JEE Mathematics Continuity and Differentiability MCQs Set 23

Answer: (c) \( (\sqrt2,\,1-\sqrt3) \)

Explanation:
1. At x = 1: 2/a = a, so a² = 2 and a = ±√2.
2. At x = √2: a = (2b² − 4b)/(√2)³ = (2b² − 4b)/(2√2), i.e. 2b² − 4b = 2√2 a.
3. For a = √2: b² − 2b − 2 = 0, so b = 1 ± √3. For a = −√2: b² − 2b + 2 = 0 has no real root.
4. So a = √2 and b = 1 ± √3; among the options, (√2, 1 − √3) is correct.

Teacher's Note:
1. Equate the left and right limits at each break point.

Answer: (d) Statement 1 is true, Statement 2 is false

Explanation:
1. For x slightly less than 2, [x] = 1, so x/[x] → 2 and √(6 − x) → 2.
2. By the sandwich theorem lim_x→2⁻ f(x) = 2, so Statement 1 is true.
3. But f(2) = 1 ≠ 2, so f is not continuous at 2 and Statement 2 is false.

Teacher's Note:
1. Squeeze the one-sided limit, then compare it with f(2).

Answer: (b) continuous as well as differentiable

Explanation:
1. |f(0)| ≤ 0 gives f(0) = 0, and |f(x)| ≤ x² gives f(x) → 0, so f is continuous at 0.
2. |f(x) − f(0)|/|x| = |f(x)|/|x| ≤ |x| → 0, so f'(0) = 0 exists.

Teacher's Note:
1. Differentiability at 0 follows from the bound |f(x)/x| ≤ |x|.

Answer: (d) a = 0, b = 0, and c is any real number

Explanation:
1. |sin x| has a corner at 0 (slopes ∓1), so it needs a = 0.
2. e^|x| has slopes ∓1 at 0 (right derivative 1, left derivative −1), so it needs b = 0.
3. |x|³ is differentiable at 0 (derivative 0), so c is arbitrary.

Teacher's Note:
1. Each absolute value of an odd-order term creates a corner unless its coefficient vanishes.

Answer: (b) 0

Explanation:
1. |f(x) − f(y)|/|x − y| ≤ |x − y| → 0 as y → x, so f'(x) = 0 for all x.
2. f is constant, equal to f(0) = 0, so f(1) = 0.

Teacher's Note:
1. An exponent greater than 1 on |x − y| forces zero derivative.

Answer: (c) differentiable at x = 1

Explanation:
1. Near x = 0 and on (−1, 0) and (0, 1): 0 ≤ x sin πx < 1, so f ≡ 0 there. So f is continuous at 0 and in (−1, 0) and differentiable in (−1, 1).
2. At x = 1 write x = 1 + h: x sin πx = −(1 + h) sin πh, which is negative for h > 0 (so f = −1) and positive for h < 0 (so f = 0).
3. f jumps at x = 1, so it is not continuous, hence not differentiable at x = 1.

Teacher's Note:
1. A greatest-integer function is locally constant except where its inside crosses an integer.

Answer: (d) \( \sin(|x|)-|x| \)

Explanation:
1. cos|x| = cos x is smooth, so the first two functions have the corner of ±|x| at 0.
2. sin|x| ≈ |x| near 0, so sin|x| + |x| ≈ 2|x| has a corner.
3. sin|x| − |x| ≈ −|x|³/6 is smooth at 0 (derivative 0).

Teacher's Note:
1. The corner of sin|x| at 0 is cancelled only by subtracting |x|.

Answer: (d) f'(x) exists for all x

Explanation:
1. [x − π] is an integer, so tan(π[x − π]) = 0 and f(x) = 0 for all x.
2. f is the zero function, so f', f'' and all derivatives exist everywhere.

Teacher's Note:
1. The numerator vanishes identically.

Answer: (c) g is not differentiable exactly at four points in (−1/2, 2)

Explanation:
1. x² − 3 runs over [−3, 1]; [x² − 3] jumps at x = 1, √2, √3 and at the end point x = 2: four discontinuities of f, so (a) and (b) are false.
2. g = f(x)(|x| + |4x − 7|). In (−1/2, 2) g is not differentiable where f jumps (the factor is non-zero): x = 1, √2, √3, and at x = 0, where |x| has a corner and f(0) = −3 ≠ 0.
3. At x = 7/4 the corner of |4x − 7| is multiplied by f = 0 (f ≡ 0 on [√3, 2)), so g is smooth there.
4. Exactly four points: 0, 1, √2, √3.

Teacher's Note:
1. A corner is hidden when it is multiplied by a function that vanishes identically near that point.

Answer: (a) 1

Explanation:
1. The term (x − 1)² sin(1/(x − 1)) is differentiable everywhere, including x = 1 (its derivative there is 0).
2. −|x| is differentiable except at x = 0.
3. So f is not differentiable only at x = 0: one point.

Teacher's Note:
1. Check each piece separately; (x − 1)² sin(1/(x − 1)) is differentiable at 1.

Answer: (a) {0, 1, 2}

Explanation:
1. [x] jumps at each integer, and |1 − x| has a corner at x = 1.
2. In the open interval (−1, 3) the integers are 0, 1, 2 and the function is differentiable at every other point.
3. So the set is {0, 1, 2}.

Teacher's Note:
1. Greatest integer jumps make a function non-differentiable at integers.

Answer: (a) {0, 1}

Explanation:
1. For x < 0: f(|x|) = −x − 1 (as |x| > 0) and |f(x)| = 1, so g = −x.
2. For 0 ≤ x ≤ 1: f(|x|) = x − 1 (value −1 at 0) and |f(x)| = 1 − x, so g = 0. For 1 < x < 2: g = 2(x − 1).
3. The slope changes from −1 to 0 at x = 0 and from 0 to 2 at x = 1, so g is continuous but not differentiable at 0 and 1.

Teacher's Note:
1. Write g piece by piece on each sub-interval.

Answer: (c) not differentiable at exactly one point

Explanation:
1. For x < 0: |f(x)| = 1 and f(|x|) = x² − 1, so g = x².
2. For 0 ≤ x ≤ 1: |x² − 1| + x² − 1 = 0; for 1 ≤ x < 2: g = 2(x² − 1).
3. At x = 0 both one-sided derivatives are 0, so g is differentiable. At x = 1 the derivative jumps from 0 to 4, so g is not differentiable at exactly one point.

Teacher's Note:
1. Check x = 0 carefully; the left piece x² meets the zero piece smoothly.

Answer: (a) equals {−2, −1, 0, 1, 2}

Explanation:
1. For |x| ≤ 2, f = |x| when |x| ≤ 1 and x² when 1 ≤ |x| ≤ 2. Corners occur at x = 0 (the |x| kink), x = ±1 (slopes 1 and 2).
2. At x = ±2 the function is continuous (value 4) but the slope changes from ±4 to ∓2.
3. So S = {−2, −1, 0, 1, 2}.

Teacher's Note:
1. Find where the maximum switches branches and check the slopes on each side.

Answer: (c) 1

Explanation:
1. |f(x) − f(y)|/|x − y| ≤ 2|x − y|^1/2 → 0, so f'(x) = 0 everywhere.
2. f is constant, f(x) = f(0) = 1, so ∫₀¹ f² dx = 1.

Teacher's Note:
1. An exponent above 1 forces a constant function.

Answer: (c) φ (the empty set)

Explanation:
1. For x > 0: f = sin x − x + 2(x − π)cos x, so f'(0⁺) = cos 0 − 1 + 2cos 0 − 0 = 2.
2. For x < 0: f = −sin x + x + 2(x − π)cos x, so f'(0⁻) = −1 + 1 + 2 − 0 = 2.
3. The one-sided derivatives agree at 0, and all other points are smooth, so K is empty.

Teacher's Note:
1. Compute the left and right derivatives at x = 0.

Answer: (a) not continuous for any a and b

Explanation:
1. At x = 1: a + b = 5. At x = 3: a + 3b = b + 15, i.e. a + 2b = 15. At x = 5: b + 25 = 30, so b = 5.
2. From b = 5 and a + b = 5 we get a = 0, but then a + 2b = 10 ≠ 15.
3. The three conditions are inconsistent, so f is not continuous for any a, b.

Teacher's Note:
1. Write the continuity condition at each of the three break points.

Answer: (d) exists and equals 4

Explanation:
1. Put t = 1/x → ∞: the limit is L = lim f(t)/t.
2. By L'Hospital, L = lim f'(t) = lim (7 − (3/4)f(t)/t) = 7 − (3/4)L.
3. So (7/4)L = 7 and L = 4.

Teacher's Note:
1. Rewrite the limit as lim f(t)/t and apply L'Hospital with the given f'.

Answer: (d) three elements

Explanation:
1. −|x| ≥ −√(1 − x²) exactly when |x| ≤ 1/√2, so f = −|x| for |x| ≤ 1/√2 and f = −√(1 − x²) for 1/√2 < |x| < 1.
2. Corners: at x = 0 (the |x| kink) and at x = ±1/√2 (slope ∓1 on the inside, ±1 on the outside).
3. K = {−1/√2, 0, 1/√2}: three elements.

Teacher's Note:
1. Find where the two branches cross and compare slopes.

Answer: (b) 4e

Explanation:
1. f'(x) = f(x) gives f(x) = Ce^x, and f(1) = 2 gives f(x) = 2e^x−1.
2. h'(1) = f'(f(1)) · f'(1) = f(2) · f(1) = 2e · 2 = 4e.

Teacher's Note:
1. Chain rule: h'(x) = f'(f(x))f'(x), and f' = f.

Answer: (c) discontinuous at x = 0

Explanation:
1. For x > 0: f = x/(2x² + x) = 1/(2x + 1) → 1.
2. For x < 0: f = x/(2x² − x) = 1/(2x − 1) → −1.
3. The one-sided limits differ (1 and −1), so f is discontinuous at 0.

Teacher's Note:
1. Remove |x| by splitting into x > 0 and x < 0.

Answer: (b) 2

Explanation:
1. F' = F and G' = −G, so F = 3eˣ and G = e^−x.
2. The equation becomes 9e^2x = 9x⁵eˣ, i.e. eˣ = x⁵, which needs x > 0.
3. h(x) = x⁵ − eˣ is negative at x = 1, positive at x = 2, and negative again for large x (x = 15), so it has exactly two roots.

Teacher's Note:
1. Solve for F and G first, then count intersections of eˣ and x⁵.

Answer: (b) 5

Explanation:
1. f is a tent of height 1 over [−1, 1]; f(x − 1) is the tent over [0, 2] with peak at 1 and f(x + 1) is the tent over [−2, 0] with peak at −1.
2. g has corners at the end points ±2, at the peaks ±1, and at 0 (where the slope changes from −1 to +1).
3. So g is not differentiable at −2, −1, 0, 1, 2: five points.

Teacher's Note:
1. Sketch the two tents and look for corners.

Answer: (d) 5

Explanation:
1. h(x) = ||x| − 1| has corners at x = 0 (peak 1) and x = ±1 (value 0, where f = 1/2 is smooth).
2. h = 1/2 when |x| = 1/2 or |x| = 3/2: the branches switch there with different slopes, giving corners at ±1/2 and ±3/2.
3. Corners: 0, ±1/2, ±3/2: five points.

Teacher's Note:
1. The maximum creates corners where the two branches cross at different slopes.

Answer: (b) 1

Explanation:
1. On (1, 2], cos(πx/2) ∈ [−1, 0), so [ ] = −1 and F = 0.
2. At x = 1: F(1) = 0, the left limit 1 − x → 0 and the right limit is 0, so F is continuous with slopes −1 (left) and 0 (right): continuous but not differentiable.
3. At x = 0: F(0) = 1 but the left limit is |sin 0| = 0, so F is discontinuous at 0, and on (−1, 0) the function −sin πx is smooth.
4. Exactly one point (x = 1).

Teacher's Note:
1. Classify each break point as continuous or not before testing differentiability.

Answer: (a) f(x) is continuous in R

Explanation:
1. [(2x − 3π)³] is an integer, so tan(π·integer) = 0 and the numerator is identically 0.
2. The denominator is at least 1, so f(x) = 0 for all x, which is continuous (indeed differentiable) on R.

Teacher's Note:
1. A constant function is continuous everywhere.

Answer: (a) n ∈ (0, 1]

Explanation:
1. Continuity: |xⁿ sin(1/x)| ≤ |x|ⁿ → 0 needs n > 0.
2. Differentiability: f(x)/x = xⁿ⁻¹ sin(1/x) has a limit only if n − 1 > 0, i.e. n > 1.
3. So f is continuous but not differentiable for 0 < n ≤ 1.

Teacher's Note:
1. Compare the exponent of x in f(x) and f(x)/x with zero.

Answer: (d) {1, 3}

Explanation:
1. x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3), so the first term is (x − 2)(x + 2)|(x − 1)(x − 2)(x − 3)|.
2. At x = 1 and x = 3 the absolute value has a simple corner and the factor x² − 4 is non-zero, so f is not differentiable. At x = 2 the extra factor (x − 2) kills the corner, and x = −2 is not a zero of the cubic.
3. x/(1 + |x|) is differentiable at 0 (both one-sided derivatives equal 1).

Teacher's Note:
1. A corner of |g(x)| at a simple zero of g is removed if multiplied by another zero factor.