NDA Limits, Continuity and Differentiability previous year questions with solutions

Answer: (d) Limit does not exist

Method. one-sided limits: x|x| changes sign at 0

For x > 0: x|x| = x², so sin²xx² = (sin xx)² → 1.
For x < 0: x|x| = -x², so the expression = -(sin xx)² → -1.
One-sided limits 1 and -1 differ, so the limit does not exist.

Answer: (a) 2

Method. 1 − cos 2x = 2 sin²x and lim sin x / x = 1

For x < 0, f(x) = 1 − cos 2xx² = 2 sin²xx² = 2(sin xx)².
As x → 0^-, sin xx → 1, so the limit is 2.

Answer: (c) 8

Method. rationalise the denominator

For x > 0: ⁠√x√(16⁠ + ⁠√x)⁠ − 4 × ⁠√(16⁠ + ⁠√x)⁠ + 4⁠√(16⁠ + ⁠√x)⁠ + 4 = ⁠√x⁠ (⁠√(16⁠ + ⁠√x)⁠ + 4)(16 + ⁠√x)⁠ − 16 = ⁠√(16⁠ + ⁠√x)⁠ + 4.
As x → 0⁺: ⁠√(16⁠ + 0) + 4 = 4 + 4 = 8.

Answer: (a) -1

Method. x|x| is continuous: substitute

f(x) = x|x| is a product of continuous functions, so the limit equals the value.
f(-1) = (-1)·|-1| = (-1)(1) = -1.

Answer: (d) 3

Method. split the integral at the sign change x = 0

f(x) = -x² for x < 0 and x² for x ≥ 0, so |f(x)| = x² throughout.
Area = ∫_-2⁰ x² dx + ∫₀¹ x² dx = 83 + 13 = 3.

Answer: (c) Both I and II

Method. write x|x| piecewise and differentiate

f(x) = -x² for x < 0 and x² for x ≥ 0, so f'(x) = -2x > 0 for x < 0 and 2x > 0 for x > 0; f'(x) = 2|x| ≥ 0, zero only at x = 0.
Hence f is increasing on (-∞, ∞): I is true.
At 0: f(h) − f(0)h = h|h|h = |h| → 0 from both sides, so f'(0) = 0 exists: II is true.

Answer: (b) 0

Method. evaluate [x] separately on each side of 0

For 0 < x < 1: [x] = 0, so f(x) = sin 0 = 0 and f(x)g(x) = 0.
For -1 < x < 0: [x] = -1, so f(x) = sin(-1) = -sin 1 and f(x)g(x) = -sin 1·|x| → 0.
Both one-sided limits are 0, so the limit is 0.

Answer: (d) Limit does not exist

Method. one-sided limits; the left side is unbounded

For 0 < x < 1: f(x) = sin 0 = 0, so f(x)g(x) = 0x = 0; right-hand limit 0.
For -1 < x < 0: f(x) = sin(-1) = -sin 1 and g(x) = |x| = -x, so f(x)g(x) = -sin 1|x| → -∞.
The left-hand limit is not finite, so the limit does not exist.

Answer: (d) Limit does not exist

Method. one-sided limits of |x − a|/⁠(x − a)

For x > 3: |x − 3| = x − 3, so |x − 3|x − 3 = 1 and the right-hand limit is 1.
For x < 3: |x − 3| = -(x − 3), so |x − 3|x − 3 = -1 and the left-hand limit is -1.
The one-sided limits differ (1 ≠ -1), so the limit does not exist.

Answer: (b) -1

Method. evaluate [x] and [|x|] just to the right of 0

h(x) = f(g(x))g(f(x)) = |[x] - 1|[|x|] - 1.
For 0 < x < 1: [x] = 0, so numerator = |0 − 1| = 1; |x| = x, [|x|] = 0, so denominator = -1.
h(x) = -1 there, so lim_{x → 0⁺} h(x) = -1.

Answer: (a) -2

Method. evaluate [x] and [|x|] just to the left of 0

h(x) = |[x] - 1|[|x|] - 1.
For -1 < x < 0: [x] = -1, so numerator = |-1 − 1| = 2; |x| ∈ (0, 1), so [|x|] = 0 and denominator = -1.
h(x) = -2 there, so lim_{x → 0^-} h(x) = -2.

Answer: (d) Limit does not exist

Method. one-sided limits of 5 − x|x − 5|

x → 5⁺: |x − 5| = x − 5, expression = -(x − 5)x − 5 = -1.
x → 5^-: |x − 5| = 5 − x, expression = 5 − x5 − x = 1.
-1 ≠ 1, so the limit does not exist.

Answer: (d) lim_{x → 0} f(x) does not exist

Method. split |x|/⁠x and take one-sided limits

f(x) = x + 1 + |x|x.
x → 0⁺: |x|x = 1, so f → 0 + 1 + 1 = 2.
x → 0^-: |x|x = -1, so f → 0 + 1 − 1 = 0.
2 ≠ 0, so the limit does not exist.

Answer: (d) 1

Method. existence of a limit: left limit = right limit

LHL at 2: 1 + 22k = 1 + 1k. RHL at 2: 2k.
1 + 1k = 2k ⇒ 2k² - k − 1 = 0 ⇒ (2k + 1)(k − 1) = 0.
k = 1 or k = -12; of the options only k = 1 fits (k = 0 is not allowed since x/⁠(2k) must be defined).

Answer: (c) 1

Method. on (1, 2), [x] = 1

For 1 < x < 2: [x] = 1 and |x| = x, so f(x) = 1x.
Right-hand limit at 1: lim_{x → 1⁺} 1x = 1.

Answer: (b) -1

Method. a finite limit of a 0-denominator ratio forces the numerator to 0

As x → -1, x² - 1 → 0.
Since the limit of the ratio is finite (-32), the numerator must also → 0: lim_{x → -1} (f(x) + 1) = 0.
Hence lim_{x → -1} f(x) = -1.

Answer: (a) 0

Method. write the numerator as a perfect square

3^x + 3^-x - 2 = (3^x/⁠2 - 3^-x/⁠2)².
So the expression = (3^x/⁠2 - 3^-x/⁠2)²x = x · (3^x/⁠2 - 3^-x/⁠2x)².
As x → 0, 3^x/⁠2 - 3^-x/⁠2x = 3^x/⁠2 - 1x − 3^-x/⁠2 - 1x → ln 32 + ln 32 = ln 3.
Hence the limit = 0 · (ln 3)² = 0.

Shortcut. L'Hôpital: derivative of numerator 3^x ln 3 − 3^-x ln 3 = 0 at x = 0.

Answer: (a) 1 only

Method. limit of a product; counter-example for the converse

(1) If lim f = A and lim g = B exist, then lim f(x)g(x) = AB exists (product rule for limits). Correct.
(2) Take a = 0, f(x) = x, g(x) = 1x: f(x)g(x) = 1 for x ≠ 0, so lim {f(x)g(x)} = 1 exists, but lim_{x → 0} 1x does not exist. Incorrect.
So 1 only.

Answer: (c) 1/⁠2

Method. rationalise, then divide by ⁠√x⁠

⁠√(x⁠ + ⁠√x)⁠ - ⁠√x⁠ = (x + ⁠√x)⁠ − x⁠√(x⁠ + ⁠√x)⁠ + ⁠√x⁠ = ⁠√x√(x⁠ + ⁠√x)⁠ + ⁠√x.⁠
Divide numerator and denominator by ⁠√x:⁠ 1⁠√(1⁠ + ⁠1√x⁠) + 1.
As x → ∞, 1⁠√x⁠ → 0, so the limit is 11 + 1 = 12.

Answer: (c) n − 1

Method. lim_{x → 1} x^m - 1x − 1 = m

Divide numerator and denominator by x − 1:
x^{n2 - 1} - 1x − 1 → n² - 1 and x^{n + 1} - 1x − 1 → n + 1.
Limit = n² - 1n + 1 = (n − 1)(n + 1)n + 1 = n − 1.

Answer: (c) 4/⁠3

Method. rationalise numerator and denominator

⁠√(f(x))⁠ − 3 = ⁠√(x²⁠ + 9) − 3 = x²⁠√(x²⁠ + 9) + 3.
⁠√(f(x)⁠ + 7) − 4 = ⁠√(x²⁠ + 16) − 4 = x²⁠√(x²⁠ + 16) + 4.
Ratio = ⁠√(x²⁠ + 16) + 4⁠√(x²⁠ + 9) + 3 → 4 + 43 + 3 = 86 = 43.

Answer: (d) Neither I nor II

Method. sign of f'(x) = 2x

f'(x) = 2x: negative for x < 0, positive for x > 0. So f decreases on (-∞, 0) and increases on (0, ∞); it is not increasing everywhere. Statement I is false.
f' changes sign from − to + at x = 0, so x = 0 is a local minimum (f(0) = 9 is the least value), not a maximum. Statement II is false.

Answer: (a) -1

Method. composite of a polynomial is continuous: substitute

f ∘ f(x) = f(x² - 1) = (x² - 1)² - 1, a polynomial, so the limit is its value at x = 1.
f(1) = 0, then f(0) = 0 − 1 = -1.
lim_{x → 1} f ∘ f(x) = -1.

Answer: (c) 4/⁠3 square units

Method. area between a parabola and the x-axis between its roots

x² - 1 = 0 at x = ±1, and f(x) ≤ 0 on [-1, 1], so the bounded region lies below the x-axis there.
Area = ∫_-1¹ (1 − x²) dx = [x − x³3]_-1¹ = 23 − (-23) = 43 square units.

Shortcut. Area under a parabolic arch = 23 × base × height = 23 × 2 × 1 = 43.

Answer: (c) 3

Method. x⁹ - 1 = (x³ - 1)(x⁶ + x³ + 1)

With y = x³: x⁹ - 1 = y³ - 1 = (y − 1)(y² + y + 1).
So for x ≠ 1 the expression = x⁶ + x³ + 1.
At x = 1: 1 + 1 + 1 = 3.

Shortcut. lim x^m - 1xⁿ - 1 = mn = 93 = 3.

Answer: (c) 6

Method. factorise x³ - 1 and rationalise ⁠√x⁠ − 1

x³ - 1 = (x − 1)(x² + x + 1) and x − 1 = (⁠√x⁠ − 1)(⁠√x⁠ + 1).
So x³ - 1⁠√x⁠ − 1 = (⁠√x⁠ + 1)(x² + x + 1).
At x = 1: 2 × 3 = 6.

Shortcut. L'Hôpital: 3x²1/⁠(2⁠√x)⁠ at x = 1 = 3 × 2 = 6.

Answer: (c) 1

Method. divide numerator and denominator by aⁿ

aⁿ + bⁿaⁿ - bⁿ = 1 + (b/⁠a)ⁿ1 − (b/⁠a)ⁿ.
Since 0 < ba < 1, (b/⁠a)ⁿ → 0 as n → ∞.
Limit = 1 + 01 − 0 = 1.

Answer: (b) 1

Method. factorise and cancel the common factor (x + 1)

x³ + x² = x²(x + 1) and x² + 3x + 2 = (x + 1)(x + 2).
For x ≠ -1 the expression = x²x + 2.
At x = -1: 11 = 1.

Answer: (d) 6

Method. split as (x − 1) + (x² - 1) + (x³ - 1)

x + x² + x³ - 3 = (x − 1) + (x² - 1) + (x³ - 1).
Divide by x − 1: 1 + (x + 1) + (x² + x + 1).
At x = 1: 1 + 2 + 3 = 6.

Shortcut. It is the derivative of x + x² + x³ at x = 1: 1 + 2 + 3 = 6.

Answer: (c) 83

Method. standard limit xⁿ - aⁿx − a → n a^{n − 1}

Left side: lim_{x → 1} x⁴ - 1x − 1 = 4 · 1³ = 4.
Right side: x³ - k³x² - k² = (x³ - k³)/⁠(x − k)(x² - k²)/⁠(x − k) → 3k²2k = 3k2 (k ≠ 0).
3k2 = 4 gives k = 83.

Answer: (d) 34⁠√(2x)⁠

Method. rationalise the numerator

Multiply top and bottom by ⁠√(2x⁠ + 3h) + ⁠√(2x)⁠:
(2x + 3h) − 2x2h (⁠√(2x⁠ + 3h) + ⁠√(2x)⁠) = 32 (⁠√(2x⁠ + 3h) + ⁠√(2x)⁠).
As h → 0 this → 32 · 2⁠√(2x)⁠ = 34⁠√(2x)⁠.

Shortcut. It is 32 × (derivative of ⁠√t⁠ at t = 2x) = 32 · 12⁠√(2x)⁠.

Answer: (d) 1/⁠4

Method. f' is the integrand; use 1 − cos x ≈ x²2

f'(x) = 1 + cos²x − 2 cos xx sin²x tan x = (1 − cos x)²x sin²x tan x.
Write it as (1 − cos xx²)² × (xsin x)² × xtan x.
As x → 0: 1 − cos xx² → 12, xsin x → 1, xtan x → 1.
Limit = (12)² = 14.

Answer: (c) 1

Method. lim_{t → 0} e^t - 1t = 1 with t = (x − m)(x − n)

Since m, n are the roots, x² - px + q = (x − m)(x − n).
Put t = (x − m)(x − n); as x → m, t → 0 (and t ≠ 0 for x near m, x ≠ m).
The expression is e^t - 1t → 1.

Answer: (c) ln 10

Method. lim_{t → 0} a^t - 1t = ln a

Write 10^{sin x} - 1tan x = 10^{sin x} - 1sin x × sin xtan x = 10^{sin x} - 1sin x × cos x.
As x → 0, sin x → 0, so the first factor → ln 10, and cos x → 1.
Limit = ln 10.

Answer: (b) 0

Method. rewrite sec θ − tan θ as cos θ1 + sin θ

Read the limit as θ → π2 (the expression is in θ).
sec θ − tan θ = 1 − sin θcos θ = (1 − sin θ)(1 + sin θ)cos θ (1 + sin θ) = cos²θcos θ (1 + sin θ) = cos θ1 + sin θ.
As θ → π2: cos θ → 0 and 1 + sin θ → 2, so the limit is 02 = 0.

UPSC prints the limit variable as x while the expression is in θ; read as θ → π/⁠2, the only sensible reading, which gives option (b).

Answer: (b) ln (ab)2

Method. standard limit a^t - 1t → ln a

Put t = x − 1 → 0. Then f(x) − 1g(x) = a^t + b^t - 22t = 12(a^t - 1t + b^t - 1t).
→ 12(ln a + ln b) = ln (ab)2.

Answer: (a) ⁠√(ab)⁠

Method. 1^∞ form: lim f^1/⁠g = e^{lim (f − 1)/⁠g}

As x → 1, f(x) → 1 and 1g(x) → ∞, a 1^∞ form.
lim f(x)^1/⁠g(x) = e^lim f(x) − 1g(x).
From the previous item, lim f(x) − 1g(x) = ln (ab)2.
So the limit = e12 ln (ab) = ⁠√(ab)⁠.

Answer: (d) Limit does not exist

Method. ⁠√(1⁠ −⁠ cos 4x) = ⁠√2⁠ |sin 2x|; compare one-sided limits

1 − cos 4x = 2 sin² 2x, so ⁠√(1⁠ −⁠ cos 4x) = ⁠√2⁠ |sin 2x|.
Expression = x⁠√2⁠ |sin 2x|.
x → 0⁺: sin 2x > 0, limit = 1⁠√2⁠ · xsin 2x → 1⁠√2⁠ · 12 = 12⁠√2.⁠
x → 0^-: |sin 2x| = -sin 2x, limit = -12⁠√2.⁠
The one-sided limits differ, so the limit does not exist.

Answer: (a) -4

Method. substitute x = π2 + t

Put x = π2 + t; then t → 0, 4x − 2π = 4t and cos x = cos(π2 + t) = -sin t.
Expression = 4t-sin t → -4.

Shortcut. L'Hôpital: 4-sin x at x = π2 = -4.

Answer: (c) sin 2x

Method. limit is the derivative of sin²x

By the definition of the derivative, the limit = ddx sin²x.
ddx sin²x = 2 sin x cos x = sin 2x.

Answer: (a) 0

Method. standard limit sin xx → 1

x³ (cosec x)² = x³sin²x = x · (xsin x)².
As x → 0, xsin x → 1, so the limit = 0 · 1 = 0.

Answer: (c) 1

Method. 0/⁠0 form, L'Hôpital's rule

At x = a both numerator and denominator are 0; differentiate each with respect to x.
Numerator: a^x ln a − a x^a-1; at x = a: a^a ln a − a^a.
Denominator: a x^a-1; at x = a: a^a.
Limit = a^a(ln a − 1)a^a = ln a − 1.
ln a − 1 = -1 ⇒ ln a = 0 ⇒ a = 1.

Shortcut. Try a = 1: 1 − xx − 1 = -1 for every x ≠ 1.

Answer: (a) -1

Method. split into sin xx and log (1 − x)x

sin x log (1 − x)x² = sin xx · log (1 − x)x.
sin xx → 1; log (1 − x)x = -log (1 − x)-x → -1 (since log (1 + u)u → 1).
Limit = 1 × (-1) = -1.

Answer: (b) 13

Method. convert degrees to radians: x° = πx180

Let t = πx180; then x° = t and 3x° = 3t (in radians), and t → 0 as x → 0.
sin ttan 3t = sin tt · 3ttan 3t · 13 → 1 · 1 · 13 = 13.

Shortcut. For small angles sin t ≈ t and tan 3t ≈ 3t, giving 13; the degree factor cancels.

Answer: (c) 24

Method. factor 1 − c³ = (1 − c)(1 + c + c²) and use 1 − cos 4x = 2 sin² 2x

1 − cos³ 4x = (1 − cos 4x)(1 + cos 4x + cos² 4x).
1 − cos 4xx² = 2 sin² 2xx² = 8 (sin 2x2x)² → 8.
1 + cos 4x + cos² 4x → 1 + 1 + 1 = 3.
Limit = 8 × 3 = 24.

Shortcut. 1 − cos³ u ≈ 3u²2 for small u; with u = 4x: 3 · 162 = 24.

No correct option. This question is defective as printed in UPSC's paper.

Method. half-angle identity 1 − cos θ = 2 sin² θ2

⁠√(1⁠ −⁠ cos θ) = ⁠√(2⁠ sin² ⁠θ2⁠) = ⁠√2⁠ |sin θ2|.
⁠√2⁠ |sin (θ/⁠2)|θ = ⁠√22⁠ · |sin (θ/⁠2)|θ/⁠2.
As θ → 0⁺ this → 1⁠√2;⁠ as θ → 0^-, |sin (θ/⁠2)|θ/⁠2 → -1, so it → -1⁠√2.⁠
The one-sided limits differ, so the two-sided limit does not exist.

As printed (θ → 0) the limit does not exist, since the one-sided limits are ±1⁠√2,⁠ and no option says so; the evident intended reading θ → 0⁺ gives 1⁠√2,⁠ option (c).

Answer: (d) -3

Method. treat as a rational function of s = sin x; factor out (2s − 1)

Put s = sin x; as x → π6, s → 12.
2s² + s − 1 = (2s − 1)(s + 1) and 2s² - 3s + 1 = (2s − 1)(s − 1).
Cancel (2s − 1): the expression = s + 1s − 1 → 3/⁠2-1/⁠2 = -3.

Answer: (a) 1/⁠2

Method. standard limits tan x / x → 1 and sin 2x / 2x → 1

tan xsin 2x = tan xx · 2xsin 2x · 12.
As x → 0 the first two factors → 1, so the limit is 12.

Shortcut. tan x2 sin x cos x = 12 cos² x → 12.

Answer: (c) l = 2π, m = 0

Method. direct substitution; bounded over unbounded

sin x and x are continuous at π2 with x ≠ 0, so l = sin (π/⁠2)π/⁠2 = 1π/⁠2 = 2π.
|cos xx| ≤ 1|x| → 0 as x → ∞, so m = 0.

Answer: (b) 12

Method. series expansion of e^x

e^x = 1 + x + x²2 + x³6 + …
e^x - (1 + x) = x²2 + x³6 + …
Divide by x²: 12 + x6 + … → 12 as x → 0.

Shortcut. L'Hôpital twice: e^x - 12x → e^x2 → 12.

Answer: (b) II only

Method. [3x] jumps exactly where 3x is an integer

At x = 13, 3x = 1 is an integer: for x slightly less, [3x] = 0; at and just after, [3x] = 1. Left limit 0 ≠ f(13) = 1, so f is not continuous there: I is false.
At x = 14, 3x = 34; for x in (0, 13), [3x] = 0, a constant. So f is differentiable at 14 with f'(14) = 0: II is true.

Answer: (a) 0.5

Method. equate the one-sided limits at x = 0

Left limit at 0: 4(5⁰) = 4. Right value/⁠limit: 8k + 0 = 8k.
Continuity at 0 needs 8k = 4, so k = 12 = 0.5.
(Each branch is continuous elsewhere.)

Answer: (c) 45 ln 5

Method. derivative of a^x is a^x ln a

At x = -1 the branch f(x) = 4(5^x) applies (x < 0).
f'(x) = 4.5^x ln 5, so f'(-1) = 4.5^-1 ln 5 = 45 ln 5.

Answer: (d) 3

Method. x − 3|x − 3| = -1 left of 3, +1 right of 3; continuity equations

LHL = -1 + a, RHL = 1 + b, f(3) = a − b.
LHL = f(3): a − 1 = a − b ⇒ b = 1.
RHL = f(3): 1 + b = a − b ⇒ a = 1 + 2b = 3.

Answer: (b) 1

Method. x − 3|x − 3| = -1 left of 3, +1 right of 3; continuity equations

LHL = -1 + a, RHL = 1 + b, f(3) = a − b.
LHL = f(3): a − 1 = a − b ⇒ b = 1.
(Check: then a = 3, and RHL = 2 = f(3).)

Answer: (b) 3

Method. continuity at x = 1 and x = 2 gives two linear equations

At x = 1: f(1) = 1 + 6 = 7 and right limit = p + q, so p + q = 7.
At x = 2: f(2) = 10 and left limit = 2p + q, so 2p + q = 10.
Subtract: p = 3.

Answer: (c) 4

Method. continuity at x = 1 and x = 2 gives two linear equations

At x = 1: p + q = 1 + 6 = 7.
At x = 2: 2p + q = 5 × 2 = 10.
So p = 3 and q = 7 − 3 = 4.

Answer: (a) 1 only

Method. [x] is constant between integers and jumps at integers

On (-1, 0), [x] = -1; on [0, 1), [x] = 0.
Statement 1: near x = 0.5, y = 0 (constant), so y' = 0. True.
Statement 2: lim_{x → 0^-} [x] = -1, lim_{x → 0⁺} [x] = 0; the limits differ, so y is not continuous at 0. False.
Only 1 is correct.

Answer: (d) 1, 2 and 3

Method. exponential function a^x (a > 0)

10^x is defined for every real x: domain (-∞, ∞). True.
10^x = e^{x ln 10} is continuous on R. True.
ddx 10^x = 10^x ln 10 exists for all x; at x = 0 it equals ln 10. True.
All three are correct.

Answer: (b) 2 only

Method. sin(1/⁠x²) oscillates near 0

Statement 1: as x → 0, 1x² → ∞ and sin(1x²) takes every value in [-1, 1] infinitely often; e.g. at x = 1⁠√(π/⁠2⁠ + 2nπ) it equals 1. So lim_{x → 0} f(x) does not exist and no value f(0) makes f continuous. False.
Statement 2: x = 2⁠√π⁠ ≠ 0, and f is a composition of continuous functions there (f = sin(π4) = 1⁠√2).⁠ True.
Only 2 is correct.

Answer: (a) 5

Method. continuity at x = 1: left limit = f(1) = right limit

Left limit: a + b; right limit: b − a; f(1) = 5.
a + b = 5 and b − a = 5 ⇒ a = 0, b = 5.
a + b = 5.

Shortcut. Continuity directly gives a + b = LHL = f(1) = 5.

Answer: (b) should be 1

Method. value at 0 must equal lim_{x → 0} sin xx

lim_{x → 0} sin xx = 1.
For continuity at 0 the function must be defined there with f(0) = 1.

Answer: (d) Neither 1 nor 2

Method. sin 1x oscillates as x → 0

Along x_n = 1nπ → 0, sin 1x_n = 0; along x_n = 12nπ + π/⁠2 → 0, sin 1x_n = 1.
Two different limiting values, so lim_{x → 0} sin 1x does not exist: statement 1 is false.
Without a limit at 0, f cannot be continuous there whatever f(0) is: statement 2 is false.

Answer: (a) 14

Method. left limit = right limit = f(0)

Away from 0 each piece is a polynomial, so only x = 0 matters.
Left limit: 2(0) + 14 = 14. Right limit: (0 + 12)² = 14.
The limits agree, so continuity needs k = 14.

Answer: (d) 0

Method. k must equal lim_{x → 0} sin x

sin x is continuous, so lim_{x → 0} f(x) = sin 0 = 0.
Continuity at 0 needs k = f(0) = 0.

Answer: (a) It is not continuous at x = 0

Method. compare lim_{x → 0} sin 2x5x with f(0)

lim_{x → 0} sin 2x5x = 25 lim sin 2x2x = 25.
f(0) = 215 ≠ 25, so f is not continuous at x = 0.
(For x ≠ 0, f is a quotient of continuous functions with non-zero denominator, so it is continuous at x = π; (b), (c), (d) are false.)

Answer: (b) 13

Method. f(0) must equal lim_{x → 0} f(x); divide through by x

Divide numerator and denominator by x: f(x) = 2 − sin^-1 xx2 + tan^-1 xx.
As x → 0, sin^-1 xx → 1 and tan^-1 xx → 1.
So f(0) = 2 − 12 + 1 = 13.

Answer: (a) f(x) is continuous at x = 0 and x = 1

Method. |x − a| is continuous everywhere; sums of continuous functions are continuous

|x| and |x − 1| are continuous for all real x, so their sum f(x) is continuous for all x.
At x = 0: f(0) = 0 + 1 = 1, and both one-sided limits are 1. At x = 1: f(1) = 1 + 0 = 1, both one-sided limits 1.
f is continuous at both points (it is only non-differentiable there).

Answer: (b) f(3) = 1.5

Method. factorise and cancel (x − 3), then f(3) = the limit

x² - 9 = (x − 3)(x + 3) and x² - 2x − 3 = (x − 3)(x + 1).
For x ≠ 3, f(x) = x + 3x + 1.
Continuity at 3 needs f(3) = lim_{x → 3} x + 3x + 1 = 64 = 1.5.

Answer: (a) 1 and 2 only

Method. sums of polynomials and the oscillating cos 1x

(1) x + x² is a polynomial, continuous everywhere, in particular at 0. Correct.
(2) As x → 0, x → 0 while cos 1x oscillates between -1 and 1, so x + cos 1x has no limit at 0; it is not defined at 0 and cannot be made continuous there. Correct.
(3) Similarly x² → 0 but cos 1x oscillates, so x² + cos 1x has no limit at 0 and is not continuous there. Incorrect.
So 1 and 2 only.

Answer: (c) tan [f(x)], where [·] is the greatest integer function, and 1f(x) are both discontinuous

Method. locate where [f(x)] jumps and where f(x) = 0 on [0, π]

On [0, π], f(x) = x2 − 1 runs from -1 up to π2 − 1 ≈ 0.57.
[f(x)] = -1 for 0 ≤ x < 2 and 0 for 2 ≤ x ≤ π, so tan [f(x)] jumps from tan (-1) to tan 0 = 0 at x = 2: discontinuous.
f(2) = 0, so 1f(x) is undefined at x = 2 and |1f(x)| → ∞ there: discontinuous.
Both are discontinuous. (f^-1(x) = 2(x + 1) is continuous, but tan [f(x)] is not, so (b) fails.)

Answer: (d) all integers except 1

Method. compare left and right limits of g(g − 1) at each integer n

Between integers g is constant, so f is continuous there; check each integer n.
Right of n: g = n, f = n(n −⁠ 1) = f(n). Left of n: g = n − 1, f → (n − 1)(n − 2).
Continuous at n ⟺ (n − 1)(n − 2) = n(n −⁠ 1) ⟺ (n − 1)(-2) = 0 ⟺ n = 1.
So f is discontinuous at every integer except 1.

Answer: (c) It is discontinuous at x = 0 but continuous at every other point.

Method. compare left and right limits at the junction points x = -2 and x = 0

Each piece is a polynomial, so f is continuous inside each interval; only x = -2 and x = 0 need checking.
At x = -2: left limit = 2(-2) + 1 = -3; f(-2) = -2 − 1 = -3; right limit = -3. Continuous.
At x = 0: left limit = 0 − 1 = -1; f(0) = 0 + 2 = 2. Left limit ≠ f(0), so discontinuous.
Hence f is discontinuous at x = 0 only.

Answer: (b) 0

Method. one-sided derivatives of |x| at 0

f(h) − f(0)h = p|h|h: equals p for h > 0 and -p for h < 0.
f'(0) exists only if p = -p, i.e. p = 0.

Answer: (c) Both I and II

Method. x³/⁠|x| = x|x|

For x ≠ 0: x³|x| = x²·x|x| = |x|·x (since x² = |x|²). At 0, f(0) = 0 = 0·|0|. So f(x) = x|x| for all x.
f(x) = -x² for x < 0 and x² for x ≥ 0: continuous everywhere, and differentiable away from 0.
At 0: f(h) − f(0)h = |h| → 0, so f'(0) = 0 exists. Both I and II are true.

Answer: (c) Both I and II

Method. differentiability implies continuity

If f is differentiable at a, then f(x) − f(a) = f(x) − f(a)x − a·(x − a) → f'(a)·0 = 0 as x → a.
So lim_{x → a} f(x) = f(a) (II), which is exactly continuity at a (I). Both are correct.

Answer: (c) 13

Method. continuity of f and f' at x = 3 give two linear equations

Continuity at x = 1: left value a·1.0 = 0 and middle value 1 − 1 = 0, true for every a. (Non-differentiability at 1 needs left derivative a(2x − 1) = a ≠ right derivative 1, i.e. a ≠ 1; it does not affect p, q.)
Continuity at x = 3: 9p + 3q + 2 = 3 − 1 = 2, so 9p + 3q = 0, i.e. q = -3p.
f' continuous at x = 3: derivative of px² + qx + 2 at 3 is 6p + q; derivative of x − 1 is 1. So 6p + q = 1.
Substitute q = -3p: 6p − 3p = 1, so p = 13 and q = -3·13 = -1.

Answer: (a) -1

Method. continuity of f and f' at x = 3 give two linear equations

Continuity at x = 1: left value a·1.0 = 0 and middle value 1 − 1 = 0, true for every a. (Non-differentiability at 1 needs left derivative a(2x − 1) = a ≠ right derivative 1, i.e. a ≠ 1; it does not affect p, q.)
Continuity at x = 3: 9p + 3q + 2 = 3 − 1 = 2, so 9p + 3q = 0, i.e. q = -3p.
f' continuous at x = 3: derivative of px² + qx + 2 at 3 is 6p + q; derivative of x − 1 is 1. So 6p + q = 1.
Substitute q = -3p: 6p − 3p = 1, so p = 13 and q = -3·13 = -1.

Answer: (c) Both I and II

Method. check the derivative at 0 from the definition

p(x) = tan(x²)·x|x| is a product of functions continuous at 0, so p is continuous at 0 (p(0) = 0): I is true.
p(h) − p(0)h = tan(h²)·h|h|h = tan(h²)·|h| → 0.0 = 0 from both sides.
So p'(0) = 0 exists: II is true.

Answer: (c) Both I and II

Method. simplify the composite: (x|x|)² = x⁴

q(x) = f(g(x)) = tan((x|x|)²) = tan(x⁴) for x near 0 (x⁴ is near 0, inside the domain).
tan(x⁴) is continuous at 0 (I true) and differentiable with q'(x) = 4x³ sec²(x⁴), so q'(0) = 0 (II true).

Answer: (c) 0

Method. near 0 only the branch x³ applies

Near x = 0 we have x² < 1, so f(x) = x³ on (-1, 1).
There f'(x) = 3x², so lim_{x → 0} f'(x) = 3.0 = 0.

Answer: (d) Neither I nor II

Method. compare one-sided limits and one-sided derivatives at x = ±1

At x = -1: f(-1) = (-1)² = 1 (since x² ≥ 1). For x slightly greater than -1, x² < 1 and f(x) = x³ → -1. Since -1 ≠ 1, f is not continuous at x = -1; I is false.
At x = 1: f(1) = 1; left branch x³ → 1 and right branch x² → 1, so f is continuous at 1.
Left derivative at 1: 3x² = 3; right derivative: 2x = 2. Since 3 ≠ 2, f is not differentiable at x = 1; II is false.

Answer: (a) 1 and 2 only

Method. |x| has its only corner at 0; [x] is constant between integers

1. For x < 0, f(x) = -x + 1, a straight line, differentiable. True.
2. 0.0001 is not an integer; on [0, 1), g(x) = -1 is constant, so g is continuous there. True.
3. On [2, 3), g(x) = 2 − 1 = 1 is constant, so g'(2.5) = 0, not 1. False.
Statements 1 and 2 only.

Answer: (a) -32

Method. one-sided values of [x] near 0

x → 0^-: f(x) → 1, [x] = -1, g(x) = -2, so h → -12.
x → 0⁺: f(x) → 1, [x] = 0, g(x) = -1, so h → -1.
Sum = -12 − 1 = -32.

Answer: (a) -13

Method. differentiability at 1: continuity plus equal one-sided derivatives

Continuity at 1: left limit a · 1 · 2 + b = 2a + b must equal f(1) = 0, so 2a + b = 0.
Left derivative: ddx(ax² + ax + b) = a(2x + 1) = 3a at x = 1; right derivative of x − 1 is 1.
3a = 1 ⇒ a = 13, b = -23.
a + b = -13.

Answer: (b) -23

Method. near x = 0 the rule ax(x + 1) + b applies

Taking f differentiable at x = 1 as in item 63 (the condition the UPSC set carries across both items), 2a + b = 0 and 3a = 1 give a = 13, b = -23. Near 0, x < 1, so f(x) = ax(x + 1) + b → b = -23.
Near 0, x < 1, so f(x) = ax(x + 1) + b → a · 0 + b = b.
lim_{x → 0} f(x) = -23.

Answer: (a) 1 only

Method. |x| is continuous everywhere but has a corner at 0

Statement 1: |x| − 1 is a sum of continuous functions, so it is continuous at x = 1. True.
Statement 2: at x = 0, right derivative = lim_{h → 0⁺} hh = 1, left derivative = lim_{h → 0^-} -hh = -1. Not differentiable. False.
Only 1 is correct.

Answer: (d) f'(0) does not exist

Method. compare left and right derivatives at 0

For x ≥ 0, f(x) = e^x; for x < 0, f(x) = e^-x.
Right derivative at 0: lim_{h → 0⁺} e^h - 1h = 1.
Left derivative at 0: lim_{h → 0^-} e^-h - 1h = -1.
1 ≠ -1, so f'(0) does not exist.

Answer: (a) 1 only

Method. e^-|x| is a composition of continuous functions; one-sided derivatives at 0 differ

|x| and e^t are continuous, so e^-|x| is continuous everywhere, including x = 0. Statement 1 true.
f(0) = 1. RHD = lim_{h → 0⁺} e^-h - 1h = -1; LHD = lim_{h → 0^-} e^h - 1h = 1.
RHD ≠ LHD, so f is not differentiable at 0. Statement 2 false.

Answer: (a) The function is not continuous at x = -3

Method. |x − 3| is continuous everywhere and differentiable except at x = 3

|x − 3| is continuous for every real x, so (a) 'not continuous at x = -3' is false; (b) is true.
For x < 3, f(x) = 3 − x with f'(x) = -1, so f is differentiable at x = 0 and at x = -3: (c), (d) true.
The statement that is not correct is (a).

Answer: (a) 0

Method. derivative at 0 from the definition; h ln |h| → 0

f'(0) = lim_{h → 0} h² ln |h| − 0h = lim_{h → 0} h ln |h|.
h ln |h| = ln |h|1/⁠h; by L'Hôpital (for h → 0⁺) 1/⁠h-1/⁠h² = -h → 0, and the same holds for h → 0^- since the expression is odd in h.
So f'(0) = 0.

Answer: (c) f(x) is discontinuous at x = 0

Method. ⁠√(x²)⁠ = |x|, so f is minus the signum function

⁠√(x²)⁠ = |x|, so for x ≠ 0, f(x) = -x|x| = -1 for x > 0 and +1 for x < 0.
Right limit at 0 = -1, left limit = 1; they differ, so lim_{x → 0} f(x) does not exist.
f is discontinuous at x = 0 (and therefore not differentiable there).

Answer: (d) 1, 2 and 3

Method. the three possible behaviours of the derivative at a point

(1) f(x) = |x| has no derivative at x = 0 (left 1, right -1). Correct.
(2) f(x) = x² has f'(0) = 0, a finite derivative. Correct.
(3) f(x) = x^1/⁠3: f(h) − f(0)h = h^-2/⁠3 → ∞; geometrically the tangent at 0 is vertical. Correct.
All three statements hold.

Answer: (a) (-1)^k (k − 1)π

Method. left-hand derivative from the definition, with [x] = k − 1 just left of k

f(k) = k sin (kπ) = 0.
For 0 < h < 1, [k − h] = k − 1, so f(k − h) = (k − 1) sin (kπ − πh).
sin (kπ − πh) = sin kπ cos πh − cos kπ sin πh = -(-1)^k sin πh.
LHD = lim_{h → 0⁺} f(k − h) − f(k)-h = lim (k − 1)(-(-1)^k sin πh)-h = (-1)^k (k − 1) lim sin πhh = (-1)^k (k − 1)π.

Answer: (c) (-∞, 0) ∪ (0, ∞)

Method. near x = 0, 1 − e^-x2 ≈ x², so f behaves like |x|

1 − e^-x2 > 0 for x ≠ 0, and √ is differentiable on (0, ∞), so f is differentiable at every x ≠ 0.
At x = 0: f(h) − f(0)h = ⁠√(1⁠ −⁠ e^-h2)h.
Since 1 − e^-h2 = h² - h⁴2 + … , ⁠√(1⁠ −⁠ e^-h2) = |h| ⁠√(1⁠ −⁠ ⁠h²2⁠ + …).
So the quotient → 1 as h → 0⁺ and → -1 as h → 0^-: f is not differentiable at 0.
Set = (-∞, 0) ∪ (0, ∞).

Answer: (b) differentiable at x = 0

Method. right-hand derivative at the end-point x = 0 of the domain [0, ∞)

⁠√x⁠ needs x ≥ 0, so the domain is [0, ∞) and x = 0 is its left end-point; f(0) = 0.
Continuity: lim_{x → 0⁺} x(⁠√x⁠ - ⁠√(x⁠ + 1)) = 0 · (0 − 1) = 0 = f(0).
Derivative: f(h) − f(0)h = ⁠√h⁠ - ⁠√(h⁠ + 1) → 0 − 1 = -1 as h → 0⁺.
So f'(0) = -1 exists (at an end-point only the one-sided derivative is required): f is differentiable at x = 0.

Answer: (a) [1, ∞)

Method. derivative of xⁿ at x = 0 from the definition

For x ≠ 0 in its domain, f'(x) = n x^{n − 1}, so the only point in question is x = 0 (for n < 0, 0 is not even in the domain).
At x = 0: f(h) − f(0)h = h^{n − 1}.
If 0 < n < 1, h^{n − 1} → ∞, so f is not differentiable at 0; this rules out (0, ∞) and (12, ∞) (e.g. n = 34).
If n ≥ 1, h^{n − 1} → 0 (n > 1) or 1 (n = 1), so f'(0) exists.
Hence every n in [1, ∞) works, reading 'differentiable for all x' as differentiable at every point of the domain of xⁿ (for non-integer n that domain is [0, ∞), and the only point in question is x = 0).