Continuity and Differentiability Class 12 NCERT Solutions Chapter 5 Ex 5.1

Concept used. A function f is said to be continuous at a point c in its domain if _{x → c} f(x) = f(c). For f to be continuous at c the limit must exist (i.e. left-hand limit equals right-hand limit) and must coincide with the value f(c). A polynomial of degree 1 takes its limit by direct substitution.

1. At x = 0. f(0) = 5(0) - 3 = -3. Compute the limit: _{x → 0} f(x) = _{x → 0}(5x - 3) = 5(0) - 3 = -3. Since _{x → 0} f(x) = f(0) = -3, f is continuous at x = 0.
2. At x = -3. f(-3) = 5(-3) - 3 = -18. _{x → -3}(5x - 3) = 5(-3) - 3 = -18. Hence _{x → -3} f(x) = f(-3), so f is continuous at x = -3.
3. At x = 5. f(5) = 5(5) - 3 = 22. _{x → 5}(5x - 3) = 5(5) - 3 = 22. Hence _{x → 5} f(x) = f(5), so f is continuous at x = 5.

f(x) = 5x - 3 is continuous at each of x = 0, -3, 5.

Concept used. f is continuous at x = c iff _{x → c} f(x) = f(c). For a polynomial, the limit at any point equals direct substitution.

1. Function value. f(3) = 2(3)² - 1 = 2 × 9 - 1 = 18 - 1 = 17.
2. Limit at x = 3. _{x → 3} f(x) = _{x → 3}(2x² - 1) = 2(3)² - 1 = 17.
3. Conclude. Since _{x → 3} f(x) = 17 = f(3), f is continuous at x = 3.

f(x) = 2x² - 1 is continuous at x = 3.

Concept used. A function is continuous on a set S if it is continuous at every point of S. Polynomials are continuous on R; a rational function p(x)/q(x) is continuous wherever q(x) ≠ 0; the modulus function |x| is continuous on R (and hence so is any horizontal translate |x - a|).

1. (a) f(x) = x - 5. This is a polynomial of degree 1, so it is continuous on R. To check explicitly at any c ∈ R: _{x → c}(x - 5) = c - 5 = f(c).
2. (b) f(x) = 1/(x - 5), x ≠ 5. The domain is R 5. For any c ≠ 5 the denominator c - 5 ≠ 0, so _{x → c} 1x - 5 = 1c - 5 = f(c). Hence f is continuous at every point of its domain R 5.
3. (c) f(x) = x² - 25x + 5, x ≠ -5. For x ≠ -5, factor the numerator: f(x) = (x - 5)(x + 5)x + 5 = x - 5. Thus on its domain R -5, f coincides with the polynomial x - 5, which is continuous everywhere. So f is continuous at every point of its domain.
4. (d) f(x) = |x - 5|. Write piecewise: f(x) = cases x - 5, & x ≥ 5, -(x - 5), & x < 5. cases For c > 5: f(c) = c - 5 and _{x → c}(x - 5) = c - 5; continuous.
   For c < 5: f(c) = 5 - c and _{x → c}(5 - x) = 5 - c; continuous.
   At c = 5: f(5) = 0. Left limit _{x → 5^-}(5 - x) = 0; right limit _{x → 5⁺}(x - 5) = 0. Both equal f(5), so f is continuous at 5 as well. Hence f is continuous on R.

(a) continuous on R;   (b) continuous on R 5;   (c) continuous on R -5;   (d) continuous on R.

Concept used. For positive integer n, xⁿ is the n-fold product of the identity function with itself. The product of continuous functions is continuous, and the identity function x ↦ x is continuous on R, hence xⁿ is continuous on R. We must show _{x → n} xⁿ = nⁿ.

1. Compute the value. f(n) = nⁿ.
2. Compute the limit. Using the algebra of limits, _{x → n} xⁿ = (_{x → n} x)ⁿ = nⁿ. (Here we used that the limit of a product is the product of the limits, applied n times.)
3. Conclude. Since _{x → n} f(x) = nⁿ = f(n), f is continuous at x = n.

f(x) = xⁿ is continuous at x = n for every positive integer n.

Concept used. For a piecewise-defined function, continuity at the join point c requires _{x → c^-} f(x) = _{x → c⁺} f(x) = f(c). At interior points of one piece, continuity follows from the continuity of that piece.

1. At x = 0. Since 0 ≤ 1, we are inside the piece f(x) = x, which is a polynomial. Hence _{x → 0} f(x) = _{x → 0} x = 0 = f(0). So f is continuous at x = 0.
2. At x = 1 (the join point). f(1) = 1 (using the first piece, since 1 ≤ 1).
   Left limit _{x → 1^-} f(x) = _{x → 1^-} x = 1 Right limit _{x → 1⁺} f(x) = _{x → 1⁺} 5 = 5 Since the left and right limits differ (1 ≠ 5), _{x → 1} f(x) does not exist; f is not continuous at x = 1.
3. At x = 2. Since 2 > 1, we are inside the piece f(x) = 5, a constant. Hence _{x → 2} f(x) = 5 = f(2). So f is continuous at x = 2.

Continuous at x = 0 and x = 2; discontinuous at x = 1.

Concept used. Each piece is a polynomial, hence continuous wherever it is defined. The only potentially discontinuous point is the join x = 2, where we apply the LHL/RHL/value test.

1. For x < 2, f(x) = 2x + 3 is a polynomial, continuous.
2. For x > 2, f(x) = 2x - 3 is a polynomial, continuous.
3. At x = 2: f(2) = 2(2) + 3 = 7.
   _{x → 2^-} f(x) = _{x → 2^-}(2x + 3) = 2(2) + 3 = 7 _{x → 2⁺} f(x) = _{x → 2⁺}(2x - 3) = 2(2) - 3 = 1 Since the left limit (7) and the right limit (1) are unequal, _{x → 2} f(x) does not exist; f is discontinuous at x = 2.

f is discontinuous only at x = 2.

Concept used. Inside each open piece the formula is continuous (modulus, polynomial). Discontinuity can only happen at the join points x = -3 and x = 3. We check each with the LHL/RHL/value test.

1. At x = -3. For x ≤ -3 we use f(x) = |x| + 3 = -x + 3 (since x ≤ -3 < 0). f(-3) = -(-3) + 3 = 6. Left limit _{x → -3^-} f(x) = _{x → -3^-}(-x + 3) = -(-3) + 3 = 6 Right limit _{x → -3⁺} f(x) = _{x → -3⁺}(-2x) = -2(-3) = 6 All three equal 6, so f is continuous at x = -3.
2. At x = 3. For x ≥ 3, f(x) = 6x + 2, so f(3) = 6(3) + 2 = 20. Left limit _{x → 3^-} f(x) = _{x → 3^-}(-2x) = -2(3) = -6 Right limit _{x → 3⁺} f(x) = _{x → 3⁺}(6x + 2) = 6(3) + 2 = 20 Left limit -6 ≠ 20, so f is discontinuous at x = 3.
3. Interior points. For x < -3, |x| + 3 = -x + 3 is a polynomial, continuous. For -3 < x < 3, -2x is a polynomial, continuous. For x > 3, 6x + 2 is a polynomial, continuous.

f is discontinuous only at x = 3.

Concept used. For x > 0, |x| = x, so |x|/x = 1. For x < 0, |x| = -x, so |x|/x = -1. Thus f(x) = cases 1, & x > 0, -1, & x < 0, 0, & x = 0. cases

1. At x = 0. f(0) = 0. Left limit _{x → 0^-} f(x) = -1 Right limit _{x → 0⁺} f(x) = 1 Since LHL = -1 ≠ 1 = RHL, _{x → 0} f(x) does not exist. Hence f is discontinuous at x = 0.
2. For x > 0. f(x) = 1 is constant, continuous.
3. For x < 0. f(x) = -1 is constant, continuous.

f is discontinuous only at x = 0.

Concept used. For x < 0, |x| = -x, so x|x| = x-x = -1. Therefore f(x) = cases -1, & x < 0, -1, & x ≥ 0. cases That is, f(x) = -1 for every real x.

1. For any c ∈ R, _{x → c} f(x) = -1 = f(c).
2. Hence f is continuous at every real c; there is no point of discontinuity.

f is continuous everywhere on R; there is no point of discontinuity.

Concept used. Polynomials are continuous inside each piece. Test the join x = 1 with LHL/RHL/value.

1. f(1) = 1 + 1 = 2 (using the first piece, x ≥ 1).
2. _{x → 1^-} f(x) = _{x → 1^-}(x² + 1) = 1² + 1 = 2
3. _{x → 1⁺} f(x) = _{x → 1⁺}(x + 1) = 1 + 1 = 2
4. LHL = RHL = f(1) = 2, so f is continuous at x = 1. Together with continuity inside each piece, f is continuous on all of R.

No point of discontinuity; f is continuous on R.

Concept used. Polynomials are continuous on R, so each piece is continuous on its open part. Only the join x = 2 needs checking.

1. f(2) = 2³ - 3 = 8 - 3 = 5.
2. _{x → 2^-} f(x) = _{x → 2^-}(x³ - 3) = 2³ - 3 = 5
3. _{x → 2⁺} f(x) = _{x → 2⁺}(x² + 1) = 2² + 1 = 5
4. All three equal 5, hence f is continuous at x = 2. Together with the interior continuity, f is continuous on R.

No point of discontinuity; f is continuous everywhere.

Concept used. Polynomials are continuous; check only the join x = 1.

1. f(1) = 1¹⁰ - 1 = 0.
2. _{x → 1^-} f(x) = 1¹⁰ - 1 = 0
3. _{x → 1⁺} f(x) = 1² = 1
4. LHL = 0 ≠ 1 = RHL, so _{x → 1} f(x) does not exist; f is discontinuous at x = 1.

f is discontinuous only at x = 1.

Concept used. Each piece is a polynomial, continuous inside its open part. Test x = 1.

1. f(1) = 1 + 5 = 6.
2. _{x → 1^-} f(x) = 1 + 5 = 6
3. _{x → 1⁺} f(x) = 1 - 5 = -4
4. LHL = 6 ≠ -4 = RHL, so f is not continuous at x = 1; the function is not continuous on R.

No, f is not a continuous function; it is discontinuous at x = 1.

Concept used. Domain is [0, 10]. Each piece is a constant, hence continuous in its open part. Test the join points x = 1 and x = 3 with LHL/RHL/value.

1. At x = 1. f(1) = 3 (from the piece 0 ≤ x ≤ 1).
   LHL. (using f = 3 to the left). _{x → 1^-} f(x) = 3 RHL _{x → 1⁺} f(x) = 4 LHL ≠ RHL, hence discontinuous at x = 1.
2. At x = 3. f(3) = 5 (from the piece 3 ≤ x ≤ 10).
   LHL _{x → 3^-} f(x) = 4 RHL _{x → 3⁺} f(x) = 5 LHL ≠ RHL, hence discontinuous at x = 3.
3. All other points (interior of the three intervals): the function is constant, hence continuous.

f is discontinuous at x = 1 and x = 3; continuous everywhere else on [0, 10].

Concept used. Each piece is a polynomial, continuous inside open intervals. Test the joins x = 0 and x = 1.

1. At x = 0. f(0) = 0.
   LHL _{x → 0^-} 2x = 0 RHL _{x → 0⁺} 0 = 0 All three equal 0, hence continuous at x = 0.
2. At x = 1. f(1) = 0 (using the second piece, 0 ≤ x ≤ 1).
   LHL _{x → 1^-} 0 = 0 RHL _{x → 1⁺} 4x = 4(1) = 4 LHL ≠ RHL, hence discontinuous at x = 1.
3. All other points: f is polynomial within open pieces, hence continuous.

f is discontinuous only at x = 1; continuous everywhere else.

Concept used. Each piece is constant or linear, hence continuous in its open part. Check x = -1 and x = 1.

1. At x = -1. f(-1) = -2.
   LHL _{x → -1^-}(-2) = -2 RHL _{x → -1⁺} 2x = 2(-1) = -2 All equal -2, so continuous at x = -1.
2. At x = 1. f(1) = 2(1) = 2 (from the second piece, -1 < x ≤ 1).
   LHL _{x → 1^-} 2x = 2(1) = 2 RHL _{x → 1⁺} 2 = 2 All equal 2, so continuous at x = 1.
3. All other points are inside open pieces of polynomials, hence continuous.

f is continuous at every point of R; no discontinuity.

Concept used. Continuity at the join x = 3 requires LHL = RHL = f(3).

1. f(3) = 3a + 1 (using the first piece, x ≤ 3).
2. LHL. _{x → 3^-}(ax + 1) = 3a + 1
3. RHL. _{x → 3⁺}(bx + 3) = 3b + 3
4. Equate LHL and RHL: 3a + 1 = 3b + 3 3a - 3b = 2 a - b = 23 a = b + 23.

a = b + 23, i.e. 3a - 3b = 2.

Concept used. Match LHL, RHL and function value at x = 0. At x = 1, the function is single-valued by the formula 4x+1, continuous as a polynomial.

1. At x = 0. f(0) = λ(0² - 2 · 0) = 0.
   LHL _{x → 0^-} λ(x² - 2x) = λ(0 - 0) = 0 RHL _{x → 0⁺}(4x + 1) = 4(0) + 1 = 1 For continuity, LHL = RHL: 0 = 1, which is impossible for any λ.
   Therefore there is no value of λ making f continuous at x = 0.
2. At x = 1. Since 1 > 0, f(x) = 4x + 1 in a neighbourhood of 1, a polynomial. Hence _{x → 1} f(x) = 4(1) + 1 = 5 = f(1). f is continuous at x = 1 for every λ.

No λ makes f continuous at x = 0. At x = 1, f is continuous for every value of λ.

Concept used. The greatest integer function [x] is defined as the unique integer n such that n ≤ x < n + 1. For any integer n: [x] = cases n - 1, & n - 1 ≤ x < n, n, & n ≤ x < n + 1. cases

1. Let n be an arbitrary integer. Evaluate g(n) = n - [n] = n - n = 0.
2. LHL at n. For x slightly less than n, [x] = n - 1, so g(x) = x - (n-1). Hence _{x → n^-} g(x) = _{x → n^-}(x - (n - 1)) = n - (n - 1) = 1.
3. RHL at n. For x slightly greater than n (but still < n + 1), [x] = n, so g(x) = x - n. Hence _{x → n⁺} g(x) = _{x → n⁺}(x - n) = n - n = 0.
4. LHL = 1 ≠ 0 = RHL, so _{x → n} g(x) does not exist; therefore g is discontinuous at every integer n.

g(x) = x - [x] is discontinuous at every integral point n ∈ Z.

Concept used. Sums, differences and products of continuous functions are continuous. x² is a polynomial; sin x is continuous on R; constants are continuous. Hence f is continuous on R.

1. Value. f(π) = π² - sin π + 5 = π² - 0 + 5 = π² + 5.
2. Limit. Using algebra of limits, _{x → π} f(x) = _{x → π} x² - _{x → π} sin x + 5 = π² - sin π + 5 = π² + 5.
3. _{x → π} f(x) = f(π), so f is continuous at x = π.

Yes, f is continuous at x = π with value π² + 5.

Concept used. sin x and cos x are continuous on R. The sum, difference and product of two continuous functions is continuous.

1. (a) sin x + cos x is a sum of two continuous functions, hence continuous on R.
2. (b) sin x - cos x is a difference of two continuous functions, hence continuous on R.
3. (c) sin x · cos x is a product of two continuous functions, hence continuous on R.

All three functions are continuous on the whole of R.

Concept used. sin x and cos x are continuous on R. A reciprocal 1/g is continuous wherever g is continuous and non-zero. The quotient g/h is continuous wherever both g and h are continuous and h ≠ 0.

1. Cosine cos x. For any c ∈ R, _{x → c} cos x = cos c (a standard limit). Hence cos x is continuous on R.
2. Cosecant csc x = 1/sin x. sin x = 0 x = nπ, n ∈ Z. On the domain R nπ : n ∈ Z, csc x is the reciprocal of a continuous, non-zero function, hence continuous.
3. Secant sec x = 1/cos x. cos x = 0 x = (2n+1)π/2, n ∈ Z. On the domain R (2n+1)π/2 : n ∈ Z, sec x is continuous.
4. Cotangent cot x = cos x/sin x. Same exclusion as csc: continuous on R nπ : n ∈ Z.

cos x: continuous on R. csc x, cot x: continuous on R nπ. sec x: continuous on R (2n+1)π/2.

Concept used. On x < 0, sin x/x is the quotient of two continuous functions with non-zero denominator, hence continuous. On x ≥ 0, x + 1 is a polynomial, continuous. Check the join x = 0.

1. f(0) = 0 + 1 = 1 (using the second piece, x ≥ 0).
2. LHL at 0. The standard limit _{x → 0} sin xx = 1 holds for both one-sided approaches. Hence _{x → 0^-} sin xx = 1.
3. RHL at 0. _{x → 0⁺}(x + 1) = 0 + 1 = 1.
4. All three equal 1, so f is continuous at x = 0. Combined with continuity inside each open piece, f is continuous on R.

No point of discontinuity; f is continuous on R.

Concept used. For x ≠ 0, f(x) = x² sin(1/x) is a product of x² (polynomial, continuous) and sin(1/x) (composition of sin and the continuous-on-R 0 function 1/x). Hence f is continuous on R 0. At x = 0 we use the Sandwich (Squeeze) theorem: since -1 ≤ sin t ≤ 1 for all t, we have -x² ≤ x² sin1x ≤ x² (x ≠ 0).

1. As x → 0, both -x² → 0 and x² → 0. By the Squeeze theorem, _{x → 0} x² sin1x = 0.
2. f(0) = 0 by definition.
3. Since _{x → 0} f(x) = 0 = f(0), f is continuous at x = 0. Together with continuity for x ≠ 0, f is continuous on R.

Yes, f is a continuous function on R.

Concept used. For x ≠ 0, sin x - cos x is the difference of two continuous functions, hence continuous. At x = 0, use LHL/RHL/value.

1. f(0) = -1.
2. Limit at 0. _{x → 0}(sin x - cos x) = sin 0 - cos 0 = 0 - 1 = -1.
3. _{x → 0} f(x) = -1 = f(0), so f is continuous at x = 0.
4. Outside 0, f is the difference of continuous functions, continuous on R 0.

f is continuous everywhere on R.

Concept used. Continuity at π/2 requires _{x → π/2} f(x) = f(π/2) = 3. We compute the limit using the substitution x = π/2 + h, then use cos(π/2 + h) = -sin h and the standard limit sin h/h → 1.

1. Let h = x - π/2, so x → π/2 becomes h → 0. Then π - 2x = π - 2(π2 + h) = -2h, cos x = cos(π2 + h) = -sin h.
2. Substitute: _{x → π/2} k cos xπ - 2x = _{h → 0} k(-sin h)-2h = _{h → 0} k sin h2h = k2 _{h → 0} sin hh = k2 · 1 = k2.
3. Continuity demands k2 = 3, i.e. k = 6.

k = 6.

Concept used. Continuity at x = 2 requires LHL = RHL = f(2).

1. f(2) = k(2)² = 4k.
2. LHL. _{x → 2^-} k x² = k(2)² = 4k
3. RHL. _{x → 2⁺} 3 = 3
4. Equate: 4k = 3 k = 34.

k = 34.

Concept used. Continuity at π requires LHL = RHL = f(π).

1. f(π) = kπ + 1.
2. LHL. _{x → π^-}(kx + 1) = kπ + 1
3. RHL. _{x → π⁺} cos x = cos π = -1
4. Equate LHL = RHL: kπ + 1 = -1 kπ = -2 k = -2π.

k = -2π.

Concept used. Continuity at 5 requires LHL = RHL = f(5).

1. f(5) = 5k + 1.
2. LHL. _{x → 5^-}(kx + 1) = 5k + 1
3. RHL. _{x → 5⁺}(3x - 5) = 3(5) - 5 = 10
4. Equate: 5k + 1 = 10 5k = 9 k = 95.

k = 95.

Concept used. Continuity at the two join points x = 2 and x = 10 gives two linear equations in a and b.

1. At x = 2. f(2) = 5. RHL: _{x → 2⁺}(ax + b) = 2a + b. Continuity requires 2a + b = 5. ...(i)
2. At x = 10. f(10) = 21. LHL: _{x → 10^-}(ax + b) = 10a + b. Continuity requires 10a + b = 21. ...(ii)
3. Subtract (i) from (ii): (10a + b) - (2a + b) = 21 - 5 8a = 16 a = 2.
4. Back-substitute into (i): 2(2) + b = 5 b = 5 - 4 = 1.

a = 2, b = 1.

Concept used. The composition of two continuous functions is continuous: if g is continuous at c and h is continuous at g(c), then h ∘ g is continuous at c.

1. Let g(x) = x² and h(t) = cos t. Then f = h ∘ g, i.e. f(x) = h(g(x)) = cos(x²).
2. g(x) = x² is a polynomial, hence continuous on R.
3. h(t) = cos t is continuous on R.
4. By the composition theorem, f = h ∘ g is continuous on R.

f(x) = cos(x²) is continuous on R.

Concept used. f = h ∘ g where g(x) = cos x (continuous on R) and h(t) = |t| (continuous on R). Composition of continuous functions is continuous.

1. g(x) = cos x is continuous on R.
2. h(t) = |t| is continuous on R (as shown by the LHL = RHL argument at 0, and identity elsewhere).
3. Therefore f(x) = |cos x| = h(g(x)) is continuous on R.

f(x) = |cos x| is continuous everywhere on R.

Concept used. Composition of continuous functions is continuous. Here the inner function is g(x) = |x| and the outer is h(t) = sin t.

1. g(x) = |x| is continuous on R (right-/left-limits at 0 both equal 0 = g(0); elsewhere |x| agrees with the polynomial ± x).
2. h(t) = sin t is continuous on R.
3. By composition, f(x) = sin |x| = h(g(x)) is continuous on R.

sin |x| is a continuous function on R.

Concept used. |x| and |x + 1| are each continuous on R, so their difference is continuous on R.

1. g(x) = |x| is continuous on R.
2. h(x) = |x + 1| is the composition of x ↦ x + 1 (continuous) and |·| (continuous), so continuous on R.
3. The difference f(x) = g(x) - h(x) is continuous on R as the difference of two continuous functions.
4. Hence f has no point of discontinuity.

f(x) = |x| - |x + 1| has no point of discontinuity; it is continuous on R.