44

  1. Left-hand limit 2k−12k - 1; right-hand limit and value k+3k + 3.

  2. 2k−1=k+32k - 1 = k + 3 gives k=4k = 4.

Worth noting

2 forgets the kk in the right-hand piece and solves 2k−1=32k - 1 = 3.

f(x)=sin⁡xx+cos⁡xf(x) = \frac{\sin x}{x} + \cos x for x≠0x \ne 0, and the value of ff at x=0x = 0 is kk. For which kk is ff continuous at x=0x = 0?

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AnswerC · 22

  1. As x→0x \to 0, sin⁡xx→1\frac{\sin x}{x} \to 1 and cos⁡x→cos⁡0=1\cos x \to \cos 0 = 1.

  2. The limit is 1+1=21 + 1 = 2, so k=2k = 2.

Worth noting

1 keeps only one of the two limits; each of sin⁡xx\frac{\sin x}{x} and cos⁡x\cos x tends to 1.

Assertion (A): No value of kk makes f(x)=∣x∣xf(x) = \frac{\lvert x \rvert}{x} for x≠0x \ne 0, f(0)=kf(0) = k, continuous at 0.

Reason (R): The left and right limits of ∣x∣x\frac{\lvert x \rvert}{x} at 0 are −1-1 and 1.

Options for question 36

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. A is true: no single value can match two different one-sided limits.

  2. R is true and is the reason: with no limit at 0, choosing f(0)f(0) cannot help.

For which value of kk is f(x)={kx2,x≤23x+4,x>2f(x) = \begin{cases} kx^{2}, & x \le 2 \\ 3x + 4, & x > 2 \end{cases} continuous at x=2x = 2?

Options for question 37

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AnswerB · 52\frac{5}{2}

  1. The pieces must meet at x=2x = 2: 4k=104 k = 10.

  2. So k=52k = \frac{5}{2}.

Worth noting

5 sets 2k=102k = 10, forgetting that x2x^{2} is 4 at x=2x = 2.

f(x)=sin⁡5x2xf(x) = \frac{\sin 5x}{2x} for x≠0x \ne 0, and f(0)=kf(0) = k. For which kk is ff continuous at x=0x = 0?

Options for question 38

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AnswerC · 52\frac{5}{2}

  1. sin⁡5x2x=52⋅sin⁡5x5x\frac{\sin 5x}{2x} = \frac{5}{2} \cdot \frac{\sin 5x}{5x}.

  2. As x→0x \to 0, sin⁡5x5x→1\frac{\sin 5x}{5x} \to 1, so the limit is 52\frac{5}{2}.

Worth noting

25\frac{2}{5} turns the fraction upside down.

For which value of kk is f(x)={x2+kx,x<12x+3,x≥1f(x) = \begin{cases} x^{2} + kx, & x < 1 \\ 2x + 3, & x \ge 1 \end{cases} continuous at x=1x = 1?

Options for question 39

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AnswerB · 44

  1. The pieces must meet at x=1x = 1: k+1=5k + 1 = 5.

  2. So k=4k = 4.

Worth noting

5 is the right-hand value; the left piece still has its x2=1x^{2} = 1 to account for.

f(x)=1+x−1xf(x) = \frac{\sqrt{1 + x} - 1}{x} for x≠0x \ne 0, and f(0)=kf(0) = k. For which kk is ff continuous at x=0x = 0?

Options for question 40

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AnswerB · 12\frac{1}{2}

  1. Multiply above and below by 1+x+1\sqrt{1 + x} + 1: the expression becomes 11+x+1\frac{1}{\sqrt{1 + x} + 1}.

  2. At x=0x = 0 that is 12\frac{1}{2}.

Worth noting

0 sets the top to 0 and stops, but the bottom is 0 too.

f(x)=e3x−1xf(x) = \frac{e^{3x} - 1}{x} for x≠0x \ne 0, and f(0)=kf(0) = k. For which kk is ff continuous at x=0x = 0?

Options for question 41

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AnswerD · 33

  1. e3x−1x=3⋅e3x−13x\frac{e^{3x} - 1}{x} = 3 \cdot \frac{e^{3x} - 1}{3x}.

  2. et−1t→1\frac{e^{t} - 1}{t} \to 1, so the limit is 3.

Worth noting

1 uses ex−1x→1\frac{e^{x} - 1}{x} \to 1 without the 3 in the power.

For which value of kk is f(x)={ksin⁡x,x≤π2cos⁡x+1,x>π2f(x) = \begin{cases} k\sin x, & x \le \frac{\pi}{2} \\ \cos x + 1, & x > \frac{\pi}{2} \end{cases} continuous at x=π2x = \frac{\pi}{2}?

Options for question 42

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AnswerC · 11

  1. The pieces must meet at x=π2x = \frac{\pi}{2}: k=1k = 1.

  2. So k=1k = 1.

Worth noting

0 takes cos⁡π2+1\cos \frac{\pi}{2} + 1 as 0, forgetting the +1+1.

For which value of kk is f(x)={3ex+k,x≤0x+7,x>0f(x) = \begin{cases} 3e^{x} + k, & x \le 0 \\ x + 7, & x > 0 \end{cases} continuous at x=0x = 0?

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AnswerA · 44

  1. The pieces must meet at x=0x = 0: k+3=7k + 3 = 7.

  2. So k=4k = 4.

Worth noting

7 forgets that 3e0=33e^{0} = 3 is already on the left.

f(x)=tan⁡4xsin⁡2xf(x) = \frac{\tan 4x}{\sin 2x} for x≠0x \ne 0, and f(0)=kf(0) = k. For which kk is ff continuous at x=0x = 0?

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AnswerC · 22

  1. tan⁡4x≈4x\tan 4x \approx 4x and sin⁡2x≈2x\sin 2x \approx 2x near 0.

  2. The limit is 42=2\frac{4}{2} = 2.

Worth noting

1 cancels tan⁡\tan against sin⁡\sin and ignores the 4 and the 2.

f(x)=log⁡(1+2x)xf(x) = \frac{\log(1 + 2x)}{x} for x≠0x \ne 0, and f(0)=kf(0) = k. For which kk is ff continuous at x=0x = 0?

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AnswerD · 22

  1. log⁡(1+2x)x=2⋅log⁡(1+2x)2x\frac{\log(1 + 2x)}{x} = 2 \cdot \frac{\log(1 + 2x)}{2x}.

  2. log⁡(1+t)t→1\frac{\log(1 + t)}{t} \to 1, so the limit is 2.

Worth noting

1 uses log⁡(1+x)x→1\frac{\log(1 + x)}{x} \to 1 without the 2 inside.

f(x)=kcos⁡xπ−2xf(x) = \frac{k \cos x}{\pi - 2x} for x≠π2x \ne \frac{\pi}{2}, and f(π2)=5f\left(\frac{\pi}{2}\right) = 5. For which kk is ff continuous at x=π2x = \frac{\pi}{2}?

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AnswerD · 1010

  1. Put x=π2−hx = \frac{\pi}{2} - h: cos⁡x=sin⁡h\cos x = \sin h and π−2x=2h\pi - 2x = 2h, so kcos⁡xπ−2x=ksin⁡h2h→k2\frac{k \cos x}{\pi - 2x} = \frac{k \sin h}{2h} \to \frac{k}{2}.

  2. Continuity needs k2=5\frac{k}{2} = 5, so k=10k = 10.

Worth noting

5 takes cos⁡xπ−2x→1\frac{\cos x}{\pi - 2x} \to 1; the 2 in π−2x\pi - 2x halves the limit.

For which value of kk is f(x)={kx+3,x≤πcos⁡x,x>πf(x) = \begin{cases} kx + 3, & x \le \pi \\ \cos x, & x > \pi \end{cases} continuous at x=πx = \pi?

Options for question 47

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AnswerA · −4π- \frac{4}{\pi}

  1. The pieces must meet at x=πx = \pi: kπ+3=cos⁡π=−1k\pi + 3 = \cos \pi = -1.

  2. So kπ=−4k\pi = -4 and k=−4πk = -\frac{4}{\pi}.

Worth noting

−2π-\frac{2}{\pi} takes cos⁡π\cos \pi as 1; it is −1-1.

f(x)=[x]+[−x]f(x) = [x] + [-x] for x≠2x \ne 2, and the value of ff at x=2x = 2 is kk. For which kk is ff continuous at x=2x = 2?

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AnswerB · −1-1

  1. Just left of 2: [x]=1[x] = 1 and [−x]=−2[-x] = -2, sum −1-1. Just right of 2: [x]=2[x] = 2 and [−x]=−3[-x] = -3, sum −1-1.

  2. Both one-sided limits are −1-1, so k=−1k = -1.

Worth noting

0 is [2]+[−2][2] + [-2], the formula's own value at 2; continuity needs the limit −1-1 there instead.

f(x)=xtan⁡x1−cos⁡xf(x) = \frac{x \tan x}{1 - \cos x} for x≠0x \ne 0, and the value of ff at x=0x = 0 is kk. For which kk is ff continuous at x=0x = 0?

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AnswerD · 22

  1. 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^{2} \frac{x}{2}, so f(x)=tan⁡xx⋅x22sin⁡2x2=tan⁡xx⋅2(x2sin⁡x2)2f(x) = \frac{\tan x}{x} \cdot \frac{x^{2}}{2\sin^{2} \frac{x}{2}} = \frac{\tan x}{x} \cdot 2\left(\frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)^{2}.

  2. As x→0x \to 0 this tends to 1×2×1=21 \times 2 \times 1 = 2, so k=2k = 2.

Worth noting

1 replaces 1−cos⁡x1 - \cos x by x2x^{2}; near 0 it behaves like x22\frac{x^{2}}{2}.

Assertion (A): f(x)=1−cos⁡2xx2f(x) = \frac{1 - \cos 2x}{x^{2}} for x≠0x \ne 0, with f(0)=1f(0) = 1, is continuous at x=0x = 0.

Reason (R): 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^{2} x for every real xx.

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AnswerD · A is false, but R is true.

  1. R is true: it is the double-angle formula cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^{2} x.

  2. A is false: by R, f(x)=2(sin⁡xx)2→2f(x) = 2\left(\frac{\sin x}{x}\right)^{2} \to 2, not 1.

Worth noting

Treating 1−cos⁡2x1 - \cos 2x as x2x^{2} near 0 gives 1; the factor 2 in 2sin⁡2x2\sin^{2} x makes the limit 2.

f(x)=2f(x) = 2 for x≤1x \le 1, f(x)=ax+bf(x) = ax + b for 1<x<31 < x < 3, and f(x)=8f(x) = 8 for x≥3x \ge 3. For which aa and bb is ff continuous everywhere?

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AnswerA · a=3, b=−1a = 3,\ b = -1

  1. At x=1x = 1: a+b=2a + b = 2. At x=3x = 3: 3a+b=83a + b = 8.

  2. Subtracting, 2a=62a = 6: a=3a = 3, b=−1b = -1.

Worth noting

a=2, b=0a = 2,\ b = 0 joins only at x=1x = 1; the middle piece must meet both ends.

For which value of kk is f(x)={3x−k,x<4k2x,x≥4f(x) = \begin{cases} 3x - k, & x < 4 \\ \frac{k}{2}x, & x \ge 4 \end{cases} continuous at x=4x = 4?

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AnswerB · 44

  1. The pieces must meet at x=4x = 4: 12−k=2k12 - k = 2 k.

  2. So k=4k = 4.

Worth noting

12 sets 3×4=k3 \times 4 = k, leaving out the −k-k on the left.

f(x)=1−cos⁡4x8x2f(x) = \frac{1 - \cos 4x}{8x^{2}} for x≠0x \ne 0, and f(0)=kf(0) = k. For which kk is ff continuous at x=0x = 0?

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AnswerB · 11

  1. 1−cos⁡4x=2sin⁡22x1 - \cos 4x = 2\sin^{2} 2x, so the expression is 2sin⁡22x8x2=14(sin⁡2xx)2\frac{2\sin^{2} 2x}{8x^{2}} = \frac{1}{4}\left(\frac{\sin 2x}{x}\right)^{2}.

  2. sin⁡2xx→2\frac{\sin 2x}{x} \to 2, so the limit is 14×4=1\frac{1}{4} \times 4 = 1.

Worth noting

8 is the limit of 1−cos⁡4xx2\frac{1 - \cos 4x}{x^{2}}; the 8 below divides it down to 1.

f(x)=x2f(x) = x^{2} for x<1x < 1, f(x)=ax+bf(x) = ax + b for 1≤x≤21 \le x \le 2, and f(x)=x+5f(x) = x + 5 for x>2x > 2. For which aa and bb is ff continuous everywhere?

Options for question 54

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AnswerA · a=6, b=−5a = 6,\ b = -5

  1. At x=1x = 1: 1=a+b1 = a + b. At x=2x = 2: 2a+b=72a + b = 7.

  2. Subtracting, a=6a = 6, so b=−5b = -5.

Worth noting

a=2, b=−1a = 2,\ b = -1 joins only at x=1x = 1.

For which values of kk is f(x)={k2x,x<2x+3k,x≥2f(x) = \begin{cases} k^{2}x, & x < 2 \\ x + 3k, & x \ge 2 \end{cases} continuous at x=2x = 2?

Options for question 55

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AnswerD · k=2k = 2 or k=−12k = -\frac{1}{2}

  1. The pieces must meet at x=2x = 2: 2k2=2+3k2k^{2} = 2 + 3k.

  2. 2k2−3k−2=(2k+1)(k−2)=02k^{2} - 3k - 2 = (2k + 1)(k - 2) = 0, so k=2k = 2 or k=−12k = -\frac{1}{2}.

Worth noting

k=2k = 2 alone drops the second root; k=−12k = -\frac{1}{2} also makes both pieces 12\frac{1}{2} at x=2x = 2.

f(x)=log⁡(1+3x)−log⁡(1−2x)xf(x) = \frac{\log(1 + 3x) - \log(1 - 2x)}{x} for x≠0x \ne 0, and the value of ff at x=0x = 0 is kk. For which kk is ff continuous at x=0x = 0?

Options for question 56

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AnswerC · 55

  1. log⁡(1+3x)x=3⋅log⁡(1+3x)3x→3\frac{\log(1 + 3x)}{x} = 3 \cdot \frac{\log(1 + 3x)}{3x} \to 3 and log⁡(1−2x)x=−2⋅log⁡(1−2x)−2x→−2\frac{\log(1 - 2x)}{x} = -2 \cdot \frac{\log(1 - 2x)}{-2x} \to -2.

  2. The limit is 3−(−2)=53 - (-2) = 5, so k=5k = 5.

Worth noting

1 takes the second limit as 2, losing the minus sign in 1−2x1 - 2x.

For which aa and bb is f(x)={sin⁡axx,x<04,x=0ebx−12x,x>0f(x) = \begin{cases} \frac{\sin ax}{x}, & x < 0 \\ 4, & x = 0 \\ \frac{e^{bx} - 1}{2x}, & x > 0 \end{cases} continuous at x=0x = 0?

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AnswerB · a=4, b=8a = 4,\ b = 8

  1. Left: sin⁡axx=a⋅sin⁡axax→a\frac{\sin ax}{x} = a \cdot \frac{\sin ax}{ax} \to a, and it must equal f(0)=4f(0) = 4, so a=4a = 4.

  2. Right: ebx−12x=b2⋅ebx−1bx→b2\frac{e^{bx} - 1}{2x} = \frac{b}{2} \cdot \frac{e^{bx} - 1}{bx} \to \frac{b}{2}, so b2=4\frac{b}{2} = 4 and b=8b = 8.

Worth noting

a=4, b=4a = 4,\ b = 4 forgets the 2 in 2x2x, which halves the right-hand limit.

Continuous functions

25 questions

At how many points in the open interval (−2,3)(-2, 3) is the greatest integer function f(x)=[x]f(x) = [x] not continuous?

Options for question 58

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AnswerB · 44

  1. [x][x] jumps by 1 at every integer and is continuous elsewhere.

  2. The integers strictly between −2-2 and 3 are −1,0,1,2-1, 0, 1, 2: 4 points.

Worth noting

6 counts −2-2 and 3 too, which the open interval leaves out.

At how many points is the signum function sgn⁡x\operatorname{sgn} x (11 for x>0x > 0, 00 at 0, −1-1 for x<0x < 0) not continuous?

Options for question 59

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AnswerB · 11

  1. Away from 0 it is constant, so continuous.

  2. At 0 it jumps from −1-1 to 1: 1 point.

Worth noting

0 forgets the jump at the origin.

What is the set of all points at which f(x)=x+1x2−5x+6f(x) = \frac{x + 1}{x^{2} - 5x + 6} is continuous?

Options for question 60

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AnswerD · R−{2,3}\mathbb{R} - \{2, 3\}

  1. A rational function is continuous at every point where its denominator is not 0.

  2. x2−5x+6=(x−2)(x−3)x^{2} - 5x + 6 = (x - 2)(x - 3) is 0 only at x=2x = 2 and x=3x = 3, so ff is continuous on R−{2,3}\mathbb{R} - \{2, 3\}.

Worth noting

R−{−1}\mathbb{R} - \{-1\} removes the zero of the numerator; only the zeros of the denominator leave the domain.

Let f(x)=x2f(x) = x^{2} and g(x)=cos⁡xg(x) = \cos x. Which of these functions is not continuous at every real xx?

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AnswerA · fg\frac{f}{g}

  1. Sums, differences, products and compositions of functions continuous everywhere are continuous everywhere.

  2. fg\frac{f}{g} needs cos⁡x≠0\cos x \ne 0: at x=π2x = \frac{\pi}{2} it is not even defined, so it is not continuous there.

Worth noting

g∘f=cos⁡(x2)g \circ f = \cos(x^{2}) is a composition of continuous functions, so it is continuous everywhere.

A courier charges C(w)=40+20[w]C(w) = 40 + 20[w] rupees for a parcel of weight ww kg, where 0<w<50 < w < 5 and [w][w] is the greatest integer not more than ww.

What does this courier charge for a parcel of 2.5 kg?

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AnswerB · ₹80

  1. [2.5]=2[2.5] = 2: only whole kilograms count.

  2. C(2.5)=40+20×2=80C(2.5) = 40 + 20 \times 2 = 80.

Worth noting

₹90 uses 20×2.520 \times 2.5, ignoring the greatest integer.

A courier charges C(w)=40+20[w]C(w) = 40 + 20[w] rupees for a parcel of weight ww kg, where 0<w<50 < w < 5 and [w][w] is the greatest integer not more than ww.

At how many points of (0,5)(0, 5) is this charge not continuous?

Options for question 63

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AnswerC · 44

  1. [w][w] jumps by 1 at each integer, so CC jumps by 20 there.

  2. The integers inside (0,5)(0, 5) are 1, 2, 3, 4: 4 points.

Worth noting

5 counts w=5w = 5 as well, which the interval leaves out.

At how many points in [0,3][0, 3] is f(x)=x−[x]f(x) = x - [x] not continuous?

Options for question 64

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AnswerC · 33

  1. x−[x]x - [x] drops from almost 1 to 0 at each integer, from the right side.

  2. On [0,3][0, 3] the break is at 1, 2 and 3; at 0 only the right side counts, and there it is continuous.

Worth noting

4 also counts 0, where the interval begins and only the right-hand limit matters.

Which of these functions are continuous at every real xx?

I. f(x)=∣x−1∣+∣x−2∣f(x) = \lvert x - 1 \rvert + \lvert x - 2 \rvert

II. f(x)=∣sin⁡x∣f(x) = \lvert \sin x \rvert

III. f(x)=[x]f(x) = [x], the greatest integer function

IV. f(x)=cos⁡∣x∣f(x) = \cos \lvert x \rvert

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AnswerD · I, II and IV only

  1. Sums and compositions of continuous functions are continuous: I, II and IV.

  2. III jumps at every integer.

Worth noting

A corner, as in ∣x−1∣\lvert x - 1 \rvert at 1, spoils differentiability, not continuity.

Which of the following statements are true?

I. If ff and gg are continuous at aa, so is f+gf + g.

II. If ff and gg are continuous at aa, so is fg\frac{f}{g}, whatever g(a)g(a) is.

III. If ff is continuous at aa, so is ∣f∣\lvert f \rvert.

IV. If ∣f∣\lvert f \rvert is continuous at aa, so is ff.

Options for question 66

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AnswerD · I and III only

  1. I and III are true. II needs g(a)≠0g(a) \ne 0.

  2. IV is false: f(x)=1f(x) = 1 for x≥0x \ge 0, −1-1 for x<0x < 0 has ∣f∣=1\lvert f \rvert = 1, continuous, while ff jumps at 0.

Worth noting

Taking the modulus can hide a jump, so IV runs the wrong way.

Read the two statements and choose the correct option.

I. f(x)=1xf(x) = \frac{1}{x} is continuous at every point of its domain.

II. f(x)=tan⁡xf(x) = \tan x is continuous at x=π2x = \frac{\pi}{2}.

Options for question 67

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AnswerA · Only I

  1. I is true: 0 is not in the domain, and at every other point 1x\frac{1}{x} is continuous.

  2. II is false: tan⁡π2\tan \frac{\pi}{2} is not defined, so continuity there cannot even be asked.

Worth noting

Continuity is checked only where the function is defined.

Read the two statements and choose the correct option.

I. If ff and gg are continuous everywhere, so is f∘gf \circ g.

II. f(x)=1x2−4f(x) = \frac{1}{x^{2} - 4} is continuous at x=2x = 2.

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AnswerA · Only I

  1. I: a composition of continuous functions is continuous. True.

  2. II: 2 is not in the domain, so ff has no value there to be continuous with. False.

Worth noting

A rational function is continuous on its domain, which leaves out the zeros of its denominator.

At how many points in the open interval (0,2)(0, 2) is f(x)=[2x]f(x) = [2x] not continuous?

Options for question 69

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AnswerC · 33

  1. [2x][2x] jumps when 2x2x is an integer, that is at x=12,1,32x = \frac{1}{2}, 1, \frac{3}{2} inside (0,2)(0, 2).

  2. 3 points.

Worth noting

1 counts only the integer x=1x = 1; the factor 2 doubles how often it jumps.

Read the two statements and choose the correct option.

I. f(x)=[x2]f(x) = [x^{2}] is continuous at x=0x = 0.

II. g(x)=[x]2g(x) = [x]^{2} is continuous at x=0x = 0.

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AnswerA · Only I

  1. I: for −1<x<1-1 < x < 1, 0≤x2<10 \le x^{2} < 1, so [x2]=0[x^{2}] = 0 all around 0, equal to f(0)f(0). True.

  2. II: just left of 0, [x]=−1[x] = -1, so [x]2=1[x]^{2} = 1; at 0 and to its right, [x]2=0[x]^{2} = 0. A jump, so false.

Worth noting

The order matters: in [x2][x^{2}] the input of the greatest integer function never crosses an integer near 0.

A courier charges C(w)=40+20[w]C(w) = 40 + 20[w] rupees for a parcel of weight ww kg, where 0<w<50 < w < 5 and [w][w] is the greatest integer not more than ww.

What is the left-hand limit of this charge at w=3w = 3?

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AnswerC · ₹80

  1. Just below 3 kg, [w]=2[w] = 2.

  2. So C(w)=40+20×2=80C(w) = 40 + 20 \times 2 = 80 there, and the left-hand limit is ₹80.

Worth noting

₹100 is C(3)C(3) itself; at exactly 3 kg the charge has already stepped up.

Let f(x)=1xf(x) = \frac{1}{x} and g(x)=1x−1g(x) = \frac{1}{x - 1}. What is the set of all points at which f∘gf \circ g is continuous?

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AnswerA · R−{1}\mathbb{R} - \{1\}

  1. f(g(x))f(g(x)) needs g(x)g(x) to be defined, so x≠1x \ne 1, and g(x)≠0g(x) \ne 0, which always holds.

  2. On that domain f(g(x))=x−1f(g(x)) = x - 1, a polynomial, so f∘gf \circ g is continuous on R−{1}\mathbb{R} - \{1\}.

Worth noting

R\mathbb{R} simplifies f(g(x))f(g(x)) to x−1x - 1 and forgets that g(1)g(1) is not defined.

Assertion (A): f(x)=x2+∣x∣f(x) = x^{2} + \lvert x \rvert is continuous at every real xx.

Reason (R): f(x)=x2+∣x∣f(x) = x^{2} + \lvert x \rvert is not differentiable at x=0x = 0.

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AnswerB · Both A and R are true, but R is not the correct explanation of A.

  1. A is true: x2x^{2} and ∣x∣\lvert x \rvert are continuous everywhere, and so is their sum.

  2. R is true as well: the slopes at 0 are −1-1 and 1. But a corner says nothing about continuity, so R does not explain A.

Worth noting

The continuity in A comes from the algebra of continuous functions, not from anything about slopes.

What is lim⁡x→0etan⁡2xx\lim_{x \to 0} e^{\frac{\tan 2x}{x}}?

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AnswerD · e2e^{2}

  1. eue^{u} is continuous, so the limit passes inside: the answer is eLe^{L} with L=lim⁡x→0tan⁡2xxL = \lim_{x \to 0} \frac{\tan 2x}{x}.

  2. tan⁡2xx=2⋅tan⁡2x2x→2\frac{\tan 2x}{x} = 2 \cdot \frac{\tan 2x}{2x} \to 2, so the limit is e2e^{2}.

Worth noting

ee takes tan⁡2xx→1\frac{\tan 2x}{x} \to 1, losing the 2 inside the tangent.

Which of these functions is not continuous at x=0x = 0?

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AnswerD · f(x)=x+[x]f(x) = x + [x]

  1. Just left of 0, [x]=−1[x] = -1, so x+[x]→−1x + [x] \to -1; at 0 and just right of it, [x]=0[x] = 0, so x+[x]→0x + [x] \to 0. A jump.

  2. x[x]→0=f(0)x[x] \to 0 = f(0) from both sides, because the factor xx tends to 0; x−∣x∣x - \lvert x \rvert and sin⁡∣x∣\sin \lvert x \rvert are built from continuous functions.

Worth noting

x[x]x[x] contains [x][x] too, but its factor xx squeezes the jump of [x][x] to nothing at 0.

At how many points in the open interval (−3,3)(-3, 3) is f(x)=sin⁡(π[x])f(x) = \sin(\pi[x]) not continuous?

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AnswerA · 00

  1. [x][x] is always an integer nn, and sin⁡(nπ)=0\sin(n\pi) = 0 for every integer nn.

  2. So f(x)=0f(x) = 0 for every xx, a constant function, continuous everywhere: 0 points.

Worth noting

5 counts each integer where [x][x] jumps, but sin⁡\sin gives the same value 0 at every one of them, so ff never jumps.

At how many points is f(x)={x+1,x<0x2,0≤x<22x,x≥2f(x) = \begin{cases} x + 1, & x < 0 \\ x^{2}, & 0 \le x < 2 \\ 2x, & x \ge 2 \end{cases} not continuous?

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AnswerB · 11

  1. Each piece is a polynomial, so only the joins at 0 and 2 can break.

  2. At 0: 0+1=10 + 1 = 1 on the left, 02=00^{2} = 0 on the right, a break. At 2: 22=4=2×22^{2} = 4 = 2 \times 2, no break. So 1 point.

Worth noting

2 assumes every join is a break; at x=2x = 2 the pieces meet.

At how many points in the open interval (0,2π)(0, 2\pi) is f(x)=[sin⁡x]f(x) = [\sin x] not continuous?

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AnswerC · 22

  1. [sin⁡x][\sin x] can change only where sin⁡x\sin x is an integer: at x=π2,π,3π2x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}.

  2. At π2\frac{\pi}{2}, ff is 0 on both sides but f=1f = 1 at the point; at π\pi it drops from 0 to −1-1; at 3π2\frac{3\pi}{2} it is −1-1 on both sides and at the point. So 2 points.

Worth noting

3 also counts 3π2\frac{3\pi}{2}; there sin⁡x\sin x only touches −1-1, so [sin⁡x]=−1[\sin x] = -1 on both sides and nothing breaks.

Which of the following statements are true?

I. If ff is continuous at aa and gg is not, then f+gf + g is not continuous at aa.

II. If neither ff nor gg is continuous at aa, then f+gf + g is not continuous at aa.

III. If ff is continuous at aa and gg is not, then fgfg is not continuous at aa.

IV. If ff is continuous at aa and f(a)≠0f(a) \ne 0, then 1f\frac{1}{f} is continuous at aa.

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AnswerB · I and IV only

  1. I is true: if f+gf + g were continuous at aa, so would be (f+g)−f=g(f + g) - f = g. IV is true: a quotient of continuous functions is continuous where the denominator is not 0.

  2. II is false: [x][x] and −[x]-[x] both jump at 0, but their sum is 0. III is false: xsgn⁡x=∣x∣x \operatorname{sgn} x = \lvert x \rvert is continuous at 0, though sgn⁡x\operatorname{sgn} x is not.

Worth noting

Two jumps can cancel in a sum, and a factor that is 0 at aa can cancel a jump in a product.

Let sgn⁡u\operatorname{sgn} u be 1, 0 or −1-1 as uu is positive, zero or negative. At how many points is f(x)=sgn⁡((x−1)2(x−4))f(x) = \operatorname{sgn}\left((x - 1)^{2}(x - 4)\right) not continuous?

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AnswerC · 22

  1. ff can break only where u=(x−1)2(x−4)u = (x - 1)^{2}(x - 4) is zero: at x=1x = 1 and x=4x = 4.

  2. At 4, uu changes sign, so ff jumps from −1-1 to 1. At 1, u<0u < 0 on both sides, so f→−1f \to -1, but f(1)=0f(1) = 0: a break as well. So 2 points.

Worth noting

1 counts only the sign change at 4; at the double root 1 the sign does not change, but the value 0 still differs from the limit −1-1.

Match each function in List I with the number of points of the open interval (0,2)(0, 2) at which it is not continuous (List II).

List IList II
A. [x][x]I. 3 points
B. [x2][x^{2}]II. No point
C. [x+12]\left[x + \frac{1}{2}\right]III. 1 point
D. [x2]\left[\frac{x}{2}\right]IV. 2 points
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AnswerD · A-III, B-I, C-IV, D-II

  1. [u][u] jumps exactly where uu crosses an integer. For 0<x<20 < x < 2, xx crosses 1, and x2x^{2} runs over (0,4)(0, 4), crossing 1, 2, 3 at x=1,2,3x = 1, \sqrt{2}, \sqrt{3}.

  2. x+12x + \frac{1}{2} runs over (12,52)\left(\frac{1}{2}, \frac{5}{2}\right) and crosses 1 and 2; x2\frac{x}{2} runs over (0,1)(0, 1) and crosses none.

Worth noting

[x2][x^{2}] also jumps at the irrational points 2\sqrt{2} and 3\sqrt{3}, where x2x^{2} is an integer.

A courier charges C(w)=40+20[w]C(w) = 40 + 20[w] rupees for a parcel of weight ww kg, where 0<w<50 < w < 5 and [w][w] is the greatest integer not more than ww.

Another courier charges D(w)=C(w)+20(w−[w])D(w) = C(w) + 20(w - [w]) rupees. At how many points of (0,5)(0, 5) is DD not continuous?

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AnswerA · 00

  1. D(w)=40+20[w]+20w−20[w]=40+20wD(w) = 40 + 20[w] + 20w - 20[w] = 40 + 20w.

  2. A linear function is continuous everywhere: 0 points.

Worth noting

4 assumes the jumps of CC survive; 20(w−[w])20(w - [w]) drops by 20 exactly where CC rises by 20, so they cancel.

Differentiability

43 questions

Which of these describes f(x)=∣x∣f(x) = \lvert x \rvert at x=0x = 0?

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AnswerD · Continuous but not differentiable

  1. ∣x∣→0=f(0)\lvert x \rvert \to 0 = f(0) from both sides: continuous.

  2. The left derivative is −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=ax = a is continuous at x=ax = a.

II. Every function continuous at x=ax = a is differentiable at x=ax = a.

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AnswerA · Only I

  1. I is true: differentiability forces the limit to equal the value.

  2. II is false: ∣x∣\lvert x \rvert 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] at x=1x = 1?

Options for question 85

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AnswerD · Neither continuous nor differentiable

  1. [x][x] jumps from 0 to 1 at x=1x = 1: not continuous.

  2. 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.

What is f(2)f(2) for this sum of two moduli?

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AnswerC · 22

  1. ∣2−1∣+∣2−3∣=1+1\lvert 2 - 1 \rvert + \lvert 2 - 3 \rvert = 1 + 1.

  2. f(2)=2f(2) = 2.

Worth noting

0 drops the modulus on 2−3=−12 - 3 = -1.

Let f(x)=∣x−1∣+∣x−3∣f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert.

At how many points is this sum of two moduli not continuous?

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AnswerA · 00

  1. Each modulus is continuous everywhere, and so is their sum.

  2. 0 points.

Worth noting

2 counts the corners, which spoil differentiability, not continuity.

At which value of xx is f(x)=∣2x−5∣f(x) = \lvert 2x - 5 \rvert not differentiable?

Options for question 88

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AnswerC · 52\frac{5}{2}

  1. A modulus has its corner where its inside is 0.

  2. 2x−5=02x - 5 = 0 at x=52x = \frac{5}{2}; there the slope changes from −2-2 to 2.

Worth noting

5 reads off the constant in 2x−52x - 5 without dividing by the 2 in front of xx.

Let f(x)=x∣x∣+∣x−1∣f(x) = x\lvert x \rvert + \lvert x - 1 \rvert.

What is f(−2)f(-2) for this function?

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AnswerB · −1-1

  1. x∣x∣=(−2)(2)=−4x\lvert x \rvert = (-2)(2) = -4 and ∣−2−1∣=3\lvert -2 - 1 \rvert = 3.

  2. f(−2)=−4+3=−1f(-2) = -4 + 3 = -1.

Worth noting

7 takes x∣x∣x\lvert x \rvert as x2=4x^{2} = 4; for negative xx it is −x2-x^{2}.

Which of these describes f(x)=∣x+2∣f(x) = \lvert x + 2 \rvert at x=0x = 0?

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AnswerD · Continuous and differentiable

  1. Near 0, x+2>0x + 2 > 0, so f(x)=x+2f(x) = x + 2.

  2. A linear function there: continuous and differentiable, with f′(0)=1f'(0) = 1.

Worth noting

The corner of ∣x+2∣\lvert x + 2 \rvert is at x=−2x = -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 not differentiable?

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AnswerC · 22

  1. Each modulus has a corner where its inside is 0: at x=0x = 0 and x=1x = 1.

  2. 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 is continuous at x=0x = 0.

Reason (R): f(x)=∣x∣f(x) = \lvert x \rvert is differentiable at x=0x = 0.

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AnswerC · A is true, but R is false.

  1. A is true: both one-sided limits are 0, equal to f(0)f(0).

  2. R is false: the one-sided derivatives are −1-1 and 1.

Which of these describes f(x)=x∣x∣f(x) = x\lvert x \rvert at x=0x = 0?

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AnswerA · Continuous and differentiable

  1. f(x)=−x2f(x) = -x^{2} for x<0x < 0 and x2x^{2} for x≥0x \ge 0; both pieces give 0 at 0.

  2. Both one-sided derivatives are 0: differentiable.

Worth noting

Multiplying ∣x∣\lvert x \rvert by xx flattens the corner: both slopes become 0.

At how many points is f(x)=∣x2−1∣f(x) = \lvert x^{2} - 1 \rvert not differentiable?

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AnswerC · 22

  1. x2−1=0x^{2} - 1 = 0 at x=±1x = \pm 1, where it changes sign.

  2. Each gives a corner: 2 points.

Worth noting

1 counts only x=1x = 1; x=−1x = -1 is a root too.

Which of these describes f(x)={x2,x≤12x−1,x>1f(x) = \begin{cases} x^{2}, & x \le 1 \\ 2x - 1, & x > 1 \end{cases} at x=1x = 1?

Options for question 95

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AnswerD · Continuous and differentiable

  1. Both pieces give 1 at x=1x = 1: continuous.

  2. The slopes are 2x=22x = 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)={x2,x≤1x,x>1f(x) = \begin{cases} x^{2}, & x \le 1 \\ x, & x > 1 \end{cases} at x=1x = 1?

Options for question 96

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AnswerA · Continuous but not differentiable

  1. Both pieces give 1 at x=1x = 1: continuous.

  2. 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 at x=3x = 3?

Options for question 97

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AnswerA · Continuous but not differentiable

  1. ∣x−3∣→0=f(3)\lvert x - 3 \rvert \to 0 = f(3) from both sides: continuous.

  2. Slopes −1-1 and 1: a corner, not differentiable.

Worth noting

∣x−3∣\lvert x - 3 \rvert is ∣x∣\lvert x \rvert moved 3 to the right, corner and all.

What are the left-hand and right-hand derivatives of f(x)=∣x−1∣+xf(x) = \lvert x - 1 \rvert + x at x=1x = 1?

Options for question 98

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AnswerC · 00 and 22

  1. Left of 1: f=1−x+x=1f = 1 - x + x = 1, slope 0.

  2. Right of 1: f=x−1+x=2x−1f = x - 1 + x = 2x - 1, slope 2.

Worth noting

−1-1 and 1 are the slopes of ∣x−1∣\lvert x - 1 \rvert alone; the +x+x adds 1 to each.

Assertion (A): f(x)=[x]f(x) = [x] is not differentiable at x=2x = 2.

Reason (R): f(x)=[x]f(x) = [x] is not continuous at x=2x = 2.

Options for question 99

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. R is true: [x][x] jumps from 1 to 2 at x=2x = 2.

  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 not differentiable?

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AnswerC · 22

  1. x(x−2)x(x - 2) changes sign at x=0x = 0 and x=2x = 2.

  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) gives a corner.

Let f(x)=∣x−1∣+∣x−3∣f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert.

At how many points is this sum of two moduli not differentiable?

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AnswerC · 22

  1. Corners at x=1x = 1 and x=3x = 3, where each inside is 0.

  2. 2 points.

Worth noting

1 counts only the midpoint, where nothing special happens.

Which of these describes f(x)=∣x2−9∣f(x) = \lvert x^{2} - 9 \rvert at x=3x = 3?

Options for question 102

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AnswerD · Continuous but not differentiable

  1. Both sides tend to 0 = f(3)f(3): continuous.

  2. Slopes −2x=−6-2x = -6 and 2x=62x = 6: a corner.

Worth noting

The modulus folds the curve up where x2−9x^{2} - 9 changes sign, leaving a corner.

Which of these describes f(x)=e∣x∣f(x) = e^{\lvert x \rvert} at x=0x = 0?

Options for question 103

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AnswerB · Continuous but not differentiable

  1. Both sides tend to e0=1=f(0)e^{0} = 1 = f(0): continuous.

  2. Slopes −1-1 on the left and 1 on the right: not differentiable.

Worth noting

∣x∣\lvert x \rvert inside a smooth function keeps its corner.

Assertion (A): No value of kk makes f(x)=∣x−2∣+kxf(x) = \lvert x - 2 \rvert + kx differentiable at x=2x = 2.

Reason (R): Adding kxkx adds the same kk to both one-sided derivatives at x=2x = 2.

Options for question 104

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. R is true: the one-sided derivatives at 2 are k−1k - 1 and k+1k + 1.

  2. They differ by 2 whatever kk 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) for f(x)={x2+3x,x≤15x−1,x>1f(x) = \begin{cases} x^{2} + 3x, & x \le 1 \\ 5x - 1, & x > 1 \end{cases}?

Options for question 105

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AnswerD · 55

  1. Values at 1: 1+3=41 + 3 = 4 and 5−1=45 - 1 = 4, so ff is continuous there.

  2. Slopes: 2x+3=52x + 3 = 5 on the left and 5 on the right; they agree, so f′(1)=5f'(1) = 5.

Worth noting

4 is f(1)f(1), the value at the join, not the slope.

Which of these functions are differentiable at x=0x = 0?

I. f(x)=∣x∣sin⁡xf(x) = \lvert x \rvert \sin x

II. f(x)=∣sin⁡x∣f(x) = \lvert \sin x \rvert

III. f(x)=cos⁡∣x∣f(x) = \cos \lvert x \rvert

IV. f(x)=x+∣x∣f(x) = x + \lvert x \rvert

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AnswerC · I and III only

  1. I is −xsin⁡x-x \sin x on the left and xsin⁡xx \sin x on the right, both with slope 0 at 0. III is cos⁡x\cos x, since cosine is even. Both are differentiable.

  2. II has slopes −cos⁡0=−1-\cos 0 = -1 and cos⁡0=1\cos 0 = 1, a corner. IV is 0 on the left and 2x2x on the right, with slopes 0 and 2.

Worth noting

A factor that is 0 at the corner, like sin⁡x\sin x in I, or an even outer function, like cos⁡\cos in III, smooths ∣x∣\lvert x \rvert out.

Let f(x)=x∣x∣+∣x−1∣f(x) = x\lvert x \rvert + \lvert x - 1 \rvert.

What is f′(0)f'(0) for this function?

Options for question 107

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AnswerB · −1-1

  1. Near 0, x−1<0x - 1 < 0, so ∣x−1∣=1−x\lvert x - 1 \rvert = 1 - x, with slope −1-1.

  2. x∣x∣x\lvert x \rvert has slope 2∣x∣=02\lvert x \rvert = 0 at 0, so f′(0)=0−1=−1f'(0) = 0 - 1 = -1.

Worth noting

1 takes the slope of ∣x−1∣\lvert x - 1 \rvert as +1+1; to the left of 1 it falls, with slope −1-1.

Match each function in List I with its left-hand and right-hand derivatives at x=0x = 0 in List II.

List IList II
A. ∣x∣\lvert x \rvertI. −2-2 and 22
B. x+∣x∣x + \lvert x \rvertII. −3-3 and −1-1
C. ∣x∣−2x\lvert x \rvert - 2xIII. 00 and 22
D. ∣sin⁡2x∣\lvert \sin 2x \rvertIV. −1-1 and 11
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AnswerB · A-IV, B-III, C-II, D-I

  1. Write each on the two sides of 0: ∣x∣\lvert x \rvert is −x-x then xx; x+∣x∣x + \lvert x \rvert is 0 then 2x2x; ∣x∣−2x\lvert x \rvert - 2x is −3x-3x then −x-x; ∣sin⁡2x∣\lvert \sin 2x \rvert is −sin⁡2x-\sin 2x then sin⁡2x\sin 2x.

  2. Their slopes at 0, left then right, are −1,1-1, 1; 0,20, 2; −3,−1-3, -1; −2,2-2, 2.

Worth noting

∣sin⁡2x∣\lvert \sin 2x \rvert is not ∣x∣\lvert x \rvert again: the 2 inside doubles both slopes.

At how many points in the open interval (0,3)(0, 3) is f(x)=[x]+∣x−2∣f(x) = [x] + \lvert x - 2 \rvert not differentiable?

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AnswerC · 22

  1. [x][x] jumps at 1 and 2, so ff is not even continuous there, and hence not differentiable.

  2. ∣x−2∣\lvert x - 2 \rvert has its corner at 2, a point already counted. So 2 points.

Worth noting

3 counts x=2x = 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.

At how many points is this function not differentiable?

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AnswerB · 11

  1. x∣x∣x\lvert x \rvert is −x2-x^{2} for x<0x < 0 and x2x^{2} for x≥0x \ge 0: slope 0 on both sides of 0, so no corner there.

  2. ∣x−1∣\lvert x - 1 \rvert has a corner at 1, and the smooth x∣x∣x\lvert x \rvert cannot remove it: 1 point.

Worth noting

2 counts x=0x = 0 too; x∣x∣x\lvert x \rvert is smooth there.

What are the left-hand and right-hand derivatives of f(x)=∣x2−4∣f(x) = \lvert x^{2} - 4 \rvert at x=2x = 2?

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AnswerD · −4-4 and 44

  1. Just left of 2, x2−4<0x^{2} - 4 < 0, so f=4−x2f = 4 - x^{2} with slope −2x=−4-2x = -4.

  2. Just right of 2, f=x2−4f = x^{2} - 4 with slope 2x=42x = 4.

Worth noting

−2-2 and 2 drop the factor x=2x = 2 in the derivative 2x2x.

For which aa and bb is f(x)={ax2+1,x≤1x+b,x>1f(x) = \begin{cases} ax^{2} + 1, & x \le 1 \\ x + b, & x > 1 \end{cases} differentiable at x=1x = 1?

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AnswerA · a=12, b=12a = \frac{1}{2},\ b = \frac{1}{2}

  1. Values must match at x=1x = 1: a+1=b+1a + 1 = b + 1.

  2. Slopes must match: 2a=12 a = 1, so a=12a = \frac{1}{2}, b=12b = \frac{1}{2}.

Worth noting

a=1, b=1a = 1,\ b = 1 joins the values but leaves slopes 2 and 1.

Match each function in List I with its behaviour at x=0x = 0 in List II.

List IList II
A. x2x^{2}I. Neither continuous nor differentiable
B. ∣x∣\lvert x \rvertII. Differentiable, with derivative 0
C. ∣x∣x\frac{\lvert x \rvert}{x} with value 0 at 0III. Differentiable, though built from ∣x∣\lvert x \rvert
D. ∣x∣3\lvert x \rvert^{3}IV. Continuous but not differentiable
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AnswerB · A-II, B-IV, C-I, D-III

  1. x2x^{2} has slope 0 at 0; ∣x∣\lvert x \rvert has a corner; ∣x∣x\frac{\lvert x \rvert}{x} jumps from −1-1 to 1.

  2. ∣x∣3\lvert x \rvert^{3} has one-sided slopes ±3x2=0\pm 3x^{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 not differentiable?

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AnswerD · 33

  1. The corners are at 1, 2 and 3.

  2. 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 aa and bb is f(x)={x3,x≤2ax+b,x>2f(x) = \begin{cases} x^{3}, & x \le 2 \\ ax + b, & x > 2 \end{cases} differentiable at x=2x = 2?

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AnswerC · a=12, b=−16a = 12,\ b = -16

  1. Values must match at x=2x = 2: 8=2a+b8 = 2 a + b.

  2. Slopes must match: 12=a12 = a, so a=12a = 12, b=−16b = -16.

Worth noting

a=4, b=0a = 4,\ b = 0 only joins the values at 2; the slope of x3x^{3} there is 3x2=123x^{2} = 12.

At how many points is f(x)=∣sin⁡x∣f(x) = \lvert \sin x \rvert on (0,2π)(0, 2\pi) not differentiable?

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AnswerB · 11

  1. sin⁡x=0\sin x = 0 inside (0,2π)(0, 2\pi) only at x=πx = \pi.

  2. There ∣sin⁡x∣\lvert \sin x \rvert has a corner: 1 point.

Worth noting

2 or 3 count 0 and 2π2\pi, which the open interval leaves out.

At how many points is f(x)=x∣x∣f(x) = x\lvert x \rvert not differentiable?

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AnswerA · 00

  1. f(x)=−x2f(x) = -x^{2} for x<0x < 0 and x2x^{2} for x≥0x \ge 0.

  2. Both slopes are 0 at 0, so it is differentiable everywhere: 0 points.

Worth noting

1 treats it like ∣x∣\lvert x \rvert; the extra factor xx removes the corner.

Let f(x)=∣x−1∣+∣x−3∣f(x) = \lvert x - 1 \rvert + \lvert x - 3 \rvert.

What is f′(2)f'(2) for this sum of two moduli?

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AnswerB · 00

  1. For 1<x<31 < x < 3: f(x)=(x−1)+(3−x)=2f(x) = (x - 1) + (3 - x) = 2, a constant.

  2. So f′(2)=0f'(2) = 0.

Worth noting

2 adds the slopes +1+1 and +1+1; between 1 and 3 the second modulus has slope −1-1.

Which of these describes f(x)=∣sin⁡x∣f(x) = \lvert \sin x \rvert at x=πx = \pi?

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AnswerD · Continuous but not differentiable

  1. Both sides tend to 0 = f(π)f(\pi): continuous.

  2. Slopes cos⁡π=−1\cos \pi = -1 on the left and −cos⁡π=1-\cos \pi = 1 on the right: a corner.

Worth noting

Where sin⁡x\sin x crosses 0, the modulus turns its slope round.

For which aa and bb is f(x)={ax+b,x<2x2−1,x≥2f(x) = \begin{cases} ax + b, & x < 2 \\ x^{2} - 1, & x \ge 2 \end{cases} differentiable at x=2x = 2?

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AnswerA · a=4, b=−5a = 4,\ b = -5

  1. Values match: 2a+b=32a + b = 3. Slopes match: a=4a = 4.

  2. So b=3−8=−5b = 3 - 8 = -5.

Worth noting

a=2, b=−1a = 2,\ b = -1 joins the values but takes the slope as 2 instead of 2x=42x = 4.

Read the two statements and choose the correct option.

I. f(x)=x2sin⁡1xf(x) = x^{2} \sin \frac{1}{x} for x≠0x \ne 0, with f(0)=0f(0) = 0, is differentiable at x=0x = 0.

II. g(x)=xsin⁡1xg(x) = x \sin \frac{1}{x} for x≠0x \ne 0, with g(0)=0g(0) = 0, is differentiable at x=0x = 0.

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AnswerA · Only I

  1. I: f(h)−f(0)h=hsin⁡1h→0\frac{f(h) - f(0)}{h} = h \sin \frac{1}{h} \to 0, so f′(0)=0f'(0) = 0. True.

  2. II: g(h)−g(0)h=sin⁡1h\frac{g(h) - g(0)}{h} = \sin \frac{1}{h}, which keeps swinging between −1-1 and 1 and has no limit. False, though gg is continuous at 0.

Worth noting

Continuity at 0 does not give differentiability: the difference quotient of gg never settles.

What are the left-hand and right-hand derivatives of f(x)=[x]sin⁡πxf(x) = [x] \sin \pi x at x=1x = 1?

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AnswerA · 00 and −π-\pi

  1. Just left of 1, [x]=0[x] = 0, so f(x)=0f(x) = 0 there: the left-hand derivative is 0.

  2. At 1 and just right of it, [x]=1[x] = 1, so f(x)=sin⁡πxf(x) = \sin \pi x, with slope πcos⁡π=−π\pi \cos \pi = -\pi at 1.

Worth noting

−π-\pi and −π-\pi treats [x][x] as 1 on both sides; to the left of 1 it is 0, which flattens ff.

Let f(x)=x∣x∣+∣x−1∣f(x) = x\lvert x \rvert + \lvert x - 1 \rvert.

What is the right-hand derivative of this function at x=1x = 1?

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AnswerD · 33

  1. For x>1x > 1, x∣x∣=x2x\lvert x \rvert = x^{2} and ∣x−1∣=x−1\lvert x - 1 \rvert = x - 1, so f(x)=x2+x−1f(x) = x^{2} + x - 1.

  2. Its slope at 1 is 2(1)+1=32(1) + 1 = 3.

Worth noting

1 is the left-hand derivative, the slope of x2+1−xx^{2} + 1 - x; the corner at 1 makes the two sides differ.

For which aa and bb is f(x)={ax2+b,x<11x,x≥1f(x) = \begin{cases} ax^{2} + b, & x < 1 \\ \frac{1}{x}, & x \ge 1 \end{cases} differentiable at x=1x = 1?

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AnswerD · a=−12, b=32a = -\frac{1}{2},\ b = \frac{3}{2}

  1. Values must match at x=1x = 1: a+b=1a + b = 1. Slopes must match: 2a=−1x2=−12a = -\frac{1}{x^{2}} = -1 at x=1x = 1.

  2. So a=−12a = -\frac{1}{2} and b=1−a=32b = 1 - a = \frac{3}{2}.

Worth noting

a=12, b=12a = \frac{1}{2},\ b = \frac{1}{2} takes the slope of 1x\frac{1}{x} at 1 as +1+1; it is −1x2=−1-\frac{1}{x^{2}} = -1.

At how many points is f(x)=∣(x−1)2(x−3)∣f(x) = \lvert (x - 1)^{2}(x - 3) \rvert not differentiable?

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AnswerB · 11

  1. u=(x−1)2(x−3)u = (x - 1)^{2}(x - 3) is 0 at x=1x = 1 and x=3x = 3. At 3 it changes sign, so ∣u∣\lvert u \rvert has a corner there, with slopes −4-4 and 4.

  2. At 1 the double root keeps u≤0u \le 0 on both sides, so f=−uf = -u near 1, a smooth curve with slope 0. So 1 point.

Worth noting

2 counts the double root 1 as well; there uu touches 0 without changing sign, so the modulus does not fold it.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.