IOE entrance Mathematics · Chapter 2
Algebra
Tap an option to check it. Wrong picks show the right answer and the hint; “Show answer” gives the worked solution.
96 questions in 4 syllabus topics · 38 are 2-mark questions.
2.1 Matrices and determinants
25 questions
1. If A is a square matrix of order 3 and |A| = 4, then |2A| equals
Show hintHide hint
Each of the 3 rows gets multiplied by 2.
Show answerHide answer
Answer: C. 32
For an n×n matrix, |kA| = kⁿ|A|. Here |2A| = 2³ × 4 = 32.
2. A square matrix A satisfying Aᵀ = −A is called
Show hintHide hint
Compare A with its transpose.
Show answerHide answer
Answer: D. skew-symmetric
By definition, Aᵀ = −A means A is skew-symmetric; Aᵀ = A would be symmetric.
3. The diagonal elements of every skew-symmetric matrix are
Show hintHide hint
Put i = j in aᵢⱼ = −aⱼᵢ.
Show answerHide answer
Answer: C. all zero
aᵢᵢ = −aᵢᵢ gives 2aᵢᵢ = 0, so every diagonal element is 0.
4. If A = [2 3; 1 4] (rows separated by ;), then A⁻¹ is
Show hintHide hint
A⁻¹ = adj A / |A|.
Show answerHide answer
Answer: B. (1/5)[4 −3; −1 2]
|A| = 8 − 3 = 5. For a 2×2 matrix, adj A swaps the diagonal entries and changes the signs of the off-diagonal ones: [4 −3; −1 2]. So A⁻¹ = (1/5)[4 −3; −1 2].
5. If A is a square matrix of order 3 with |A| = 5, then |adj A| is
Show hintHide hint
Use |adj A| = |A|ⁿ⁻¹.
Show answerHide answer
Answer: C. 25
|adj A| = |A|ⁿ⁻¹ = 5² = 25 for n = 3.
6. The value of the determinant |1 2 3; 4 5 6; 7 8 9| (rows separated by ;) is
Show hintHide hint
Look at the differences between successive rows.
Show answerHide answer
Answer: C. 0
R₃ − R₂ = (3, 3, 3) = R₂ − R₁, so the operation R₃ → R₃ − 2R₂ + R₁ gives a zero row. The determinant is 0.
7. If two rows of a determinant are interchanged, its value
Show hintHide hint
Recall the effect of a row swap.
Show answerHide answer
Answer: A. changes sign only
Interchanging any two rows (or columns) multiplies the determinant by −1.
8. A square matrix A is said to be singular if
Show hintHide hint
Singular matrices have no inverse.
Show answerHide answer
Answer: A. |A| = 0
A singular matrix has zero determinant, so it has no inverse. A⁻¹ = Aᵀ defines an orthogonal matrix.
9. If A = [1 1; 0 1] (rows separated by ;), then A¹⁰ equals
2 marksShow hintHide hint
Compute A² and A³ and spot the pattern.
Show answerHide answer
Answer: A. [1 10; 0 1]
A² = [1 2; 0 1], A³ = [1 3; 0 1], and in general Aⁿ = [1 n; 0 1]. So A¹⁰ = [1 10; 0 1].
10. If A and B are invertible matrices of the same order, then (AB)⁻¹ equals
Show hintHide hint
Multiply each option by AB and see which gives I.
Show answerHide answer
Answer: A. B⁻¹A⁻¹
(AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I, so (AB)⁻¹ = B⁻¹A⁻¹ (reversal law).
11. For any square matrix A, the matrix A + Aᵀ is always
Show hintHide hint
Take the transpose of A + Aᵀ.
Show answerHide answer
Answer: B. symmetric
(A + Aᵀ)ᵀ = Aᵀ + A = A + Aᵀ, so it is symmetric. (A − Aᵀ is skew-symmetric.)
12. A square matrix A is called idempotent if
Show hintHide hint
The Latin word idem means "the same".
Show answerHide answer
Answer: B. A² = A
Idempotent means A² = A. A² = I is involutory, A² = O (or Aᵏ = O) is nilpotent, and Aᵀ = A⁻¹ is orthogonal.
13. The value of the determinant |x, x + 1; x − 1, x| (entries in a row separated by commas, rows by ;) is
Show hintHide hint
Use ad − bc.
Show answerHide answer
Answer: B. 1
x·x − (x + 1)(x − 1) = x² − (x² − 1) = 1.
14. For the system AX = B of three equations, Cramer's rule gives D = 0 while at least one of D₁, D₂, D₃ is non-zero. The system then
Show hintHide hint
Think about Dᵢ/D when D = 0.
Show answerHide answer
Answer: A. has no solution
x = D₁/D etc. With D = 0 and some Dᵢ ≠ 0, no finite values satisfy the equations, so the system is inconsistent.
15. Using Cramer's rule, the solution of 2x + y = 5 and x − y = 1 is
Show hintHide hint
x = Dx/D, y = Dy/D.
Show answerHide answer
Answer: A. x = 2, y = 1
D = 2(−1) − 1(1) = −3, Dx = 5(−1) − 1(1) = −6, Dy = 2(1) − 5(1) = −3. So x = −6/−3 = 2 and y = −3/−3 = 1.
16. In the determinant |1 2 3; 4 5 6; 7 8 10| (rows separated by ;), the cofactor of the element 2 (first row, second column) is
Show hintHide hint
Cofactor = (−1)ⁱ⁺ʲ × minor.
Show answerHide answer
Answer: D. 2
Minor M₁₂ = 4×10 − 6×7 = 40 − 42 = −2. Cofactor C₁₂ = (−1)¹⁺² M₁₂ = −(−2) = 2.
17. If A = [1 2; 3 4] (rows separated by ;), then A² − 5A equals
2 marksShow hintHide hint
Square A first, then subtract.
Show answerHide answer
Answer: D. 2I
A² = [7 10; 15 22] and 5A = [5 10; 15 20], so A² − 5A = [2 0; 0 2] = 2I. (Check: the characteristic equation is λ² − 5λ − 2 = 0.)
18. The value of k for which the matrix [2 k 1; 1 3 2; 3 1 1] (rows separated by ;) is singular is
2 marksShow hintHide hint
Singular means the determinant is zero.
Show answerHide answer
Answer: B. 6/5
Expanding along R₁: 2(3 − 2) − k(1 − 6) + 1(1 − 9) = 2 + 5k − 8 = 5k − 6. Setting 5k − 6 = 0 gives k = 6/5.
19. If A = [cosθ −sinθ; sinθ cosθ] (rows separated by ;), then A² equals
2 marksShow hintHide hint
Multiply and use the double-angle formulae.
Show answerHide answer
Answer: C. [cos2θ −sin2θ; sin2θ cos2θ]
A² has (1,1) entry cos²θ − sin²θ = cos2θ and (2,1) entry 2sinθcosθ = sin2θ, so A² = [cos2θ −sin2θ; sin2θ cos2θ]: a rotation by 2θ.
20. The value of the determinant |1 a a²; 1 b b²; 1 c c²| (rows separated by ;) is
2 marksShow hintHide hint
Subtract R₁ from the other rows and factorise.
Show answerHide answer
Answer: C. (a − b)(b − c)(c − a)
R₂ − R₁ and R₃ − R₁ give rows (0, b − a, b² − a²) and (0, c − a, c² − a²). Taking out (b − a)(c − a) leaves |1 b + a; 1 c + a| = c − b. Value = (b − a)(c − a)(c − b) = (a − b)(b − c)(c − a).
21. If A is a non-singular 3×3 matrix with |A| = 3, then |adj(adj A)| equals
2 marksShow hintHide hint
Apply |adj B| = |B|ⁿ⁻¹ twice.
Show answerHide answer
Answer: C. 81
|adj A| = |A|ⁿ⁻¹ = 3² = 9. Applying the rule again to adj A: |adj(adj A)| = |adj A|ⁿ⁻¹ = 9² = 81 (= |A|^((n−1)²)).
22. The system x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = 12 has no solution when λ equals
2 marksShow hintHide hint
Find when D = 0, then check consistency.
Show answerHide answer
Answer: A. 3
D = |1 1 1; 1 2 3; 1 2 λ| = λ − 3, which is zero when λ = 3. Then the last two equations read x + 2y + 3z = 10 and x + 2y + 3z = 12, which contradict each other, so there is no solution.
23. For the system x + y + z = 6, x − y + z = 2, 2x + y − z = 1, the coefficient determinant D and the value of z (by Cramer's rule) are
2 marksShow hintHide hint
z = Dz/D, where Dz replaces the z-column by the constants.
Show answerHide answer
Answer: C. D = 6, z = 3
D = 1(1 − 1) − 1(−1 − 2) + 1(1 + 2) = 6. Dz = |1 1 6; 1 −1 2; 2 1 1| = −3 + 3 + 18 = 18. So z = 18/6 = 3 (the full solution is x = 1, y = 2, z = 3).
24. If A = [1 2 0; 0 1 3; 0 0 1] (rows separated by ;), the element in the first row and third column of A⁻¹ is
2 marksShow hintHide hint
Use the cofactor of the (3,1) element, or A⁻¹ = I − N + N².
Show answerHide answer
Answer: A. 6
Write A = I + N with N = [0 2 0; 0 0 3; 0 0 0]. Since N³ = O, A⁻¹ = I − N + N², and N² has only the (1,3) entry 2×3 = 6. So A⁻¹ = [1 −2 6; 0 1 −3; 0 0 1].
25. If a square matrix A satisfies A² − 4A + I = O, then A⁻¹ equals
2 marksShow hintHide hint
Multiply the equation by A⁻¹.
Show answerHide answer
Answer: A. 4I − A
A² − 4A = −I gives A(4I − A) = I, so A⁻¹ = 4I − A.
2.2 Complex numbers and polynomial equations
26 questions
26. The value of k for which one root of the equation 3x² + 7x + 6 − k = 0 is equal to zero is
IOE model question 2080Show hintHide hint
Put x = 0 in the equation.
Show answerHide answer
Answer: D. 6
If x = 0 is a root, substituting gives 6 − k = 0, so k = 6.
27. The modulus of the complex number 3 − 4i is
Show hintHide hint
|a + ib| = √(a² + b²).
Show answerHide answer
Answer: A. 5
|3 − 4i| = √(3² + 4²) = √25 = 5.
28. The principal argument of −1 + i√3 is
Show hintHide hint
Find the quadrant first.
Show answerHide answer
Answer: B. 2π/3
The point lies in the second quadrant with tan α = √3/1, so α = π/3 and arg = π − π/3 = 2π/3.
29. The value of i¹⁰³ is
Show hintHide hint
Powers of i repeat every 4.
Show answerHide answer
Answer: C. −i
103 = 4×25 + 3, so i¹⁰³ = i³ = −i.
30. If ω is a non-real cube root of unity, then 1 + ω + ω² equals
Show hintHide hint
Factorise ω³ − 1.
Show answerHide answer
Answer: C. 0
ω³ = 1 and ω ≠ 1 gives (ω − 1)(ω² + ω + 1) = 0, so 1 + ω + ω² = 0.
31. If ω is a non-real cube root of unity, then (1 + ω²)³ equals
Show hintHide hint
Use 1 + ω + ω² = 0.
Show answerHide answer
Answer: A. −1
1 + ω² = −ω, so (1 + ω²)³ = (−ω)³ = −ω³ = −1.
32. The conjugate of 1/(2 + i) is
Show hintHide hint
Rationalise first, then conjugate.
Show answerHide answer
Answer: C. (2 + i)/5
1/(2 + i) = (2 − i)/((2 + i)(2 − i)) = (2 − i)/5. Its conjugate is (2 + i)/5.
33. The polar form of 1 + i is
Show hintHide hint
Find r and θ separately.
Show answerHide answer
Answer: B. √2(cos π/4 + i sin π/4)
r = √(1 + 1) = √2 and θ = tan⁻¹(1) = π/4 (first quadrant).
34. The value of (cos π/12 + i sin π/12)⁶ is
Show hintHide hint
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ.
Show answerHide answer
Answer: A. i
By De Moivre's theorem it equals cos(6π/12) + i sin(6π/12) = cos π/2 + i sin π/2 = i.
35. For any complex number z, the product z·z̄ equals
Show hintHide hint
Write z = x + iy.
Show answerHide answer
Answer: C. |z|²
If z = x + iy, z·z̄ = (x + iy)(x − iy) = x² + y² = |z|². (z + z̄ = 2 Re z and z − z̄ = 2i Im z.)
36. The roots of the equation 2x² + 3x + 5 = 0 are
Show hintHide hint
Check the sign of b² − 4ac.
Show answerHide answer
Answer: D. complex conjugates (non-real)
Discriminant = 3² − 4×2×5 = 9 − 40 = −31 < 0, so the roots are non-real complex conjugates.
37. If α and β are the roots of x² − 7x + 10 = 0, then 1/α + 1/β equals
Show hintHide hint
Use sum and product of roots.
Show answerHide answer
Answer: A. 7/10
1/α + 1/β = (α + β)/αβ = 7/10.
38. The quadratic equation whose roots are 2 and −3 is
Show hintHide hint
x² − (sum)x + product = 0.
Show answerHide answer
Answer: D. x² + x − 6 = 0
Sum = −1, product = −6, so x² − (−1)x + (−6) = x² + x − 6 = 0.
39. A quadratic equation with real coefficients has 2 + 3i as one root. The equation is
Show hintHide hint
Non-real roots of real equations occur in conjugate pairs.
Show answerHide answer
Answer: C. x² − 4x + 13 = 0
The other root is 2 − 3i. Sum = 4, product = 4 + 9 = 13, giving x² − 4x + 13 = 0.
40. The equation x² + kx + 9 = 0 has equal roots when k equals
Show hintHide hint
Set the discriminant to zero.
Show answerHide answer
Answer: C. ±6
Equal roots need k² − 4×9 = 0, so k² = 36 and k = ±6.
41. If α, β, γ are the roots of 2x³ − 4x² + 6x − 8 = 0, then αβγ equals
Show hintHide hint
Product of roots of a cubic is −d/a.
Show answerHide answer
Answer: D. 4
For ax³ + bx² + cx + d = 0, αβγ = −d/a = −(−8)/2 = 4. (The sum α + β + γ = −b/a = 2.)
42. The square roots of 5 + 12i are
2 marksShow hintHide hint
Use x² − y² = 5, 2xy = 12 and x² + y² = 13.
Show answerHide answer
Answer: A. ±(3 + 2i)
Let (x + iy)² = 5 + 12i: x² − y² = 5 and xy = 6 (same sign). Also x² + y² = |5 + 12i| = 13, so x² = 9, y² = 4, giving ±(3 + 2i). Check: (3 + 2i)² = 9 − 4 + 12i.
43. If α and β are the roots of x² − 3x + 1 = 0, then α³ + β³ equals
2 marksShow hintHide hint
Express α³ + β³ in terms of α + β and αβ.
Show answerHide answer
Answer: D. 18
α + β = 3, αβ = 1. α³ + β³ = (α + β)³ − 3αβ(α + β) = 27 − 9 = 18.
44. The value of (1 + i)⁸ is
2 marksShow hintHide hint
Square 1 + i first.
Show answerHide answer
Answer: B. 16
(1 + i)² = 2i, so (1 + i)⁸ = (2i)⁴ = 16 i⁴ = 16.
45. If ω is a non-real cube root of unity, then (1 + ω − ω²)(1 − ω + ω²) equals
2 marksShow hintHide hint
Replace 1 + ω and 1 + ω² using 1 + ω + ω² = 0.
Show answerHide answer
Answer: C. 4
1 + ω = −ω², so the first factor is −2ω². 1 + ω² = −ω, so the second is −2ω. Product = 4ω³ = 4.
46. The modulus and principal argument of (1 + i)/(1 − i) are respectively
2 marksShow hintHide hint
Multiply numerator and denominator by 1 + i.
Show answerHide answer
Answer: D. 1 and π/2
(1 + i)/(1 − i) = (1 + i)²/2 = 2i/2 = i, whose modulus is 1 and argument is π/2.
47. One root of x² + px + 12 = 0 is 4, and the equation x² + px + q = 0 has equal roots. The value of q is
2 marksShow hintHide hint
Find p first, then set the discriminant to zero.
Show answerHide answer
Answer: C. 49/4
16 + 4p + 12 = 0 gives p = −7. Equal roots of x² − 7x + q = 0 need 49 − 4q = 0, so q = 49/4.
48. If the roots of x² − 6x + k = 0 differ by 2, then k equals
2 marksShow hintHide hint
Use (α − β)² = (α + β)² − 4αβ.
Show answerHide answer
Answer: A. 8
(α − β)² = (α + β)² − 4αβ = 36 − 4k = 4, so k = 8 (roots 4 and 2).
49. If ω is a non-real cube root of unity, then (2 − ω)(2 − ω²) equals
2 marksShow hintHide hint
Expand and use ω + ω² = −1, ω³ = 1.
Show answerHide answer
Answer: B. 7
(2 − ω)(2 − ω²) = 4 − 2(ω + ω²) + ω³ = 4 − 2(−1) + 1 = 7.
50. The least positive integer n for which ((1 + i)/(1 − i))ⁿ = 1 is
2 marksShow hintHide hint
Simplify the base to a power of i.
Show answerHide answer
Answer: B. 4
(1 + i)/(1 − i) = i, and iⁿ = 1 first for n = 4.
51. If α and β are the roots of x² − 4x + 2 = 0, the equation whose roots are α² and β² is
2 marksShow hintHide hint
Find the new sum and product.
Show answerHide answer
Answer: B. x² − 12x + 4 = 0
α + β = 4, αβ = 2. α² + β² = 16 − 4 = 12 and α²β² = 4, so the equation is x² − 12x + 4 = 0.
2.3 Sequences, series, permutations and combinations
25 questions
52. The 10th term of the arithmetic progression 3, 7, 11, 15, … is
Show hintHide hint
Use tₙ = a + (n − 1)d.
Show answerHide answer
Answer: A. 39
Here a = 3 and d = 4, so t₁₀ = a + 9d = 3 + 36 = 39.
53. The sum of the first n odd natural numbers 1 + 3 + 5 + … + (2n − 1) equals
Show hintHide hint
Apply Sₙ = (n/2)[2a + (n − 1)d].
Show answerHide answer
Answer: B. n²
It is an AP with a = 1, d = 2: Sₙ = (n/2)[2 + (n − 1)2] = n².
54. The first term of a geometric progression of positive terms is 3 and its 5th term is 48. The common ratio is
Show hintHide hint
The nth term of a GP is arⁿ⁻¹.
Show answerHide answer
Answer: A. 2
ar⁴ = 48 with a = 3 gives r⁴ = 16, so r = 2 (positive terms).
55. The arithmetic mean and geometric mean of two positive numbers are 10 and 8 respectively. Their harmonic mean is
Show hintHide hint
Relate A, G and H for two numbers.
Show answerHide answer
Answer: B. 6.4
For two positive numbers G² = AH, so H = G²/A = 64/10 = 6.4.
56. The sum to infinity of the series 1 + 1/3 + 1/9 + 1/27 + … is
Show hintHide hint
S∞ = a/(1 − r) for |r| < 1.
Show answerHide answer
Answer: D. 3/2
Infinite GP with a = 1, r = 1/3: S∞ = a/(1 − r) = 1/(2/3) = 3/2.
57. The value of 1² + 2² + 3² + … + 10² is
Show hintHide hint
Use the formula for Σn².
Show answerHide answer
Answer: A. 385
Σn² = n(n + 1)(2n + 1)/6 = 10 × 11 × 21/6 = 385.
58. If 1 + 2 + 3 + … + n = 210, then 1³ + 2³ + 3³ + … + n³ equals
Show hintHide hint
Σn³ is related to (Σn).
Show answerHide answer
Answer: D. 44100
Σn³ = [n(n + 1)/2]² = (Σn)² = 210² = 44100.
59. The sum of the first n terms of the series 1·2 + 2·3 + 3·4 + … is
2 marksShow hintHide hint
Write the kth term as k² + k and use Σn², Σn.
Show answerHide answer
Answer: C. n(n + 1)(n + 2)/3
tₖ = k(k + 1) = k² + k, so Sₙ = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(2n + 4)/6 = n(n + 1)(n + 2)/3. Check n = 1: 1·2·3/3 = 2 ✓.
60. Three numbers in AP have sum 15 and product 80. The largest of the three numbers is
2 marksShow hintHide hint
Take the numbers as a − d, a, a + d.
Show answerHide answer
Answer: B. 8
Take a − d, a, a + d: 3a = 15 so a = 5; 5(25 − d²) = 80 gives d² = 9, d = ±3. Numbers are 2, 5, 8; largest is 8.
61. The 10th term of the harmonic progression 1/2, 1/5, 1/8, … is
Show hintHide hint
Terms of an HP are reciprocals of an AP.
Show answerHide answer
Answer: B. 1/29
Reciprocals 2, 5, 8, … form an AP with d = 3; its 10th term is 2 + 27 = 29, so the HP term is 1/29.
62. The sum of the first n terms of an AP is 3n² + 5n. The common difference of the AP is
2 marksShow hintHide hint
tₙ = Sₙ − Sₙ₋₁.
Show answerHide answer
Answer: C. 6
tₙ = Sₙ − Sₙ₋₁ = 3[n² − (n − 1)²] + 5 = 6n + 2, so t₁ = 8, t₂ = 14 and d = 6.
63. Three geometric means are inserted between 2 and 162. The product of these three geometric means is
2 marksShow hintHide hint
Find r from the 5-term GP 2, _, _, _, 162.
Show answerHide answer
Answer: D. 5832
2r⁴ = 162 gives r = 3, so the means are 6, 18, 54 and their product is 5832 (= 18³, the cube of the middle mean).
64. If a, b, c are in geometric progression, then
Show hintHide hint
The ratio of consecutive terms is constant.
Show answerHide answer
Answer: C. b² = ac
In a GP, b/a = c/b, so b² = ac. (2b = a + c is for AP and b = 2ac/(a + c) is for HP.)
65. If A, G and H are the arithmetic, geometric and harmonic means of two distinct positive numbers, then
Show hintHide hint
Pick two unequal positive numbers, compute all three means and compare them.
Show answerHide answer
Answer: B. A > G > H
For distinct positive numbers A > G > H, with equality only when the numbers are equal; also G² = AH.
66. The sum of the first n terms of the series 0.7 + 0.77 + 0.777 + … is
2 marksShow hintHide hint
Write 0.77… as (7/9)(1 − 10⁻ᵏ).
Show answerHide answer
Answer: C. (7/81)(9n − 1 + 10⁻ⁿ)
Sₙ = (7/9)[0.9 + 0.99 + …] = (7/9)[n − (0.1 + 0.01 + …)] = (7/9)[n − (1 − 10⁻ⁿ)/9] = (7/81)(9n − 1 + 10⁻ⁿ). Check n = 1: (7/81)(8.1) = 0.7 ✓.
67. The sum 1/(1·2) + 1/(2·3) + 1/(3·4) + … + 1/(n(n + 1)) equals
Show hintHide hint
Split each term into partial fractions.
Show answerHide answer
Answer: A. n/(n + 1)
1/(k(k + 1)) = 1/k − 1/(k + 1); the sum telescopes to 1 − 1/(n + 1) = n/(n + 1).
68. In an AP, the pth term is q and the qth term is p (p ≠ q). The (p + q)th term is
2 marksShow hintHide hint
Find d first by subtracting the two equations.
Show answerHide answer
Answer: A. 0
Subtracting a + (p − 1)d = q and a + (q − 1)d = p gives d = −1, then a = p + q − 1. So t₍p+q₎ = a + (p + q − 1)d = 0.
69. The number of distinct arrangements of all the letters of the word LETTER is
Show hintHide hint
Divide by the factorials of repeated letters.
Show answerHide answer
Answer: A. 180
6 letters with E twice and T twice: 6!/(2! 2!) = 720/4 = 180.
70. If ⁿC₃ = ⁿC₇, then the value of ⁿC₂ is
Show hintHide hint
Use ⁿCᵣ = ⁿCₙ₋ᵣ.
Show answerHide answer
Answer: C. 45
ⁿCᵣ = ⁿCₛ with r ≠ s means r + s = n, so n = 10 and ¹⁰C₂ = 45.
71. The ratio ⁿPᵣ / ⁿCᵣ is equal to
Show hintHide hint
Each selection of r objects can be arranged in a number of ways.
Show answerHide answer
Answer: B. r!
ⁿPᵣ = n!/(n − r)! and ⁿCᵣ = n!/(r!(n − r)!), so the ratio is r!.
72. The number of diagonals of a polygon with 10 sides is
Show hintHide hint
Count all vertex pairs, then remove the sides.
Show answerHide answer
Answer: D. 35
Lines joining two vertices = ¹⁰C₂ = 45; subtract the 10 sides to get 35 diagonals. (Also n(n − 3)/2 = 35.)
73. A committee of 5 is to be chosen from 6 men and 4 women. In how many ways can it be formed so that it contains exactly 2 women?
2 marksShow hintHide hint
Select women and men separately and multiply.
Show answerHide answer
Answer: B. 120
Choose 2 women in ⁴C₂ = 6 ways and 3 men in ⁶C₃ = 20 ways: 6 × 20 = 120.
74. How many four-digit numbers with all digits distinct can be formed using the digits 0 to 9?
2 marksShow hintHide hint
The thousands place cannot be 0.
Show answerHide answer
Answer: C. 4536
First digit: 9 choices (not 0); then 9, 8, 7 choices for the rest: 9 × 9 × 8 × 7 = 4536.
75. In how many ways can 6 persons sit around a round table if two particular persons must always sit together?
2 marksShow hintHide hint
Circular arrangements of n units: (n − 1)!.
Show answerHide answer
Answer: B. 48
Treat the pair as one unit: 5 units around a table in (5 − 1)! = 24 ways; the pair can swap in 2 ways: 48.
76. There are 12 points in a plane, of which exactly 5 are collinear and no other three are collinear. The number of triangles formed with these points as vertices is
2 marksShow hintHide hint
Subtract the triples chosen from the collinear points.
Show answerHide answer
Answer: B. 210
¹²C₃ − ⁵C₃ = 220 − 10 = 210, since three collinear points do not form a triangle.
2.4 Binomial theorem, exponential and logarithmic series
20 questions
77. The number of terms in the expansion of (x + y)¹⁰ is
Show hintHide hint
Count r = 0, 1, …, n.
Show answerHide answer
Answer: B. 11
(x + y)ⁿ has n + 1 terms, so 11 terms.
78. The sum of all the binomial coefficients in the expansion of (1 + x)⁸ is
Show hintHide hint
Substitute x = 1.
Show answerHide answer
Answer: C. 256
Put x = 1: C₀ + C₁ + … + C₈ = 2⁸ = 256.
79. The term independent of x in the expansion of (x² + 1/x)⁶ is
2 marksShow hintHide hint
Set the power of x in the general term to zero.
Show answerHide answer
Answer: B. 15
Tᵣ₊₁ = ⁶Cᵣ (x²)⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x¹²⁻³ʳ. 12 − 3r = 0 gives r = 4, so the term is ⁶C₄ = 15.
80. The middle term in the expansion of (x + 1/x)¹⁰ is
Show hintHide hint
For even n the middle term is T₍n/2 + 1₎.
Show answerHide answer
Answer: B. 252
n = 10 is even, so the middle term is T₆ = ¹⁰C₅ x⁵ (1/x)⁵ = 252.
81. The coefficient of x⁵ in the expansion of (1 + 2x)⁸ is
2 marksShow hintHide hint
Do not forget the power of 2.
Show answerHide answer
Answer: B. 1792
Tᵣ₊₁ = ⁸Cᵣ (2x)ʳ; for r = 5 the coefficient is ⁸C₅ × 2⁵ = 56 × 32 = 1792.
82. The (r + 1)th term in the expansion of (a + b)ⁿ is
Show hintHide hint
Check your choice with r = 0.
Show answerHide answer
Answer: B. ⁿCᵣ aⁿ⁻ʳ bʳ
The general term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ; for r = 0 it gives the first term aⁿ.
83. In the expansion of (1 + x)ⁿ, the sum C₀ + C₂ + C₄ + … of the even-indexed binomial coefficients equals
Show hintHide hint
Use x = 1 and x = −1.
Show answerHide answer
Answer: C. 2ⁿ⁻¹
Adding the results for x = 1 (sum 2ⁿ) and x = −1 (sum 0) gives 2(C₀ + C₂ + …) = 2ⁿ, so the sum is 2ⁿ⁻¹.
84. If the coefficients of the 5th and 9th terms in the expansion of (1 + x)ⁿ are equal, then n is
Show hintHide hint
The (r + 1)th term has coefficient ⁿCᵣ.
Show answerHide answer
Answer: C. 12
The 5th and 9th terms have coefficients ⁿC₄ and ⁿC₈; equal means 4 + 8 = n, so n = 12.
85. For the expansion of (1 + x)ⁿ, the value of C₁ + 2C₂ + 3C₃ + … + nCₙ is
2 marksShow hintHide hint
Differentiate the expansion, or use rCᵣ = n × ⁿ⁻¹Cᵣ₋₁.
Show answerHide answer
Answer: B. n·2ⁿ⁻¹
Differentiate (1 + x)ⁿ = ΣCᵣxʳ: n(1 + x)ⁿ⁻¹ = ΣrCᵣxʳ⁻¹; put x = 1 to get n·2ⁿ⁻¹. (Check n = 2: 2 + 2 = 4 ✓.)
86. The coefficient of x⁴ in the expansion of (x − 2/x)⁸ is
2 marksShow hintHide hint
Find r from the power of x, and track the sign of (−2)ʳ.
Show answerHide answer
Answer: B. 112
Tᵣ₊₁ = ⁸Cᵣ x⁸⁻ʳ (−2/x)ʳ = ⁸Cᵣ (−2)ʳ x⁸⁻²ʳ. 8 − 2r = 4 gives r = 2: 28 × 4 = 112.
87. The value of (√2 + 1)⁴ + (√2 − 1)⁴ is
2 marksShow hintHide hint
Expand both by the binomial theorem; terms with odd r cancel and terms with even r double.
Show answerHide answer
Answer: B. 34
Odd-power terms cancel: 2[(√2)⁴ + ⁴C₂(√2)² + 1] = 2[4 + 12 + 1] = 34.
88. When n is odd, the number of middle terms in the expansion of (a + b)ⁿ is
Show hintHide hint
Count the total number of terms first.
Show answerHide answer
Answer: C. 2
There are n + 1 (an even number of) terms, so there are two middle terms, T₍(n+1)/2₎ and T₍(n+3)/2₎.
89. The sum of the series 1/1! + 1/2! + 1/3! + … to ∞ is
Show hintHide hint
Compare with the exponential series for e.
Show answerHide answer
Answer: D. e − 1
e = 1 + 1/1! + 1/2! + …, so the given series (missing the first term 1) equals e − 1.
90. The sum 1 − 1/1! + 1/2! − 1/3! + 1/4! − … to ∞ equals
Show hintHide hint
Put a suitable value of x in eˣ.
Show answerHide answer
Answer: C. 1/e
This is eˣ = 1 + x + x²/2! + … with x = −1, giving e⁻¹ = 1/e.
91. The expansion logₑ(1 + x) = x − x²/2 + x³/3 − … is valid for
Show hintHide hint
Think about what happens at x = 1 and x = −1.
Show answerHide answer
Answer: C. −1 < x ≤ 1
The series converges for −1 < x ≤ 1; at x = −1 it becomes −(1 + 1/2 + 1/3 + …), which diverges.
92. The sum of the series 1 − 1/2 + 1/3 − 1/4 + … to ∞ is
Show hintHide hint
Use the logarithmic series.
Show answerHide answer
Answer: C. logₑ 2
Put x = 1 in logₑ(1 + x) = x − x²/2 + x³/3 − …, giving logₑ 2.
93. The coefficient of x³ in the expansion of e^(2x) is
Show hintHide hint
The general term of eᵏˣ is (kx)ⁿ/n!.
Show answerHide answer
Answer: D. 4/3
e^(2x) = Σ(2x)ⁿ/n!, so the coefficient of x³ is 2³/3! = 8/6 = 4/3.
94. The sum of the series 1 + 2/1! + 3/2! + 4/3! + … to ∞ is
2 marksShow hintHide hint
Split the numerator n + 1.
Show answerHide answer
Answer: D. 2e
General term (n + 1)/n! = n/n! + 1/n!; Σn/n! = Σ1/(n − 1)! = e and Σ1/n! = e, so the sum is 2e.
95. The sum 1/1! + 1/3! + 1/5! + 1/7! + … to ∞ equals
2 marksShow hintHide hint
Subtract the series for e⁻¹ from that of e.
Show answerHide answer
Answer: D. (e² − 1)/(2e)
e − e⁻¹ = 2(1/1! + 1/3! + 1/5! + …), so the sum is (e − 1/e)/2 = (e² − 1)/(2e).
96. The sum of the series 1/2 + 1/(3·2³) + 1/(5·2⁵) + 1/(7·2⁷) + … to ∞ is
2 marksShow hintHide hint
Use the series for logₑ[(1 + x)/(1 − x)].
Show answerHide answer
Answer: B. ½ logₑ 3
logₑ[(1 + x)/(1 − x)] = 2(x + x³/3 + x⁵/5 + …). With x = 1/2 the series equals ½ logₑ[(3/2)/(1/2)] = ½ logₑ 3.