IOE entrance Mathematics · Chapter 5
Calculus
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87 questions in 4 syllabus topics · 34 are 2-mark questions.
5.1 Limits, continuity and L’Hospital’s rule
20 questions
1. lim (x→0) sin(3x)/x is equal to
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Use lim (θ→0) sinθ/θ = 1.
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Answer: C. 3
Write sin(3x)/x = 3·sin(3x)/(3x). As x→0, sin(3x)/(3x) → 1, so the limit is 3.
2. lim (x→0) (1 − cos x)/x² equals
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Convert 1 − cos x into a sine of half angle.
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Answer: D. 1/2
1 − cos x = 2sin²(x/2), so the expression is 2sin²(x/2)/x² = ½·[sin(x/2)/(x/2)]² → ½.
3. The value of lim (x→∞) (1 + 2/x)^x is
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Recall the limit that defines e.
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Answer: A. e²
Standard limit: lim (x→∞) (1 + a/x)^x = eᵃ. With a = 2 the limit is e².
4. lim (x→0) (2ˣ − 1)/x is equal to
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Use the standard limit for (aˣ − 1)/x.
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Answer: A. ln 2
Standard limit: lim (x→0) (aˣ − 1)/x = ln a. Here a = 2, so the value is ln 2.
5. A function is defined by f(x) = ax + b for x < 1, f(1) = 7, and f(x) = 2ax + b − 3 for x > 1. If f is continuous at x = 1, then
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Equate the left limit, right limit and f(1).
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Answer: B. a = 3, b = 4
Left limit a + b = 7; right limit 2a + b − 3 = 7, i.e. 2a + b = 10. Subtracting gives a = 3 and then b = 4.
6. lim (x→3) (x⁵ − 243)/(x − 3) is
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243 = 3⁵; use the standard (xⁿ − aⁿ)/(x − a) limit.
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Answer: A. 405
lim (x→a) (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹. Here n = 5, a = 3: 5 × 3⁴ = 5 × 81 = 405.
7. Which of the following is NOT an indeterminate form?
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Think of what (small positive number)^(huge number) does.
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Answer: A. 0^∞
0/0, 1^∞ and ∞ − ∞ are indeterminate. A base tending to 0 raised to a power tending to +∞ always tends to 0, so 0^∞ is not indeterminate.
8. lim (x→∞) (3x² + 2x)/(5x² − 1) equals
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Compare the leading terms of equal degree.
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Answer: B. 3/5
Divide numerator and denominator by x²: (3 + 2/x)/(5 − 1/x²) → 3/5.
9. If f(x) = (x² − 9)/(x − 3) for x ≠ 3 and f(3) = k, then f is continuous at x = 3 when k equals
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Factorise x² − 9 and cancel.
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Answer: A. 6
For x ≠ 3, f(x) = x + 3, so lim (x→3) f(x) = 6. Continuity needs k = 6.
10. The function f(x) = |x| at x = 0 is
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Compare the left and right hand derivatives.
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Answer: B. continuous but not differentiable
lim (x→0)|x| = 0 = f(0), so f is continuous. Left derivative is −1 and right derivative is +1, so it is not differentiable at 0.
11. Using L'Hospital's rule, lim (x→0) (x − sin x)/x³ equals
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Apply L'Hospital's rule repeatedly until the form is no longer 0/0.
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Answer: B. 1/6
0/0 form. Differentiate: (1 − cos x)/(3x²) → still 0/0 → sin x/(6x) → 1/6.
12. lim (x→0) (eˣ − 1 − x)/x² is
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Use L'Hospital twice or the series of eˣ.
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Answer: C. 1/2
0/0 form. L'Hospital: (eˣ − 1)/(2x) → again 0/0 → eˣ/2 → 1/2. (Or use eˣ = 1 + x + x²/2 + …)
13. lim (x→0) (sin x°)/x, where x° means x degrees, is
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Convert degrees to radians first.
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Answer: B. π/180
x° = πx/180 radians, so sin(πx/180)/x = (π/180)·sin(πx/180)/(πx/180) → π/180.
14. lim (x→∞) [√(x² + 4x) − x] is equal to
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Multiply and divide by the conjugate.
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Answer: C. 2
Rationalise: (x² + 4x − x²)/(√(x² + 4x) + x) = 4x/(√(x² + 4x) + x) = 4/(√(1 + 4/x) + 1) → 4/2 = 2.
15. lim (x→0⁺) x·ln x equals
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Rewrite the product as a quotient before applying L'Hospital.
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Answer: C. 0
0·∞ form. Write as ln x/(1/x), an ∞/∞ form. L'Hospital: (1/x)/(−1/x²) = −x → 0.
16. The value of lim (x→0⁺) xˣ is
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Take logarithms: this is a 0⁰ form.
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Answer: C. 1
xˣ = e^(x ln x). Since x ln x → 0 as x→0⁺, the limit is e⁰ = 1.
17. If f(x) = ax + 1 for x ≤ 1 and f(x) = 3x² − 1 for x > 1 is continuous at x = 1, then a is
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Equate f(1) with the right-hand limit.
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Answer: C. 1
Left value a + 1 must equal right limit 3 − 1 = 2, so a = 1.
18. For the greatest integer function [x], the left-hand and right-hand limits at x = 0 are respectively
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Try values like −0.01 and 0.01.
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Answer: C. −1 and 0
Just left of 0 (e.g. −0.01), [x] = −1; just right of 0 (e.g. 0.01), [x] = 0. So LHL = −1, RHL = 0 and the limit does not exist.
19. lim (x→0) (tan x − sin x)/x³ equals
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Factor out tan x from the numerator.
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Answer: C. 1/2
tan x − sin x = tan x(1 − cos x). So the expression = (tan x/x)·((1 − cos x)/x²) → 1 × ½ = ½.
20. lim (x→∞) [x/(x + 1)]ˣ is
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For 1^∞ forms, the limit is e^(lim (f − 1)·g).
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Answer: D. 1/e
This is a 1^∞ form. x/(x+1) = 1 − 1/(x+1); the limit is e^(lim x·(−1/(x+1))) = e^(−1) = 1/e.
5.2 Derivatives and their applications
26 questions
21. The derivative of sech⁻¹x is valid for
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Where is 1/(x√(1 − x²)) defined?
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Answer: B. x < 1
The official options don't include the exact condition 0 < x < 1; “x < 1” is the closest.
d/dx (sech⁻¹x) = −1/(x√(1 − x²)), which is defined only for 0 < x < 1. Of the options, x < 1 is the one that fits (strictly the condition is 0 < x < 1).
22. The derivative of x² sin x with respect to x is
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Use the product rule (uv)′ = u′v + uv′.
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Answer: D. 2x sin x + x² cos x
Product rule: (x²)′ sin x + x²(sin x)′ = 2x sin x + x² cos x.
23. d/dx (sin⁻¹x) for |x| < 1 is
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Differentiate sin y = x implicitly.
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Answer: B. 1/√(1 − x²)
If y = sin⁻¹x then sin y = x, cos y·y′ = 1, y′ = 1/cos y = 1/√(1 − x²).
24. If y = e^(3x), then d²y/dx² equals
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Each differentiation brings down a factor 3.
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Answer: A. 9e^(3x)
dy/dx = 3e^(3x), d²y/dx² = 9e^(3x).
25. If y = √(x + √(x + √(x + … to ∞))), then dy/dx equals
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The infinite expression inside equals y itself.
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Answer: B. 1/(2y − 1)
y² = x + y. Differentiating: 2y·y′ = 1 + y′, so y′(2y − 1) = 1 and y′ = 1/(2y − 1).
26. Geometrically, the value of dy/dx at a point on the curve y = f(x) represents
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Think of a secant line becoming a tangent.
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Answer: B. the slope of the tangent to the curve at that point
The derivative is the limit of the slopes of secants, i.e. the slope (tan ψ) of the tangent at that point. The normal has slope −1/(dy/dx).
27. A particle moves in a straight line so that s = t³ − 6t² + 9t (s in m, t in s). What is its acceleration at the later instant when it is momentarily at rest?
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Set v = ds/dt = 0, then use a = dv/dt.
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Answer: C. 6 m/s²
v = 3t² − 12t + 9 = 3(t − 1)(t − 3) = 0 at t = 1 and t = 3. a = 6t − 12, so at t = 3, a = 6 m/s².
28. The slope of the tangent to y = x³ − 2x at the point (1, −1) is
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Evaluate dy/dx at x = 1.
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Answer: D. 1
dy/dx = 3x² − 2 = 3 − 2 = 1 at x = 1.
29. The equation of the tangent to the curve y = x² at the point (2, 4) is
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Use y − y₁ = m(x − x₁) with m = dy/dx.
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Answer: C. y = 4x − 4
Slope = 2x = 4. Tangent: y − 4 = 4(x − 2), i.e. y = 4x − 4. (x + 4y = 18 is the normal.)
30. The equation of the normal to the curve xy = 4 at the point (2, 2) is
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Normal slope = −1/(slope of tangent).
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Answer: B. y = x
Implicitly, y + x·y′ = 0, so y′ = −y/x = −1 at (2, 2). Normal slope = 1, so y − 2 = x − 2, i.e. y = x.
31. The radius of a circle increases at 2 cm/s. The rate of increase of its area when the radius is 5 cm is
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Differentiate A = πr² with respect to time.
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Answer: B. 20π cm²/s
A = πr², dA/dt = 2πr·dr/dt = 2π × 5 × 2 = 20π cm²/s.
32. Air is pumped into a spherical balloon so that its volume increases at 36π cm³/s. How fast is the radius increasing when the radius is 3 cm?
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Differentiate V = (4/3)πr³ with respect to t.
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Answer: C. 1 cm/s
V = (4/3)πr³, dV/dt = 4πr²·dr/dt. So 36π = 4π × 9 × dr/dt, giving dr/dt = 1 cm/s.
33. A 5 m ladder leans against a vertical wall. Its foot slides away from the wall at 3 m/s. How fast is the top sliding down when the foot is 3 m from the wall?
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Differentiate x² + y² = 25 with respect to time.
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Answer: B. 2.25 m/s
x² + y² = 25; at x = 3, y = 4. Differentiating: x·dx/dt + y·dy/dt = 0 → 3 × 3 + 4·dy/dt = 0 → dy/dt = −9/4 m/s. The top slides down at 2.25 m/s.
34. The largest interval on which the function f(x) = x³ − 3x² − 9x + 5 is strictly decreasing is
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Find where f′(x) < 0.
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Answer: A. (−1, 3)
f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), which is negative exactly for −1 < x < 3. So the largest interval of decrease is (−1, 3); (1, 3) is only part of it.
35. Which of the following functions is strictly increasing for all real x?
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Check whether f′(x) > 0 for every x.
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Answer: C. x³ + x
For x³ + x, f′(x) = 3x² + 1 > 0 for all x. The others have derivatives (2x + 1, 3x² − 1, cos x) that become negative somewhere.
36. The maximum value of f(x) = −x² + 4x + 1 is
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Set f′(x) = 0 and evaluate f there.
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Answer: C. 5
f′(x) = −2x + 4 = 0 at x = 2, f″ = −2 < 0 so it is a maximum. f(2) = −4 + 8 + 1 = 5.
37. The local maximum value of f(x) = x³ − 3x is
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Use the second-derivative test at the critical points.
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Answer: D. 2
f′(x) = 3x² − 3 = 0 at x = ±1. f″(x) = 6x is negative at x = −1, so x = −1 gives the local maximum f(−1) = −1 + 3 = 2.
38. Two positive numbers have sum 20. The maximum possible value of their product is
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Write the product in one variable and differentiate.
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Answer: A. 100
P = x(20 − x), P′ = 20 − 2x = 0 at x = 10. Maximum product = 10 × 10 = 100.
39. An open box is made from a square sheet of side 18 cm by cutting equal squares from each corner and folding up the sides. For maximum volume, the side of each square cut off should be
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Express the volume in terms of the cut size x and set dV/dx = 0.
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Answer: C. 3 cm
V = x(18 − 2x)². V′ = (18 − 2x)(18 − 6x) = 0 gives x = 3 (x = 9 gives zero volume). So x = 3 cm, V = 3 × 12² = 432 cm³.
40. The local maximum value of f(x) = x²e^(−x) is
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Factor f′(x) and check its sign change.
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Answer: C. 4/e²
f′(x) = e^(−x)(2x − x²) = xe^(−x)(2 − x). f′ changes from + to − at x = 2, so the local maximum is f(2) = 4e^(−2) = 4/e². (At x = 0 there is a local minimum, value 0.)
41. If y = A cos 2x + B sin 2x, where A and B are constants, then
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Differentiate twice and compare with y.
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Answer: B. d²y/dx² + 4y = 0
y′ = −2A sin 2x + 2B cos 2x, y″ = −4A cos 2x − 4B sin 2x = −4y. Hence y″ + 4y = 0.
42. The derivative of xˣ (x > 0) is
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Take logarithms before differentiating.
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Answer: B. xˣ(1 + ln x)
ln y = x ln x, so y′/y = ln x + 1, giving y′ = xˣ(1 + ln x).
43. If x = a cos t and y = a sin t, then dy/dx is
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dy/dx = (dy/dt)/(dx/dt).
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Answer: C. −cot t
dx/dt = −a sin t, dy/dt = a cos t, so dy/dx = (a cos t)/(−a sin t) = −cot t.
44. For the curve x³ + y³ = 3xy, the value of dy/dx at the point (3/2, 3/2) is
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Differentiate implicitly and collect the y′ terms.
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Answer: D. −1
3x² + 3y²y′ = 3y + 3xy′ → y′ = (y − x²)/(y² − x). At (3/2, 3/2): (3/2 − 9/4)/(9/4 − 3/2) = (−3/4)/(3/4) = −1.
45. If y = sin⁻¹(2x/(1 + x²)) for |x| < 1, then dy/dx equals
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Substitute x = tan θ.
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Answer: A. 2/(1 + x²)
Put x = tan θ: 2x/(1 + x²) = sin 2θ, so y = 2θ = 2tan⁻¹x (valid for |x| < 1). Hence dy/dx = 2/(1 + x²).
46. The point on the curve y = x² − 4x + 5 at which the tangent is parallel to the x-axis is
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Set dy/dx = 0.
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Answer: A. (2, 1)
Tangent parallel to the x-axis means dy/dx = 2x − 4 = 0, so x = 2 and y = 4 − 8 + 5 = 1.
5.3 Integration and area
25 questions
47. ∫ dx/(x² + 9) is equal to
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Standard form ∫ dx/(x² + a²).
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Answer: B. (1/3) tan⁻¹(x/3) + C
Using ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a) + C with a = 3 gives (1/3) tan⁻¹(x/3) + C.
48. ∫ dx/√(4 − x²) equals
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Compare with ∫ dx/√(a² − x²).
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Answer: A. sin⁻¹(x/2) + C
∫ dx/√(a² − x²) = sin⁻¹(x/a) + C; with a = 2 there is no extra factor, so the answer is sin⁻¹(x/2) + C.
49. ∫ eˣ (sin x + cos x) dx is
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Look for the pattern eˣ[f(x) + f′(x)].
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Answer: A. eˣ sin x + C
The integrand has the form eˣ[f(x) + f′(x)] with f(x) = sin x, and ∫ eˣ[f + f′] dx = eˣ f(x) + C. So the answer is eˣ sin x + C.
50. ∫ x eˣ dx equals
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Integrate by parts taking u = x.
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Answer: A. (x − 1) eˣ + C
By parts with u = x, dv = eˣ dx: x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = (x − 1) eˣ + C.
51. ∫ log x dx (x > 0) is
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Take log x as the first function and 1 as the second.
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Answer: A. x log x − x + C
Write log x as 1·log x and integrate by parts: x log x − ∫ x·(1/x) dx = x log x − x + C.
52. ∫ 2x/(1 + x²) dx equals
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∫ f′(x)/f(x) dx = log|f(x)| + C.
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Answer: B. log(1 + x²) + C
The numerator 2x is the derivative of the denominator 1 + x², so the integral is log(1 + x²) + C.
53. ∫ tan x dx is equal to
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Write tan x as sin x/cos x and put t = cos x.
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Answer: D. log|sec x| + C
tan x = sin x/cos x and the numerator is −(cos x)′, so ∫ tan x dx = −log|cos x| + C = log|sec x| + C.
54. ∫ dx/(x² − 1) equals
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Split 1/((x − 1)(x + 1)) into partial fractions.
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Answer: C. (1/2) log|(x − 1)/(x + 1)| + C
1/(x² − 1) = (1/2)[1/(x − 1) − 1/(x + 1)], so the integral is (1/2) log|(x − 1)/(x + 1)| + C.
55. ∫ sin(log x)/x dx equals
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Substitute t = log x.
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Answer: B. −cos(log x) + C
Put t = log x, dt = dx/x: ∫ sin t dt = −cos t + C = −cos(log x) + C.
56. ∫ x/((x − 1)(x − 2)) dx equals
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Find A and B by putting x = 1 and x = 2 (cover-up rule).
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Answer: C. 2 log|x − 2| − log|x − 1| + C
x/((x − 1)(x − 2)) = A/(x − 1) + B/(x − 2) with A = 1/(1 − 2) = −1 and B = 2/(2 − 1) = 2. Integrating gives −log|x − 1| + 2 log|x − 2| + C.
57. The value of ∫₀^(π/2) sin²x dx is
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Use sin²x = (1 − cos 2x)/2.
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Answer: B. π/4
sin²x = (1 − cos 2x)/2, so the integral is (1/2)[x − (sin 2x)/2] from 0 to π/2 = (1/2)(π/2) = π/4.
58. For any positive integer n, ∫₀^(π/2) sinⁿx/(sinⁿx + cosⁿx) dx equals
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Use ∫₀^a f(x) dx = ∫₀^a f(a − x) dx and add the two forms.
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Answer: C. π/4
Call it I. Replacing x by π/2 − x gives I = ∫ cosⁿx/(cosⁿx + sinⁿx) dx. Adding, 2I = ∫₀^(π/2) 1 dx = π/2, so I = π/4.
59. ∫₋₁¹ (x⁵ + x³ + 1) dx is equal to
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Separate the odd part of the integrand.
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Answer: C. 2
x⁵ + x³ is odd, so its integral over [−1, 1] is 0. The remaining ∫₋₁¹ 1 dx = 2.
60. The value of ∫₀^π x sin x dx is
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Integrate by parts with u = x, or use the property with f(π − x).
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Answer: C. π
By parts: [−x cos x]₀^π + ∫₀^π cos x dx = (−π·(−1) − 0) + [sin x]₀^π = π + 0 = π.
61. ∫₀² |x − 1| dx equals
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Break the interval where x − 1 changes sign.
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Answer: D. 1
Split at x = 1: ∫₀¹ (1 − x) dx + ∫₁² (x − 1) dx = 1/2 + 1/2 = 1.
62. If [x] denotes the greatest integer not exceeding x, then ∫₀³ [x] dx equals
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Split [0, 3] into unit intervals on which [x] is constant.
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Answer: A. 3
[x] = 0 on [0,1), 1 on [1,2), 2 on [2,3). So the integral is 0×1 + 1×1 + 2×1 = 3.
63. The area bounded by the curve y = x², the x-axis and the lines x = 0 and x = 3 is
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Area = ∫ y dx between the given limits.
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Answer: D. 9 sq. units
Area = ∫₀³ x² dx = [x³/3]₀³ = 27/3 = 9 sq. units.
64. The area of the region enclosed between the curves y = x² and y = x is
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Find the intersection points, then integrate (upper − lower).
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Answer: D. 1/6 sq. unit
They meet at x = 0 and x = 1, where x ≥ x². Area = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
65. Using integration, the area enclosed by the ellipse x²/9 + y²/4 = 1 is
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By symmetry, area = 4∫₀^a (b/a)√(a² − x²) dx; use ∫₀^a √(a² − x²) dx = πa²/4.
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Answer: D. 6π sq. units
Area of an ellipse x²/a² + y²/b² = 1 is πab; here a = 3, b = 2, so area = 6π.
66. The area bounded by the parabola y² = 4x and the line x = 1 is
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Double the area of the upper half y = 2√x.
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Answer: C. 8/3 sq. units
By symmetry about the x-axis, area = 2∫₀¹ 2√x dx = 4 × (2/3)[x^(3/2)]₀¹ = 8/3.
67. The area enclosed between the parabola y² = 4x and the line y = 2x is
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Intersect the curves, then integrate (parabola − line).
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Answer: A. 1/3 sq. unit
Intersection: (2x)² = 4x gives x = 0, 1. On [0,1], 2√x ≥ 2x, so area = ∫₀¹ (2√x − 2x) dx = 4/3 − 1 = 1/3.
68. The total area enclosed between the curve y = sin x and the x-axis from x = 0 to x = 2π is
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Area below the x-axis must be counted as positive.
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Answer: D. 4 sq. units
∫₀^(2π) sin x dx = 0 because the part below the axis cancels. Area must be taken positive: ∫₀^π sin x dx + |∫_π^(2π) sin x dx| = 2 + 2 = 4.
69. The value of ∫₀¹ tan⁻¹x dx is
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Take tan⁻¹x as the first function and 1 as the second.
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Answer: B. π/4 − (1/2) log 2
By parts: [x tan⁻¹x]₀¹ − ∫₀¹ x/(1 + x²) dx = π/4 − (1/2)[log(1 + x²)]₀¹ = π/4 − (1/2) log 2.
70. ∫₀¹ x(1 − x)⁹⁹ dx equals
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Replace x by 1 − x to avoid expanding the power.
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Answer: A. 1/10100
Using ∫₀¹ f(x) dx = ∫₀¹ f(1 − x) dx: ∫₀¹ (1 − x)x⁹⁹ dx = 1/100 − 1/101 = 1/10100.
71. ∫ dx/(1 + eˣ) equals
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Add and subtract eˣ in the numerator.
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Answer: D. x − log(1 + eˣ) + C
1/(1 + eˣ) = 1 − eˣ/(1 + eˣ). Integrating: x − log(1 + eˣ) + C.
5.4 Differential equations
16 questions
72. The integrating factor of the differential equation dy/dx + (x/(1 + x))y = 1 + x is
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Write x/(1 + x) as 1 − 1/(1 + x) before integrating.
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Answer: B. eˣ/(1 + x)
IF = e^(∫x/(1 + x) dx) = e^(∫(1 − 1/(1 + x)) dx) = e^(x − ln(1 + x)) = eˣ/(1 + x).
73. The order and degree of the differential equation (d²y/dx²)³ + (dy/dx)⁴ + y = 0 are respectively
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Degree is the power of the highest-order derivative.
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Answer: B. 2 and 3
The highest derivative is d²y/dx², so order = 2; its power is 3, so degree = 3.
74. The order and degree of the differential equation d²y/dx² = √(1 + (dy/dx)²) are
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Make the equation free of radicals before reading the degree.
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Answer: D. order 2, degree 2
Squaring to remove the radical: (d²y/dx²)² = 1 + (dy/dx)². The highest derivative is second order and appears squared, so order 2, degree 2.
75. For which of the following differential equations is the degree NOT defined?
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Check whether every derivative appears only as a polynomial.
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Answer: C. d²y/dx² + sin(dy/dx) = 0
Degree is defined only when the equation is a polynomial in the derivatives. sin(dy/dx) is not a polynomial in dy/dx, so the degree is not defined. Functions of x alone (sin x, eˣ) do not matter.
76. The differential equation whose general solution is y = A cos 2x + B sin 2x (A, B arbitrary constants) is
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Differentiate twice and eliminate A and B.
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Answer: A. d²y/dx² + 4y = 0
Differentiating twice gives y″ = −4A cos 2x − 4B sin 2x = −4y, so y″ + 4y = 0. Two constants means a second-order equation.
77. The general solution of dy/dx = e^(x − y) is
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Separate the variables using e^(x − y) = eˣ e^(−y).
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Answer: C. eʸ = eˣ + C
dy/dx = eˣ/eʸ, so eʸ dy = eˣ dx. Integrating gives eʸ = eˣ + C.
78. The general solution of dy/dx = (1 + y²)/(1 + x²) is
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Separate variables, integrate, then use the tan(A − B) formula.
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Answer: B. y − x = C(1 + xy)
Separating: dy/(1 + y²) = dx/(1 + x²), so tan⁻¹y − tan⁻¹x = k. Taking tan of both sides: (y − x)/(1 + xy) = tan k = C, i.e. y − x = C(1 + xy).
79. If dy/dx = 2xy and y(0) = 3, then y(1) equals
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Separate variables and use the initial condition to find the constant.
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Answer: C. 3e
dy/y = 2x dx gives log|y| = x² + c, so y = Ae^(x²). y(0) = 3 gives A = 3, so y = 3e^(x²) and y(1) = 3e.
80. Which of the following is a homogeneous differential equation?
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Check whether replacing x, y by λx, λy leaves the right side unchanged.
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Answer: C. dy/dx = (x² + y²)/(xy)
dy/dx = f(x, y) is homogeneous when f(λx, λy) = f(x, y). For (x² + y²)/(xy) both numerator and denominator are of degree 2, so it is homogeneous; the others mix degrees.
81. The general solution of dy/dx = y/x + tan(y/x) is
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It is homogeneous; substitute y = vx.
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Answer: D. sin(y/x) = Cx
Put y = vx: v + x dv/dx = v + tan v, so cot v dv = dx/x. Integrating, log|sin v| = log|x| + log C, i.e. sin(y/x) = Cx.
82. The integrating factor of dy/dx + y/x = x² (x > 0) is
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IF = e^(∫P dx).
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Answer: D. x
For dy/dx + Py = Q, IF = e^(∫P dx) = e^(∫dx/x) = e^(log x) = x.
83. The general solution of dy/dx + y = eˣ is
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It is linear with P = 1; multiply by the integrating factor.
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Answer: C. y = eˣ/2 + Ce^(−x)
IF = eˣ, so d(yeˣ)/dx = e^(2x). Integrating, yeˣ = e^(2x)/2 + C, i.e. y = eˣ/2 + Ce^(−x).
84. The general solution of dy/dx + y tan x = sec x is
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Find IF = e^(∫tan x dx) first.
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Answer: A. y = sin x + C cos x
IF = e^(∫tan x dx) = sec x. Then y sec x = ∫ sec²x dx = tan x + C, so y = sin x + C cos x.
85. The differential equation M(x, y) dx + N(x, y) dy = 0 is exact if and only if
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Differentiate M with respect to the other variable.
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Answer: A. ∂M/∂y = ∂N/∂x
The test for exactness is ∂M/∂y = ∂N/∂x (the mixed partials of the potential function must agree).
86. The general solution of the exact equation (2xy + 3) dx + (x² + 4y) dy = 0 is
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∫M dx (y constant) + ∫(terms of N without x) dy = C.
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Answer: C. x²y + 3x + 2y² = C
Check: ∂M/∂y = 2x = ∂N/∂x. Integrate M w.r.t. x: x²y + 3x. Add the terms of N free of x integrated w.r.t. y: ∫4y dy = 2y². Solution x²y + 3x + 2y² = C.
87. The differential equation (3x² + ky) dx + (2x + y²) dy = 0 is exact when k equals
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Apply the exactness test.
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Answer: C. 2
Exactness needs ∂M/∂y = ∂N/∂x, i.e. k = 2.