Exercise 5.1 answers: Continuity and Differentiability

Open NCERT p. 116One way to think about it

1. To prove f is continuous at a point, show that the limit of f as x approaches that point equals f at that point.
2. f(x) = 5x − 3 is a polynomial, so its limit at any point is found by simply substituting the point.
3. At x = 0: limit = 5(0) − 3 = −3, and f(0) = −3. They match, so f is continuous at x = 0.
4. At x = −3: limit = 5(−3) − 3 = −18, and f(−3) = −18. They match, so f is continuous at x = −3.
5. At x = 5: limit = 5(5) − 3 = 22, and f(5) = 22. They match, so f is continuous at x = 5.
6. Since the limit equals the value at each of these three points, f is continuous there.

In shortf(x) = 5x − 3 is continuous at x = 0, at x = −3 and at x = 5, because at each point the limit of f equals f's value there.

Watch this explained “Three routine checks, one shape”, 4:00 into A function is continuous where its limit and its value agree

Open NCERT p. 116Matches NCERT’s answer

1. f(x) = 2x² − 1 is a polynomial, so its limit at x = 3 is found by substitution.
2. limit as x → 3 of f(x) = 2(3)² − 1 = 2(9) − 1 = 17.
3. f(3) = 2(3)² − 1 = 17.
4. The limit (17) equals f(3) (17), so f is continuous at x = 3.

AnswerYes, f is continuous at x = 3.

Watch this explained “Three routine checks, one shape”, 4:00 into A function is continuous where its limit and its value agree

Open NCERT p. 116Matches NCERT’s answer

(a) f(x) = x − 5

1. f(x) = x − 5 is a polynomial, defined for every real number.
2. At any point c, limit as x → c of f(x) = c − 5 = f(c).
3. The limit always equals the value, so f is continuous everywhere.

AnswerYes, f(x) = x − 5 is continuous (at every real number).

(b) f(x) = 1/(x−5), x ≠ 5

1. f(x) = 1/(x−5) is defined for every real number except x = 5.
2. At any point c in the domain (c ≠ 5), limit as x → c of f(x) = 1/(c−5) = f(c).
3. So the limit equals the value at every point where f is defined.

AnswerYes, f is continuous at every point of its domain, i.e. for all x ≠ 5.

(c) f(x) = (x²−25)/(x+5), x ≠ −5

1. For x ≠ −5, (x²−25)/(x+5) = (x−5)(x+5)/(x+5) = x − 5, so f behaves like the polynomial x − 5 on its whole domain.
2. At any point c ≠ −5, limit as x → c of f(x) = c − 5 = f(c).
3. So the limit equals the value at every point of the domain.

AnswerYes, f is continuous at every point of its domain, i.e. for all x ≠ −5.

(d) f(x) = |x − 5|

1. f(x) = |x − 5| is defined for every real number.
2. For x ≥ 5, f(x) = x − 5, a polynomial. For x < 5, f(x) = 5 − x, also a polynomial.
3. At x = 5 itself: left-hand limit = 5 − 5 = 0, right-hand limit = 5 − 5 = 0, and f(5) = 0. All three match.
4. So the limit equals the value at every real number.

AnswerYes, f(x) = |x − 5| is continuous (at every real number).

Watch this explained “The function that makes the point”, 10:42 into A function is continuous where its limit and its value agree

Open NCERT p. 116One way to think about it

1. To prove: f(x) = xⁿ is continuous at x = n.
2. f(x) = xⁿ is a polynomial (a single power of x), so it is defined for every real number.
3. By the rule for limits of a product, limit as x → n of xⁿ = (limit as x → n of x)ⁿ = nⁿ.
4. f(n) = nⁿ.
5. The limit (nⁿ) equals f(n) (nⁿ), so f is continuous at x = n.

In shortf(x) = xⁿ is continuous at x = n, because the limit of xⁿ as x approaches n equals nⁿ, which is exactly f(n).

Watch this explained “Every polynomial, in one line”, 6:51 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. At x = 0: nearby, x ≤ 1 holds on both sides, so f(x) = x there. limit = 0, and f(0) = 0. They match, so f is continuous at x = 0.
2. At x = 1: the left-hand limit uses f(x) = x, giving 1. The right-hand limit uses f(x) = 5, giving 5. The two sides disagree, so the limit does not exist at x = 1, and f is not continuous there.
3. At x = 2: nearby, x > 1 holds on both sides, so f(x) = 5 there. limit = 5, and f(2) = 5. They match, so f is continuous at x = 2.

AnswerContinuous at x = 0: yes. Continuous at x = 1: no. Continuous at x = 2: yes.

Watch this explained “Two rules, and one input to argue about”, 0:00 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. Away from x = 2, f is one of two straight-line pieces, both of which are continuous, so the only place a break can happen is the join, x = 2.
2. f(2) = 2(2) + 3 = 7 (using the first piece, since it applies at x = 2).
3. Left-hand limit at x = 2 = 2(2) + 3 = 7. Right-hand limit at x = 2 = 2(2) − 3 = 1.
4. The left-hand and right-hand limits (7 and 1) do not match, so f is discontinuous at x = 2.

Answerf is discontinuous only at x = 2.

Watch this explained “Two rules, and one input to argue about”, 0:00 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. Each of the three pieces is continuous on its own interval, so only the two joins, x = −3 and x = 3, need checking.
2. At x = −3: f(−3) = |−3| + 3 = 6. Left-hand limit = |−3| + 3 = 6. Right-hand limit = −2(−3) = 6. All three agree, so f is continuous at x = −3.
3. At x = 3: f(3) = 6(3) + 2 = 20 (using the third piece, since it applies at x = 3). Left-hand limit = −2(3) = −6. Right-hand limit = 6(3) + 2 = 20.
4. The left-hand limit (−6) does not match f(3) (20), so f is discontinuous at x = 3.

Answerf is discontinuous only at x = 3.

Watch this explained “A case split that stitches shut”, 6:07 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. Away from x = 0, |x|/x is 1 when x is positive and −1 when x is negative, so f is continuous everywhere except possibly at x = 0.
2. Left-hand limit at x = 0: as x approaches 0 from the left, |x|/x = −1, so the limit is −1.
3. Right-hand limit at x = 0: as x approaches 0 from the right, |x|/x = 1, so the limit is 1.
4. The left-hand and right-hand limits (−1 and 1) do not match, so f is discontinuous at x = 0, whatever value is assigned there.

Answerf is discontinuous only at x = 0.

Watch this explained “Move the value, and watch nothing happen”, 1:43 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. For x < 0, |x| = −x, so x/|x| = x/(−x) = −1. So the first piece also always equals −1.
2. This means f(x) = −1 for every real x — the two pieces are secretly the same constant function.
3. A constant function has the same value everywhere, so its limit at any point equals its value there.
4. f is continuous at every real number, so it has no points of discontinuity.

Answerf has no points of discontinuity — it is continuous everywhere (it is really just the constant function −1).

Watch this explained “Four that look alarming and are clean”, 13:49 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. Away from x = 1, f is one of two continuous pieces, so only the join at x = 1 needs checking.
2. f(1) = 1 + 1 = 2 (using the first piece, since it applies at x = 1).
3. Left-hand limit at x = 1 = 1² + 1 = 2. Right-hand limit at x = 1 = 1 + 1 = 2.
4. The left-hand limit, right-hand limit, and value (all 2) agree, so f is continuous at x = 1 too.

Answerf has no points of discontinuity — it is continuous everywhere.

Watch this explained “Four items of exactly the same shape”, 2:40 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. Away from x = 2, f is one of two continuous polynomial pieces, so only the join at x = 2 needs checking.
2. f(2) = 2³ − 3 = 5 (using the first piece, since it applies at x = 2).
3. Left-hand limit at x = 2 = 2³ − 3 = 5. Right-hand limit at x = 2 = 2² + 1 = 5.
4. The left-hand limit, right-hand limit, and value (all 5) agree, so f is continuous at x = 2 too.

Answerf has no points of discontinuity — it is continuous everywhere.

Watch this explained “Four items of exactly the same shape”, 2:40 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. Away from x = 1, f is one of two continuous polynomial pieces, so only the join at x = 1 needs checking.
2. f(1) = 1¹⁰ − 1 = 0 (using the first piece, since it applies at x = 1).
3. Left-hand limit at x = 1 = 1¹⁰ − 1 = 0. Right-hand limit at x = 1 = 1² = 1.
4. The left-hand limit (0) does not match the right-hand limit (1), so f is discontinuous at x = 1.

Answerf is discontinuous only at x = 1.

Watch this explained “Four items of exactly the same shape”, 2:40 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 116Matches NCERT’s answer

1. The function: f(x) = x + 5 if x ≤ 1, and f(x) = x − 5 if x > 1.
2. Away from x = 1, f is one of two continuous polynomial pieces, so only the join at x = 1 needs checking.
3. f(1) = 1 + 5 = 6 (using the first piece, since it applies at x = 1).
4. Left-hand limit at x = 1 = 1 + 5 = 6. Right-hand limit at x = 1 = 1 − 5 = −4.
5. The left-hand limit (6) does not match the right-hand limit (−4), so f is discontinuous at x = 1, hence not a continuous function.

AnswerNo, f is not continuous (it breaks at x = 1).

Watch this explained “Four items of exactly the same shape”, 2:40 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 117Matches NCERT’s answer

1. Each piece is a constant, so it is continuous on its own interval; only the two joins, x = 1 and x = 3, need checking.
2. At x = 1: f(1) = 3 (first piece applies at x = 1). Left-hand limit = 3. Right-hand limit = 4. These do not match, so f is discontinuous at x = 1.
3. At x = 3: f(3) = 5 (third piece applies at x = 3). Left-hand limit = 4. Right-hand limit = 5. These do not match, so f is discontinuous at x = 3.
4. At every other point, f is a constant piece, so it is continuous there.

Answerf is discontinuous at x = 1 and x = 3, and continuous at every other point of [0, 10].

Watch this explained “Two rules, and one input to argue about”, 0:00 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 117Matches NCERT’s answer

1. Each piece is a polynomial, continuous on its own interval, so only the two joins, x = 0 and x = 1, need checking.
2. At x = 0: f(0) = 0 (middle piece applies at x = 0). Left-hand limit = 2(0) = 0. Right-hand limit = 0. All three agree, so f is continuous at x = 0.
3. At x = 1: f(1) = 0 (middle piece applies at x = 1). Left-hand limit = 0. Right-hand limit = 4(1) = 4. These do not match, so f is discontinuous at x = 1.
4. At every other point, f is a single polynomial piece, so it is continuous there.

Answerf is discontinuous only at x = 1; it is continuous everywhere else, including at x = 0.

Watch this explained “A case split that stitches shut”, 6:07 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 117Matches NCERT’s answer

1. Each piece is a constant or a straight line, so f is continuous inside each piece's interval; only the two joins, x = −1 and x = 1, need checking.
2. At x = −1: f(−1) = −2 (the first piece applies, since −1 ≤ −1). Left-hand limit = −2. Right-hand limit = 2(−1) = −2. All three agree, so f is continuous at x = −1.
3. At x = 1: f(1) = 2(1) = 2 (the middle piece applies, since −1 < 1 ≤ 1). Left-hand limit = 2(1) = 2. Right-hand limit = 2. All three agree, so f is continuous at x = 1.
4. So f has no break at either join, and none anywhere else.

Answerf has no points of discontinuity: it is continuous at every real number, including x = −1 and x = 1.

Watch this explained “Four that look alarming and are clean”, 13:49 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 117Matches NCERT’s answer

1. Away from x = 3, each piece is a straight line, so continuity only needs checking at the join, x = 3.
2. f(3) = 3a + 1 (using the first piece, since it applies at x = 3).
3. Left-hand limit at x = 3 = 3a + 1. Right-hand limit at x = 3 = 3b + 3.
4. For continuity, the two one-sided limits must be equal: 3a + 1 = 3b + 3, which simplifies to 3a − 3b = 2.

Answer3a − 3b = 2 (equivalently, a = b + 2/3).

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 117Checked by computer

1. f(0) = λ(0² − 2·0) = λ·0 = 0, whatever λ is, because the bracket itself is 0 at x = 0.
2. Left-hand limit at x = 0 = λ(0² − 2·0) = 0 as well, for the same reason — it does not depend on λ.
3. Right-hand limit at x = 0 = 4(0) + 1 = 1, which is fixed and also does not depend on λ.
4. Since the left-hand limit is always 0 and the right-hand limit is always 1, no choice of λ can make them equal, so f can never be continuous at x = 0.
5. For x = 1, the whole neighbourhood of x = 1 lies in x > 0, where f(x) = 4x + 1, a polynomial. Left-hand limit = right-hand limit = f(1) = 5.
6. So f is continuous at x = 1 for every value of λ.

AnswerNo value of λ makes f continuous at x = 0 (the two sides give 0 and 1 no matter what λ is). At x = 1, f is continuous for every λ.

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 117One way to think about it

1. Let n be any integer. g(n) = n − [n] = n − n = 0, since [n] = n for an integer n.
2. Left-hand limit at x = n: for x just below n (say x = n − h, h → 0⁺), [x] = n − 1, so g(x) = (n − h) − (n − 1) = 1 − h → 1.
3. Right-hand limit at x = n: for x just above n (say x = n + h, h → 0⁺), [x] = n, so g(x) = (n + h) − n = h → 0.
4. The left-hand limit (1) and the right-hand limit (0) do not match, so the limit of g at x = n does not exist, and g is discontinuous at x = n.
5. Since n was any integer, g is discontinuous at every integral point.

In shortg(x) = x − [x] is discontinuous at every integer, because the left-hand limit is always 1 and the right-hand limit is always 0 there, so the two never agree.

Watch this explained “Infinitely many, and none of them worse”, 10:58 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 117Matches NCERT’s answer

1. x² is a polynomial and sin x is a standard trigonometric function; both are continuous at every real number, including x = π.
2. The difference and sum of continuous functions is continuous, so f(x) = x² − sin x + 5 is continuous at x = π.
3. Check directly: limit as x → π of f(x) = π² − sin π + 5 = π² − 0 + 5 = π² + 5.
4. f(π) = π² − sin π + 5 = π² + 5, which matches the limit.

AnswerYes, f is continuous at x = π.

Watch this explained “Four parts, and one condition”, 0:45 into Combining continuous functions, and why a composite survives too

Open NCERT p. 117One way to think about it

(a) f (x) = sin x + cos x

1. sin x is continuous for every real x.
2. cos x is continuous for every real x.
3. The sum of two continuous functions is continuous.

In shortf(x) = sin x + cos x is continuous for every real x.

(b) f (x) = sin x − cos x

1. sin x is continuous for every real x.
2. cos x is continuous for every real x.
3. The difference of two continuous functions is continuous.

In shortf(x) = sin x − cos x is continuous for every real x.

(c) f (x) = sin x . cos x

1. sin x is continuous for every real x.
2. cos x is continuous for every real x.
3. The product of two continuous functions is continuous.

In shortf(x) = sin x . cos x is continuous for every real x.

Watch this explained “Four parts, and one condition”, 0:45 into Combining continuous functions, and why a composite survives too

Open NCERT p. 117One way to think about it

1. cos x is continuous at every real x (the book proves this alongside sin x).
2. cosec x = 1/sin x is defined only where sin x ≠ 0, that is x ≠ nπ (n an integer). At every such x it is a quotient of two continuous functions with a non-zero denominator, so it is continuous there.
3. sec x = 1/cos x is defined only where cos x ≠ 0, that is x ≠ (2n + 1)π/2. By the same quotient rule it is continuous at every such x.
4. cot x = cos x/sin x is defined only where sin x ≠ 0, that is x ≠ nπ, and by the quotient rule it is continuous at every such x.
5. The excluded points nπ and (2n + 1)π/2 are not in the domains of these functions, so they are not points of discontinuity: the functions are simply not defined there. So each function is continuous at every point of its domain.

In shortAll four are continuous functions: cos x is continuous at every real x; cosec x and cot x are continuous at every x ≠ nπ; sec x is continuous at every x ≠ (2n + 1)π/2 (n an integer). Each is continuous wherever it is defined.

Watch this explained “The other four ratios”, 9:29 into Combining continuous functions, and why a composite survives too

Open NCERT p. 117Matches NCERT’s answer

1. The function: f(x) = (sin x)/x if x < 0, and f(x) = x + 1 if x ≥ 0.
2. Away from x = 0, each branch is built from continuous functions (sin x, x, and constants), so f is continuous everywhere except possibly at x = 0.
3. Left-hand limit at x = 0: as x → 0⁻, sin x / x → 1.
4. Right-hand limit at x = 0: as x → 0⁺, x + 1 → 1.
5. Value at x = 0: since 0 ≥ 0, f(0) = 0 + 1 = 1.
6. All three numbers are 1, so f is continuous at x = 0 too.

Answerf has no points of discontinuity — it is continuous for every real x.

Watch this explained “Four that look alarming and are clean”, 13:49 into Where the standard functions stay unbroken, and where they jump

Open NCERT p. 117Matches NCERT’s answer

1. The function: f(x) = x² sin(1/x) if x ≠ 0, and f(0) = 0.
2. For x ≠ 0, f is a product of continuous functions (x² and sin(1/x), and 1/x is defined there), so f is continuous there.
3. sin(1/x) always lies between −1 and 1, so |f(x)| = x² |sin(1/x)| ≤ x².
4. As x → 0, x² → 0, so by squeezing f(x) between −x² and x², the limit of f(x) as x → 0 is also 0.
5. f(0) is given as 0, which matches this limit.

AnswerYes, f is continuous everywhere, including at x = 0.

Watch this explained “Where both theorems stop”, 13:09 into Combining continuous functions, and why a composite survives too

Open NCERT p. 118Matches NCERT’s answer

1. The function: f(x) = sin x − cos x if x ≠ 0, and f(0) = −1.
2. For x ≠ 0, f is sin x − cos x, a difference of continuous functions, so it is continuous there.
3. As x → 0, sin x − cos x → sin 0 − cos 0 = 0 − 1 = −1.
4. f(0) is given as −1, which matches this limit.

AnswerYes, f is continuous at x = 0 (and so continuous everywhere).

Watch this explained “One equation, three demands”, 0:53 into A function is continuous where its limit and its value agree

Open NCERT p. 118Matches NCERT’s answer

1. The function: f(x) = k cos x/(π − 2x) if x ≠ π/2, and f(π/2) = 3; the point is x = π/2.
2. Write x = π/2 + h, so h → 0 as x → π/2.
3. cos x becomes cos(π/2 + h) = −sin h, and π − 2x becomes π − 2(π/2 + h) = −2h.
4. So the formula becomes k(−sin h)/(−2h) = (k/2)(sin h / h).
5. As h → 0, sin h / h → 1, so the limit of f(x) as x → π/2 is k/2.
6. For continuity, this limit must equal f(π/2) = 3, so k/2 = 3.

Answerk = 6.

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 118Matches NCERT’s answer

1. The function: f(x) = kx² if x ≤ 2, and f(x) = 3 if x > 2; the point is x = 2.
2. Since x = 2 falls in the first rule, f(2) = k(2)² = 4k.
3. As x → 2⁺, f(x) = 3, so the right-hand limit is 3.
4. For continuity, f(2) must equal this limit: 4k = 3.

Answerk = 3/4.

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 118Matches NCERT’s answer

1. The function: f(x) = kx + 1 if x ≤ π, and f(x) = cos x if x > π; the point is x = π.
2. Since x = π falls in the first rule, f(π) = kπ + 1.
3. As x → π⁺, f(x) = cos x → cos π = −1.
4. For continuity, kπ + 1 = −1.

Answerk = −2/π.

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 118Matches NCERT’s answer

1. The function: f(x) = kx + 1 if x ≤ 5, and f(x) = 3x − 5 if x > 5; the point is x = 5.
2. Since x = 5 falls in the first rule, f(5) = 5k + 1.
3. As x → 5⁺, f(x) = 3x − 5 → 3(5) − 5 = 10.
4. For continuity, 5k + 1 = 10.

Answerk = 9/5.

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 118Matches NCERT’s answer

1. The function: f(x) = 5 if x ≤ 2, ax + b if 2 < x < 10, and 21 if x ≥ 10.
2. At x = 2: the left-hand value is 5, and the right-hand limit from the middle rule is a(2) + b. These must be equal: 2a + b = 5.
3. At x = 10: the limit from the middle rule as x → 10⁻ is a(10) + b, and the right-hand value is 21. These must be equal: 10a + b = 21.
4. Subtracting the first equation from the second: 8a = 16, so a = 2.
5. Substituting back into 2a + b = 5: 4 + b = 5, so b = 1.

Answera = 2, b = 1.

Watch this explained “Choosing a constant, and failing to”, 13:35 into A function is continuous where its limit and its value agree

Open NCERT p. 118One way to think about it

1. x² is a polynomial, so it is continuous for every real x.
2. cos t is continuous for every real t.
3. f(x) = cos(x²) feeds the output of x² into cos, so it is a composite of two continuous functions.

In shortf(x) = cos(x²) is continuous for every real x.

Watch this explained “The closure the algebra cannot reach”, 10:44 into Combining continuous functions, and why a composite survives too

Open NCERT p. 118One way to think about it

1. cos x is continuous for every real x.
2. The modulus function |t| is continuous for every real t.
3. f(x) = |cos x| feeds the output of cos x into the modulus function, so it is a composite of two continuous functions.

In shortf(x) = |cos x| is continuous for every real x.

Watch this explained “The closure the algebra cannot reach”, 10:44 into Combining continuous functions, and why a composite survives too

Open NCERT p. 118Checked by computer

1. |x| is continuous for every real x.
2. sin t is continuous for every real t.
3. sin|x| feeds the output of |x| into sin, so it is a composite of two continuous functions and is continuous everywhere.

AnswerYes, sin|x| is continuous for every real x.

Watch this explained “The closure the algebra cannot reach”, 10:44 into Combining continuous functions, and why a composite survives too

Open NCERT p. 118Matches NCERT’s answer

1. |x| is continuous for every real x (a composite of x and the modulus function).
2. |x + 1| is continuous for every real x, for the same reason.
3. f(x) is the difference of two continuous functions, so it is continuous everywhere.

Answerf has no points of discontinuity — it is continuous for every real x.

Watch this explained “Four parts, and one condition”, 0:45 into Combining continuous functions, and why a composite survives too