JEE Mathematics Continuity and Differentiability MCQs Set 21

Answer: (a) is −2

Explanation:
1. For x ≠ 1, f′(x) = 1/(x − 1) for both x > 1 and x < 1.
2. f′(1/2) = 1/(−1/2) = −2.

Teacher's Note:
1. d/dx ln|u| = u′/u holds on both sides of the zero of u.

Answer: (a) μ = 0

Explanation:
1. |x|² = x² is smooth.
2. μ|x| has right derivative μ and left derivative −μ at 0, which agree only when μ = 0.

Teacher's Note:
1. Only the odd power |x| creates a corner.

Answer: (c) non-existent

Explanation:
1. For x slightly below 1, [x] = 0 so f = 0; at and above 1, f = 1/x.
2. The function jumps at x = 1 (0 on the left, 1 on the right), so it is not continuous and f′(1) does not exist.

Teacher's Note:
1. Differentiability needs continuity first.

Answer: (a) 2

Explanation:
1. For x slightly greater than π/4, cos 2x < 0 so f = −cos 2x.
2. f′ = 2 sin 2x = 2 sin(π/2) = 2.

Teacher's Note:
1. Remove the modulus using the sign just to the right of the point.

Answer: (d) none of these

Explanation:
1. At x = 1: f(1⁻) = 0 − 0 = 0, f(1) = 0, f(1⁺) = 1 − 1 = 0, so f is continuous at 1; hence the first option is false.
2. At x = 0: f(0⁻) = 0 − 1 = −1 ≠ f(0) = 0, and at x = 2 f(2⁻) = 3 − 1 = 2 ≠ 0, so f is discontinuous at 0, 2, 3, …
3. f is also discontinuous at non-integers like √2, so the second and third options are false.

Teacher's Note:
1. Test x = 0, 1, 2 and √2 one by one.
2. f(x) = 0 on (0, 1), so f is continuous there.

Answer: (c) 0

Explanation:
1. x and |x − x²| are both continuous on [−1, 1].
2. Their difference is continuous, so there are no points of discontinuity.

Teacher's Note:
1. A combination of continuous functions is continuous.

Answer: (d) none of these

Explanation:
1. At x = √3: √(1 + 3) = 2 and √3·√3 − 1 = 2, so f is continuous.
2. At x = 4: left limit 4√3 − 1 ≈ 5.93, but f(4) = 4, so f is discontinuous.
3. At x = 5: left limit [x] → 4 and |1 − 5| = 4, so f is continuous.
4. Exactly one point (x = 4) of discontinuity, which is none of the options.

Teacher's Note:
1. Check the junctions √3, 4 and 5.
2. [x] is constant on [4, 5).

Answer: (a) 0

Explanation:
1. The integrand t sin(1/t) is bounded and tends to 0 as t → 0, so it is integrable on [0, x].
2. An integral of an integrable function is continuous in its upper limit, so f has no discontinuities.

Teacher's Note:
1. Integral functions of bounded integrands are continuous (indeed Lipschitz).

Answer: (a) f(x) = x for at least one 0 ≤ x ≤ 1

Explanation:
1. Let g(x) = f(x) − x. Then g(0) = f(0) ≥ 0 and g(1) = f(1) − 1 ≤ 0.
2. By the intermediate value theorem g(x) = 0 for some x, so f(x) = x for at least one x.

Teacher's Note:
1. Apply the intermediate value property to f(x) − x.
2. f(x) + x = 0 can hold only if f(0) = 0, which need not be true.

Answer: (b) 10

Explanation:
1. A continuous function that takes only rational values cannot pass through irrational values, so by the intermediate value theorem it is constant.
2. Hence f(1.5) = f(2) = 10.

Teacher's Note:
1. Between two different rational values a continuous function would take irrational values.

Answer: (a) non-existent

Explanation:
1. h(x) = [sin x + 1]. At x = π/2, sin x = 1 so h(π/2) = 2, but for x near π/2 (x ≠ π/2), sin x < 1 and h(x) = 1.
2. The function is discontinuous at π/2, so h′(π/2) does not exist.

Teacher's Note:
1. Compute h(x) = [1 + sin x] near the maximum of sine.

Answer: (b) 3

Explanation:
1. |x − 1/2| fails to be differentiable at x = 1/2, and |x − 1| at x = 1.
2. tan x is not defined at x = π/2 ≈ 1.57, which lies in [0, 2], so f is not differentiable there either.
3. Total: 3 points.

Teacher's Note:
1. Count corners of the moduli and the point where the tangent function is undefined.

Answer: (c) {0}

Explanation:
1. At x = 1: [f(x) − f(1)]/(x − 1) = (x − 1)cos(1/(x − 1)) − (|x| − 1)/(x − 1) → 0 − 1, so f is differentiable at 1.
2. |x| is not differentiable at 0, and the other term is smooth there.
3. The set is {0}.

Teacher's Note:
1. Check x = 1 by the definition of the derivative.
2. The only corner is that of |x| at 0.

Answer: (b) 0

Explanation:
1. Let h = x − π/2: 1 − sin x = 1 − cos h ≈ h²/2 and sin 2x = −sin 2h ≈ −2h.
2. The ratio ≈ −h/4 → 0.

Teacher's Note:
1. Shift the variable to h = x − π/2 and expand.

Answer: (a) 0

Explanation:
1. cos(sin x) = 1 − sin²x/2 + sin⁴x/24 and cos x = 1 − x²/2 + x⁴/24 − …; using sin²x ≈ x² − x⁴/3 gives cos(sin x) − cos x ≈ x⁴/6.
2. Dividing by x² gives ≈ x²/6 → 0, so a = 0.

Teacher's Note:
1. Expand both cosines to fourth order.
2. The x² terms cancel.

Answer: (c) \( -\frac12 \)

Explanation:
1. λ = lim as x → π/4 of (1 − tan x)/(4x − π) = −sec²(π/4)/4 (L'Hôpital).
2. = −2/4 = −1/2.

Teacher's Note:
1. The limit is a derivative: 1 − tan x has derivative −sec²x.

Answer: (b) 1

Explanation:
1. Let h = x − π/2: ln f = ln(cos h)/(−2h) ≈ (−h²/2)/(−2h) = h/4 → 0.
2. So f(x) → e⁰ = 1.

Teacher's Note:
1. Take the logarithm of the 1^∞ form.