JEE Mathematics Continuity and Differentiability MCQs Set 22

Answer: (d) for no value of f(0)

Explanation:
1. As x → 0, sin(1/x) oscillates between −1 and 1 and has no limit.
2. So no choice of f(0) gives continuity.

Teacher's Note:
1. Continuity at a point needs the limit to exist.

Answer: (d) (1, 1)

Explanation:
1. Continuity at 1 needs lim as x → 1⁻ of f = p − q equal to 2 (the right limit and f(1) are both 2).
2. (2, 0), (1, −1) and (4, 2) all give p − q = 2.
3. (1, 1) gives 0, so continuity fails.

Teacher's Note:
1. Equate the left limit p − q with the common value 2.

Answer: (d) b = −1, c = 1

Explanation:
1. Continuity at 1: 1 + b + c = 1, so b + c = 0.
2. Derivative match at 1: 1 = 2 + b, so b = −1.
3. Then c = 1.

Teacher's Note:
1. Two conditions at the junction: equal values and equal slopes.

Answer: (a) All of 1, 2 and 3

Explanation:
1. Continuity at 2: a + 2b = e².
2. Equal slopes at 2: b = e².
3. Then a = e² − 2e² = −e², so a + b = 0. All three statements hold.

Teacher's Note:
1. Match both value and slope at x = 2.

Answer: (b) 1 and 4 only

Explanation:
1. f(x) = |x| for x ∈ [−π, π], extended periodically (a zig-zag wave).
2. It is continuous everywhere.
3. At x = 0 and at x = π the slopes change between +1 and −1, so f is not differentiable there.

Teacher's Note:
1. Sketch the zig-zag: the graph is continuous with corners at multiples of π.

Answer: (a) 1 and 4 only

Explanation:
1. f(x) = 0 for x ≥ 0 and 2x for x < 0.
2. Both pieces are continuous and meet at 0, so f is continuous.
3. Right slope 0 and left slope 2, so f is not differentiable at 0.

Teacher's Note:
1. Write the function piecewise.
2. Compare the slopes on either side.

Answer: (c) 1 and 3 only

Explanation:
1. For x slightly below 2: |2 − x| → 0 and (2 + x) = 4, so f(2 − 0) = 4 = f(2).
2. For x slightly above 2: (2 + x) = 5, so f(2 + 0) = 5 ≠ 4.
3. f is discontinuous at 2 and therefore not differentiable: statements 1 and 3.

Teacher's Note:
1. The ceiling function jumps at x = 2 for the term (2 + x).

Answer: (d) 1, 3 and 4 only

Explanation:
1. For x < 0, x² > x so h = x; for 0 ≤ x ≤ 1, x² ≤ x so h = x²; for x > 1, x² > x so h = x.
2. h is continuous with corners at x = 0 and x = 1, so h′ = 1 for x > 1 and h is not differentiable at exactly two points.

Teacher's Note:
1. Compare x and x² on each interval: x < 0, 0 < x < 1, x > 1.

Answer: (a) continuous but not differentiable

Explanation:
1. e^(−|x|) is continuous.
2. For x > 0 the slope at 0⁺ is −1 and for x < 0 the slope at 0⁻ is +1, so it is not differentiable at 0.

Teacher's Note:
1. Compare the one-sided derivatives.

Answer: (a) 3 and 4 only

Explanation:
1. x² − 3x + 2 = (x − 1)(x − 2) changes sign at 1 and 2, giving corners there.
2. cos|x| = cos x is smooth, so x = 0 and x = −1 are fine.

Teacher's Note:
1. Identify where the expression inside the modulus changes sign.

Answer: (d) 3 and 4 only

Explanation:
1. At x = 1: values 1 and 1 but slopes 6 and 3, so continuous and not differentiable.
2. At x = −1: x⁶ → 1 from the left but x³ = −1 at and to the right of −1, so f is discontinuous there.

Teacher's Note:
1. Compare values and slopes at x = ±1.

Answer: (b) 1 and 4 only

Explanation:
1. Both pieces tend to 0 at x = 0, so f is continuous.
2. The right slope is cos 0 = 1 and the left slope is −cos 0 = −1, so f is not differentiable.

Teacher's Note:
1. f(x) = |sin x|.

Answer: (a) 1 and 3 only

Explanation:
1. Each |x|ⁿ is continuous at 0, so f is continuous.
2. For n = 1 the term a₁|x| has slopes a₁ and −a₁ at 0, so a₁ ≠ 0 spoils differentiability (statement 2 is false). For n = 2 and for n ≥ 3, |x|ⁿ is differentiable at 0.
3. If every odd-indexed coefficient is zero, f is a polynomial in x² and is differentiable: statement 3 is true.

Teacher's Note:
1. Only the n = 1 term |x| fails to be differentiable at 0.

Answer: (d) 2 and 3 only

Explanation:
1. f(x) = 2x for x ≥ 0 and 0 for x < 0.
2. It is continuous everywhere with a corner at 0 (slopes 0 and 2).

Teacher's Note:
1. Write the function piecewise.

Answer: (c) 1 and 3 only

Explanation:
1. For 0 < x < 1, xⁿ → 0 and f = 1.
2. At x = 1, f = 0. For x > 1, xⁿ → ∞ and f = −1.
3. The function jumps at 1, so it is discontinuous and not differentiable at 1.

Teacher's Note:
1. Evaluate the limit separately for x < 1, x = 1 and x > 1.

Answer: (b) f(x) is continuous at x = 0

Explanation:
1. For |x| small, 0 ≤ tan²x < 1, so [tan²x] = 0 near x = 0.
2. f is constant 0 near 0: the limit exists, f is continuous and f′(0) = 0.

Teacher's Note:
1. Greatest integer of a small positive number is 0.

Answer: (c) 1 and 3 only

Explanation:
1. On [1, 2), f = ln 1 = 0; on [2, 3), f = ln 2. At x = 2 the graph jumps from 0 to ln 2.
2. At x = 3 it jumps from ln 2 to ln 3.
3. These two breaks are the only places where there is no definite tangent.

Teacher's Note:
1. [x] is constant on each unit interval, so f is flat between the jumps.