NDA Differentiation previous year questions with solutions

Answer: (c) Both I and II

Method. Cauchy equation: f'(x) is constant = f'(0)

f(0 + 0) = 2f(0) gives f(0) = 0, and f'(x) = lim_h→0 f(x + h) − f(x)h = lim_h→0 f(h)h = f'(0) for every x.
I: f'(0) = lim_h→0 h g(h)h = lim_h→0 g(h) = g(0) (g continuous). So f'(x) = g(0): correct.
II: f'(0) = lim_h→0 h² g(h)h = lim_h→0 h g(h) = 0 · g(0) = 0. So f'(x) = 0: correct.
Both I and II.

Answer: (a) -1⁠√24⁠

Method. the limit is the definition of f'(1)

The limit is f'(1).
f'(x) = -2x2⁠√(25⁠ −⁠ x²) = -x⁠√(25⁠ −⁠ x²).
f'(1) = -1⁠√24.⁠
(Note 4⁠√3⁠ = ⁠√48,⁠ so (c) is a different number.)

Answer: (d) abf(x)

Method. first-principles derivative with f(x + h) = f(x)f(h)

f'(x) = lim_{h → 0} f(x + h) − f(x)h = f(x) lim_{h → 0} f(h) − 1h.
f(h) − 1 = h g(h) φ(h), so f(h) − 1h = g(h) φ(h) → ab.
Hence f'(x) = ab f(x).

Answer: (b) f'(x) is an odd function

Method. differentiate the identity f(-x) = f(x)

f even: f(-x) = f(x) for all x.
Differentiate both sides (chain rule on the left): -f'(-x) = f'(x).
So f'(-x) = -f'(x): f' is an odd function.

Answer: (b) ddx (l₃) = l₂

Method. l₁ and l₂ are the same derivative; l₃ is an antiderivative of it

l₁ = ddx (e^{sin x}) = e^{sin x} cos x.
l₂ is the first-principles definition of the same derivative, so l₂ = e^{sin x} cos x = l₁; (a) is false.
l₃ = ∫ e^{sin x} cos x dx = e^{sin x} + C(put t = sin x); so l₂ ≠ l₃ and (d) is false.
ddx (l₃) = e^{sin x} cos x = l₂, so (b) holds.

Answer: (c) -12⁠√2⁠

Method. the limit is the definition of F′(1)

The limit lim_{x → 1} F(x) − F(1)x − 1 is F′(1).
F′(x) = -2x2⁠√(9⁠ −⁠ x²) = -x⁠√(9⁠ −⁠ x²).
F′(1) = -1⁠√8⁠ = -12⁠√2.⁠

Answer: (a) f(5)f'(0)

Method. first-principles derivative using f(x + y) = f(x)f(y)

Put x = y = 0: f(0) = f(0)², so f(0) = 1 (f(0) = 0 would force f(x) = f(x)f(0) = 0 for every x, the trivial function, for which the result below also holds).
f'(5) = lim_{h → 0} f(5 + h) − f(5)h = lim_{h → 0} f(5)f(h) − f(5)h = f(5) lim_{h → 0} f(h) − f(0)h.
The last limit is f'(0), so f'(5) = f(5)f'(0).

Answer: (a) n2^n-1 - 2ⁿ + 1

Method. sum the GP, then quotient rule

f(x) = xⁿ - 1x − 1 (x ≠ 1).
f'(x) = n x^n-1(x − 1) − (xⁿ - 1)(x − 1)².
At x = 2: f'(2) = n 2^n-1 · 1 − (2ⁿ - 1)1 = n2^n-1 - 2ⁿ + 1.
Check n = 2: f = x + 1, f'(2) = 1 = 4 − 4 + 1.

Answer: (d) 4y

Method. chain rule; the inner derivative simplifies to the function over ⁠√(x²⁠ + 4)

y = (x + ⁠√(x²⁠ + 4))⁴.
ddx(x + ⁠√(x²⁠ + 4)) = 1 + x⁠√(x²⁠ + 4) = x + ⁠√(x²⁠ + 4)⁠√(x²⁠ + 4).
dydx = 4(x + ⁠√(x²⁠ + 4))³ · x + ⁠√(x²⁠ + 4)⁠√(x²⁠ + 4) = 4y⁠√(x²⁠ + 4).
So ⁠√(x²⁠ + 4) dydx = 4y.

Answer: (d) 2

Method. chain rule for a composite function

h'(x) = f'(2f(x) + 2) · 2f'(x).
At x = 0: 2f(0) + 2 = 2(-1) + 2 = 0.
h'(0) = f'(0) · 2f'(0) = 1 · 2 · 1 = 2.

Answer: (a) -4

Method. chain rule for a square: g' = 2h h'

g'(x) = 2h(x) h'(x).
h(0) = f(2f(0) + 2) = f(0) = -1, and h'(0) = 2 (previous item: f'(0) · 2f'(0)).
g'(0) = 2(-1)(2) = -4.

Answer: (c) 3x² cos (ln (2x − 3x))

Method. chain rule with f' given

Let u = 2x − 3x = 2 − 3x, so dudx = 3x².
dydx = f'(u) · dudx = cos (ln u) · 3x²
= 3x² cos (ln (2x − 3x)).

Answer: (c) ⁠√(x²⁠ - 16)

Method. differentiate term by term; recognise ∫⁠√(x²⁠ - a²) dx with a = 4

ddx[x⁠√(x²⁠ - 16)2] = ⁠√(x²⁠ - 16)2 + x²2⁠√(x²⁠ - 16).
ddx[8 ln|x + ⁠√(x²⁠ - 16)|] = 8x + ⁠√(x²⁠ - 16)·(1 + x⁠√(x²⁠ - 16)) = 8⁠√(x²⁠ - 16).
dydx = ⁠√(x²⁠ - 16)2 + x² - 162⁠√(x²⁠ - 16) = ⁠√(x²⁠ - 16)2 + ⁠√(x²⁠ - 16)2 = ⁠√(x²⁠ - 16).

Shortcut. y is the standard antiderivative x2⁠√(x²⁠ - a²) − a²2 ln|x + ⁠√(x²⁠ - a²)| with a² = 16, so y' = ⁠√(x²⁠ - 16).

Answer: (b) 0

Method. chain rule; the inner logarithm vanishes at x = 0

Let u = x² - x + 1x² + x + 1; y = (ln u)².
dydx = 2 ln u · 1u dudx.
At x = 0, u = 1 and ln u = 0, while dudx is finite there.
So dydx = 0 at x = 0.

Answer: (d) 2a = b

Method. factor 1 + x⁴ + x⁸ = (1 + x² + x⁴)(1 − x² + x⁴)

1 + x⁴ + x⁸ = (1 + x⁴)² - x⁴ = (1 + x² + x⁴)(1 − x² + x⁴).
So 1 + x⁴ + x⁸1 − x² + x⁴ = 1 + x² + x⁴.
ddx(1 + x² + x⁴) = 2x + 4x³, so a = 2, b = 4.
Hence b = 2a, i.e. 2a = b.

Answer: (d) Cannot be determined due to insufficient data

Method. set f'(2) = 0 and see what it fixes

f'(x) = -mx² + 2n.
f'(2) = 0 gives -m4 + 2n = 0, i.e. m = 8n.
Then m + 8n = 16n, which depends on n; the condition fixes only m − 8n = 0.
So m + 8n cannot be determined from the data.

Answer: (b) 1

Method. collapse the product with (1 − x)(1 + x)(1 + x²)… = 1 − x³²

Multiply and divide by (1 − x): y = 1 − x³²1 − x = 1 + x + x² + … + x³¹ (x ≠ 1).
dydx = 1 + 2x + 3x² + … + 31x³⁰.
At x = 0: dydx = 1.

Shortcut. Product rule at x = 0: every factor equals 1 there and only (1 + x) has non-zero derivative (1), so y'(0) = 1.

Answer: (c) 2

Method. find the polynomial f from degree comparison, then differentiate

If deg f = n, deg f(f(x)) = n² = 4, so n = 2. Let f(x) = ax² + bx + c.
Leading term of f(f(x)) is a·a²x⁴, so a³ = 1, a = 1.
The x³ term of (x² + bx + c)² is 2bx³ and nothing else gives x³, so b = 0.
Then f(f(x)) = (x² + c)² + c = x⁴ + 2cx² + c² + c, so c = 0 and f(x) = x².
f'(x) = 2x, f'(1) = 2.

Answer: (a) cos 1 − sin 1e

Method. chain rule with ln x inside

ddx[sin(ln x) + cos(ln x)] = cos(ln x)x − sin(ln x)x = cos(ln x) − sin(ln x)x.
At x = e, ln e = 1: cos 1 − sin 1e.

Answer: (b) m²

Method. slope = dy/⁠dx at the y-intercept

On the y-axis x = 0, so P = (0, m e⁰) = (0, m).
dydx = m · m e^mx = m² e^mx.
At x = 0: slope = m².

Answer: (c) sin^-1(1⁠√(1⁠ + m⁴))

Method. angle with the y-axis is the complement of the inclination

Slope at P is m² (previous item), so the tangent makes angle α = tan^-1 m² with the x-axis.
The angle with the y-axis is β = π2 − α, so tan β = cot α = 1m², with 0 < β < π2 since m > 0.
In a right triangle with opposite side 1 and adjacent side m², hypotenuse = ⁠√(1⁠ + m⁴), so sin β = 1⁠√(1⁠ + m⁴).
β = sin^-1 (1⁠√(1⁠ + m⁴)).
((a) and (d) both equal α, the angle with the x-axis.)

Answer: (d) y = m²x + m

Method. point-slope form of the tangent

P = (0, m) and slope at P = m².
y − m = m² (x − 0), i.e. y = m²x + m.

Answer: (d) (- sin x) cos (cos x)

Method. chain rule

f'(x) = cos (cos x) · ddx (cos x) = cos (cos x) · (-sin x).
So f'(x) = (- sin x) cos (cos x).

Answer: (a) a e^2ax

Method. u u' + v v' = (1/⁠2) d/⁠dx (u² + v²)

u dudx + v dvdx = 12 ddx (u² + v²).
u² + v² = e^2ax (sin² bx + cos² bx) = e^2ax.
12 ddx (e^2ax) = 12 · 2a e^2ax = a e^2ax.

Answer: (c) 2e

Method. simplify e^{ln x} = x, then differentiate term by term

e^{ln x} = x (x > 0), so f(x) = e^{tan x} + ln (sec x) − x.
f'(x) = e^{tan x} sec² x + sec x tan xsec x − 1 = e^{tan x} sec² x + tan x − 1.
At x = π4: tan x = 1, sec² x = 2, so f'(π4) = 2e + 1 − 1 = 2e.

Answer: (c) 2e^π2

Method. product rule with chain rule

dydx = 2x e^x2 sin 2x + 2 e^x2 cos 2x.
At x = π: sin 2π = 0, cos 2π = 1.
dydx = 0 + 2 e^π2 = 2e^π2.

Answer: (c) 10x log₁₀ e5x² + 3

Method. change of base, then chain rule

log₁₀ (5x² + 3) = ln (5x² + 3)ln 10 = log₁₀ e · ln (5x² + 3).
ddx = log₁₀ e · 10x5x² + 3 = 10x log₁₀ e5x² + 3.
(Option (d) has log_e 10 = 1log₁₀ e, the reciprocal factor.)

Answer: (b) -π180 cosec (x°) cot (x°)

Method. convert degrees to radians, then chain rule

x° = πx180 radians, so cosec(x°) = cosec(πx180).
ddx cosec(πx180) = -cosec(πx180) cot(πx180) · π180
= -π180 cosec(x°) cot(x°).

Answer: (a) -1

Method. cos^-1(cos x) = -x for -π ≤ x ≤ 0

Principal value of cos^-1 lies in [0, π]. For -π ≤ x ≤ 0, -x ∈ [0, π] and cos(-x) = cos x.
So y = cos^-1(cos x) = -x near x = -π4.
Slope = dydx = -1.

Answer: (a) 2x

Method. simplify with sec² θ = 1 + tan² θ before differentiating

Let θ = tan^-1 x, so tan θ = x.
sec² (tan^-1 x) = 1 + tan² θ = 1 + x².
ddx (1 + x²) = 2x.

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

Method. derivative of cos^-1(sin x); its sign depends on the sign of cos x

y = cos^-1 (sin x), dydx = -cos x⁠√(1⁠ −⁠ sin² x) = -cos x|cos x|.
For -π2 < x < π2 (cos x > 0): slope = -1, tan θ = -1, θ = 3π4.
For π2 < x < 3π2 (cos x < 0): slope = +1, tan θ = 1, θ = π4.
The question names no x, so both 3π4 (b) and π4 (c) occur.

The slope of y = cos^-1(sin x) is -cos x/⁠|cos x|, i.e. -1 where cos x > 0 and +1 where cos x < 0, so with no point specified both (b) 3π/⁠4 and (c) π/⁠4 are correct; the evident intended reading (x in the principal interval -π/⁠2 < x < π/⁠2, where y = π/⁠2 − x) gives (b).

Answer: (a) -12⁠√x⁠

Method. write the fraction as tan(α - ⁠√x)⁠ with tan α = 5/⁠2

Divide numerator and denominator by 2: 5 − 2 tan ⁠√x2⁠ + 5 tan ⁠√x⁠ = 52 − tan ⁠√x1⁠ + 52 tan ⁠√x.⁠
With tan α = 52, this is tan α − tan ⁠√x1⁠ + tan α tan ⁠√x⁠ = tan (α - ⁠√x).⁠
So y = α - ⁠√x⁠ (up to an added constant multiple of π), with α constant.
dydx = -12⁠√x.⁠

Answer: (a) 0

Method. sec^-1 t = cos^-1(1/⁠t), then sin^-1 u + cos^-1 u = π/⁠2

sec^-1 (x + 1x − 1) = cos^-1 (x − 1x + 1).
So … = π2 wherever both terms are defined (x ≥ 0, x ≠ 1)..
y is constant, so dydx = 0.

Answer: (a) -21 + x² for all |x| < 1

Method. substitute x = tan θ and use the principal-value range of sin^-1

Put x = tan θ, θ ∈ (-π2, π2): 2x1 + x² = sin 2θ.
y = cos^-1 (sin 2θ) = π2 − sin^-1 (sin 2θ).
For |x| < 1, 2θ ∈ (-π2, π2), so sin^-1 (sin 2θ) = 2θ and y = π2 − 2 tan^-1 x.
Hence dydx = -21 + x² for |x| < 1.
(For |x| > 1, sin^-1 (sin 2θ) = ±π − 2θ, giving +21 + x²; so (b) is false.)

Answer: (c) y²1 − xy

Method. take logs, then implicit differentiation

(e^y)^x = e^xy = y, so xy = ln y.
Differentiate: y + xdydx = 1ydydx.
dydx(1y − x) = y, so dydx = y1 − xyy = y²1 − xy.

The stated domain (0, 10] is not attainable: since x = ln y / y, real y exists only for 0 < x ≤ 1/⁠e; the formula holds wherever the curve exists.

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

Method. previous result evaluated at x = 1, using e^y₀ = y₀

At x = 1 the relation reads e^y₀ = y₀.
But e^y > y for every real y (e^y - y has minimum value 1 at y = 0), so no real y₀ exists and the derivative at x = 1 is undefined.
Intended working: from the previous item, dydx = y₀²1 − y₀; replacing y₀ by e^y₀ in one factor and in the denominator gives y₀ e^y₀1 − e^y₀.

Defective as printed: at x = 1 the relation e^xy = y needs e^y₀ = y₀, which no real y₀ satisfies (e^y - y ≥ 1 for all real y; indeed x = ln y / y never exceeds 1/⁠e), so the curve has no point with x = 1. Substituting e^y₀ = y₀ formally into y₀²/⁠(1 − y₀) gives option (d).

Answer: (d) is independent of both p and q

Method. logarithmic differentiation of an implicit relation

Take logs of the moduli: (p + q) ln|x + y| = p ln|x| + q ln|y| (the derivative of ln|u| is u'/⁠u, so the rest of the working is unchanged).
Differentiate: (p + q)(1 + y')x + y = px + qy y'.
Collect terms: p + qx + y − px = y'(qy − p + qx + y), i.e. qx − pyx(x + y) = y' · qx − pyy(x + y).
Hence dydx = yx, which contains neither p nor q.

Shortcut. Both sides are homogeneous of degree p + q, so the relation forces y/⁠x to be constant; then dydx = yx.

Answer: (a) yx

Method. result of the previous item

From the previous item, dydx = yx for every positive p, q.
In particular, with p + q = 10, dydx = yx.

Answer: (b) dydx − xy(1 + 2ln x) = 0

Method. (x^x)^x = x^x2; logarithmic differentiation

y = x^x2, so ln y = x² ln x.
1y dydx = 2x ln x + x = x(1 + 2 ln x).
dydx = xy(1 + 2 ln x), i.e. dydx − xy(1 + 2 ln x) = 0.

Answer: (a) -1

Method. take logs, differentiate implicitly, substitute (1, 1)

Take ln: y ln x + x ln y = 0.
Differentiate: dydx ln x + yx + ln y + xy dydx = 0.
At (1, 1): ln 1 = 0, so 0 + 1 + 0 + dydx = 0.
dydx = -1.

Answer: (b) 1

Method. (x^x)^x = x^x2; logarithmic differentiation

y = x^x2, so ln y = x² ln x.
1y dydx = 2x ln x + x²·1x = x(1 + 2 ln x).
At x = 1: y = 1, so dydx = 1.1·(1 + 0) = 1.

Answer: (a) -1

Method. logarithmic differentiation of a product

ln y = ln cos x + ln cos 4x + ln cos 8x (at x = π4, y = 1⁠√2·(-1)·1⁠ ≠ 0).
1y dydx = -tan x − 4 tan 4x − 8 tan 8x.
At x = π4: tan π4 = 1, tan π = 0, tan 2π = 0.
1y dydx = -1 − 0 − 0 = -1.

Answer: (b) - θ dφ

Method. differentiate the relation in u = θφ; u must be constant

Write u = θφ, so e^u - 4u = c.
The function e^u - 4u takes the value c only at isolated points, so u = θφ is a constant.
Hence d(θφ) = φ dθ + θ dφ = 0 (equivalently, differentiating gives (e^u - 4) du = 0).
So φ dθ = - θ dφ.

Answer: (b) -mynx

Method. logarithmic differentiation

Take ln: m ln x + n ln y = (m + n) ln a.
Differentiate: mx + ny dydx = 0.
dydx = -mynx.

Answer: (a) 0

Method. take logs and make y explicit, then the quotient rule

Take ln of both sides: y ln x = x − y, so y(1 + ln x) = x and y = x1 + ln x.
dydx = (1 + ln x)·1 − x·1x(1 + ln x)² = ln x(1 + ln x)².
At x = 1: ln 1 = 0, so dydx = 01 = 0.

Answer: (b) 1

Method. differentiate y' = ln x (1 + ln x)^-2 once more

From the previous item, dydx = ln x · (1 + ln x)^-2.
d²ydx² = 1x(1 + ln x)^-2 + ln x · (-2)(1 + ln x)^-3 · 1x.
At x = 1, ln 1 = 0: d²ydx² = 1 · 1 + 0 = 1.

Answer: (a) 2^-π6 (6 ln 2 - ⁠√3π)6⁠

Method. remove the moduli near x = -π/⁠6, then logarithmic differentiation

Near x = -π6: x < 0 and sin x < 0, so |x| = -x and |sin x| = -sin x; y = (-sin x)^-x.
ln y = -x ln (-sin x).
1y dydx = -ln (-sin x) − x · -cos x-sin x = -ln (-sin x) − x cot x.
At x = -π6: -sin x = 12, so -ln 12 = ln 2; cot (-π6) = -⁠√3,⁠ so -x cot x = -π6 · ⁠√3⁠ = -⁠√3π6.⁠
1y dydx = ln 2 − ⁠√3π6⁠ = 6 ln 2 - ⁠√3π6.⁠
y = (12)^π/⁠6 = 2^-π6, so dydx = 2^-π6 (6 ln 2 - ⁠√3π)6.⁠

Answer: (a) -y² tan x1 − y ln(cos x)

Method. infinite power tower: y = (cos x)^y, then logarithmic differentiation

The tower repeats itself, so y = (cos x)^y.
ln y = y ln(cos x).
Differentiate: 1y dydx = dydx ln(cos x) + y · -sin xcos x = dydx ln(cos x) − y tan x.
Multiply by y: dydx (1 − y ln(cos x)) = -y² tan x.
dydx = -y² tan x1 − y ln(cos x).

Answer: (c) 16(y² + 4)(x² + 4)

Method. parametric differentiation, then express in x and y

dxdθ = sec θ tan θ + sin θ = tan θ (sec θ + cos θ).
dydθ = 4sec⁴ θ tan θ + 4cos³ θ sin θ = 4tan θ (sec⁴ θ + cos⁴ θ).
dydx = 4(sec⁴ θ + cos⁴ θ)sec θ + cos θ.
Since sec θ cos θ = 1: (sec θ + cos θ)² = (sec θ − cos θ)² + 4 = x² + 4, and (sec⁴ θ + cos⁴ θ)² = (sec⁴ θ − cos⁴ θ)² + 4 = y² + 4.
(dydx)² = 16(y² + 4)(x² + 4).

Answer: (c) -16x

Method. differentiate the relation from the previous item implicitly

From the previous item: (x² + 4)(dydx)² = 16(y² + 4).
Differentiate in x: 2x(dydx)² + 2(x² + 4)dydxd²ydx² = 32ydydx.
Divide by 2dydx: (x² + 4)d²ydx² − 16y = -xdydx.
So the expression = (x² + 4y² + 4) dydx (-xdydx) = -x · (x² + 4)(dydx)²y² + 4 = -x · 16 = -16x.

Answer: (a) -1

Method. derivative of one function with respect to another: (du/⁠dx) ÷ (dv/⁠dx)

Let u = sin²x and v = cos²x.
dudx = 2 sin x cos x, dvdx = -2 cos x sin x.
dudv = 2 sin x cos x-2 sin x cos x = -1.

Shortcut. u = 1 − v, so dudv = -1 at once.

Answer: (a) e^ex

Method. substitute u = e^x

Let u = e^x; the function is e^u.
d(e^u)du = e^u = e^ex.

Answer: (b) 1

Method. parametric differentiation: dxdy = dx/⁠dtdy/⁠dt

dxdt = e^t(cos t − sin t), dydt = e^t(sin t + cos t).
dxdy = cos t − sin tsin t + cos t.
At t = 0: 1 − 00 + 1 = 1.

Answer: (a) xe^xex^e

Method. derivative of u with respect to v = du/⁠dxdv/⁠dx

ddx e^x = e^x; ddx x^e = e x^{e − 1} (power rule, e is a constant).
d(e^x)d(x^e) = e^xe x^{e − 1} = x e^xe x^e.

Answer: (a) -1

Method. tan^-1 x + cot^-1 x = π2

Let u = tan^-1 x and v = cot^-1 x. Then u + v = π2, so u = π2 − v.
dudv = -1.

Shortcut. dudx ÷ dvdx = 11 + x² ÷ (-11 + x²) = -1.

Answer: (a) sin x 2^{(sin x)2} ln 4

Method. substitute t = sin x and differentiate a^u by the chain rule

Let t = sin x; the function is 2^t2.
ddt 2^t2 = 2^t2 ln 2 · 2t = t · 2^t2 · 2 ln 2 = t · 2^t2 ln 4.
Putting t = sin x: sin x · 2^{(sin x)2} ln 4.

Answer: (b) II only

Method. derivative of ln|x| is 1/⁠x on both sides of 0

For x > 0: ln|x| = ln x, derivative 1x. So II is correct.
For x < 0: ln|x| = ln(-x), derivative 1-x · (-1) = 1x, not -1x. So I is wrong.

Answer: (d) It does not exist

Method. the modulus of a function that changes sign with non-zero slope has a corner

At x = π4, sin(π4 − x) = 0 and it changes sign there.
For x slightly less than π4: y = sin(π4 − x), y' = -cos(π4 − x) → -1.
For x slightly more than π4: y = -sin(π4 − x), y' = cos(π4 − x) → 1.
Left derivative -1 ≠ right derivative 1, so dydx does not exist at x = π4.

Answer: (b) 0

Method. simplify the modulus before differentiating

For x < 0, |x| = -x, so x|x| = x-x = -1.
The derivative of the constant -1 is 0.

Answer: (a) 0

Method. open the modulus on the interval containing the point (y = f(x))

For 3 < x < 4: |x − 3| = x − 3 and |x − 4| = 4 − x.
So f(x) = (x − 3) + (4 − x) = 1 on (3, 4).
x = 3.5 lies in (3, 4), so dydx = 0 there.

Answer: (d) Neither I nor II

Method. compare left and right derivatives at the corners

On [0, 3): f(x) = (3 − x) + (4 − x) = 7 − 2x, slope -2.
On (3, 4): f(x) = 1, slope 0.
On (4, 5]: f(x) = (x − 3) + (x − 4) = 2x − 7, slope 2.
At x = 3: Lf' = -2 ≠ Rf' = 0; at x = 4: Lf' = 0 ≠ Rf' = 2.
Not differentiable at either point, so neither I nor II.

Answer: (a) -2

Method. remove the modulus signs using the sign of ln x on (0, 1)

For 0 < x < 1: |x| = x and ln x < 0, so f(x) = |ln x| = -ln x.
f'(x) = -1x.
f'(0.5) = -10.5 = -2.

Answer: (a) -2

Method. remove the modulus by the sign of ln x

At x = 0.5, ln x < 0, so near this point f(x) = -ln x.
f'(x) = -1x; f'(0.5) = -10.5 = -2.

Answer: (c) 12

Method. remove the modulus by the sign of ln x

At x = 2, ln x > 0, so near this point f(x) = ln x.
f'(x) = 1x; f'(2) = 12.

Answer: (d) -1x lnx

Method. f∘f(x) = |ln(ln x)| with 0 < ln x < 1 on (1, 2)

For 1 < x < 2: ln x > 0, so f(x) = ln x, and 0 < ln x < ln 2 < 1.
f(f(x)) = |ln(ln x)|, and ln(ln x) < 0 because ln x < 1, so f(f(x)) = -ln(ln x).
ddx[-ln(ln x)] = -1ln x·1x = -1x ln x.

Answer: (d) 0

Method. greatest integer function is constant on the interval

For -4 < x < -3, x + 1 lies in (-3, -2), so y = [x + 1] = -3 throughout.
y is constant on the interval, so dydx = 0 at x = -3.5.

Answer: (a) cos x + sin x

Method. write 1 − sin 2x as a perfect square and fix the sign on the interval

1 − sin 2x = sin² x + cos² x − 2 sin x cos x = (sin x − cos x)².
⁠√(1⁠ −⁠ sin 2x) = |sin x − cos x|.
For π4 < x < π2, sin x > cos x, so ⁠√(1⁠ −⁠ sin 2x) = sin x − cos x.
ddx (sin x − cos x) = cos x + sin x.

Answer: (c) f'(x) = - 2x for 0 < x < 1

Method. differentiate piece by piece and test the joining point x = 1

On 0 < x < 1: f(x) = 1 − x², so f'(x) = -2x; (a) is wrong in sign.
At x = 1: left derivative = -2(1) = -2; right derivative from ln x is 11 = 1.
They differ, so f'(1) does not exist and (b), which includes x = 1, is wrong.
(d) is wrong since f' is not zero (e.g. -2x on (0, 1)).
Hence f'(x) = -2x for 0 < x < 1.

Answer: (b) 0

Method. differentiate a determinant row-wise (only row 1 depends on x)

Only the first row contains x, so d²fdx² =  6 -cos x sin x 6 -1 0 q q² q³ .
At x = 0 the first row is (6, -1, 0), identical to the second row.
A determinant with two equal rows is 0, so f''(0) = 0.
(Check by expansion: f(x) = -3q³x² - 6q³ cos x − (6q² + q) sin x, f''(0) = -6q³ + 6q³ = 0.)

Answer: (c) Both 1 and 2

Method. show h'(x) = 0 for all x, so h is constant

h'(x) = 2f(x)f'(x) + 2g(x)g'(x).
f'(x) = g(x) and g'(x) = f''(x) = -f(x).
h'(x) = 2f(x)g(x) − 2g(x)f(x) = 0 for every x.
So h'(3) = 0 (statement 1 true) and h is constant, giving h(1) = h(2) (statement 2 true).

Answer: (d) d²ydx² + 4y = 0

Method. differentiate twice and compare with y

dydx = -2a sin 2x + 2b cos 2x.
d²ydx² = -4a cos 2x − 4b sin 2x = -4y.
So d²ydx² + 4y = 0.

Answer: (c) x² d²ydx² + x dydx + y = 0

Method. differentiate twice, clearing the 1/⁠x factor first

dydx = cos (ln x)x, so x dydx = cos (ln x).
Differentiate again: dydx + x d²ydx² = -sin (ln x)x = -yx.
Multiply by x: x² d²ydx² + x dydx + y = 0.

Answer: (c) -(d²ydx²) (dydx)^-3

Method. derivative of the inverse function, differentiated a second time

dxdy = (dydx)^-1.
d²xdy² = ddx[(dydx)^-1] · dxdy = -(dydx)^-2 d²ydx² · (dydx)^-1.
So d²xdy² = -(d²ydx²) (dydx)^-3.