NDA Binomial Theorem previous year questions with solutions

Answer: (b) II only

Method. x² + x^−⁠2 + 2 = (x + 1x)²

(x² + x^−⁠2 + 2)⁴ = [(x + 1x)²]⁴ = (x + 1x)⁸.
This has 8 + 1 = 9 terms (powers x⁸, x⁶, …, x^−⁠8), so I is wrong.
The greatest coefficient is the middle one, ⁸C₄ = 70, so II is correct.

Answer: (b) 3

Method. the 4th term is a constant, so its power of x is zero; that fixes n, then m

T₄ = ⁿC₃(mx)^n−⁠3(1x)³ = ⁿC₃m^n−⁠3x^n−⁠6.
T₄ = 52 has no x, so n − 6 = 0 ⇒ n = 6.
⁶C₃m³ = 20m³ = 52 ⇒ m³ = 18 ⇒ m = 12.
mn = 12 × 6 = 3.

Answer: (d) 10

Method. general term T_r+1 with r = 5, then solve x²(log₁₀ x)⁵ = 100

T₆ = ⁸C₅(x^−⁠8/⁠3)³(x² log₁₀ x)⁵ = 56 · x^−⁠8 · x¹⁰(log₁₀ x)⁵ = 56x²(log₁₀ x)⁵.
56x²(log₁₀ x)⁵ = 5600 ⇒ x²(log₁₀ x)⁵ = 100.
x = 10: 10² × 1⁵ = 100. (holds)
For x = 6, 8, 9: x² < 100 and 0 < log₁₀ x < 1, so the product is below 100.

Answer: (a) 9

Method. combine into one trinomial raised to the 4th power, a homogeneous polynomial of degree 8

(3x − y)⁴(x + 3y)⁴ = [(3x − y)(x + 3y)]⁴ = (3x² + 8xy − 3y²)⁴.
This is homogeneous of degree 8, so its terms are x⁸, x⁷y, …, y⁸: at most 9 terms.
Expanding, the coefficients from x⁸ down to y⁸ are 81, 864, 3132, 3552, −⁠2330, −⁠3552, 3132, −⁠864, 81; none is zero, so there are 9 terms.

Answer: (b) 10

Method. combine into one binomial (1 + a)(1 − a) = 1 − a²

(1 + 2x)⁹(1 − 2x)⁹ = (1 − 4x²)⁹.
A binomial to the 9th power has 9 + 1 = 10 terms.

Answer: (c) C(2n, n)

Method. (n + 1)th term from the end in a (2n + 1)-term expansion is the middle term

(x + 1x)²ⁿ has 2n + 1 terms.
(n + 1)th from the end = (2n + 1 − (n + 1) + 1)th = (n + 1)th from the beginning, the middle term.
T_n+1 = ²ⁿC_n x^2n−⁠n(1x)ⁿ = ²ⁿC_n = C(2n, n).

Answer: (d) 43

Method. the trinomial is a perfect square, so the expansion is one binomial to the 42nd power

a²b² + b²a² + 2 = (ab + ba)².
The expression is (ab + ba)⁴².
Number of terms = 42 + 1 = 43.

Answer: (c) 21

Method. rewrite each trinomial as a perfect square, count terms, subtract the common constant

1 + 2x + x² = (1 + x)² and 1 + 4y + 4y² = (1 + 2y)².
The expression is (1 + x)¹⁰ + (1 + 2y)¹⁰.
Each expansion has 11 terms; the two constant terms (1 and 1) combine into one, and no other terms are alike.
Number of terms = 11 + 11 − 1 = 21.

Answer: (b) 5

Method. simplify to a single binomial (a − b)(a + b) = a² − b²

(2x − 3y)²(2x + 3y)² = [(2x − 3y)(2x + 3y)]² = (4x² − 9y²)².
So the expression is (4x² − 9y²)⁴.
A binomial to the 4th power has 4 + 1 = 5 terms.

Answer: (b) 9

Method. match the first three terms with 1 + ⁿC₁ax + ⁿC₂a²x²

(1 + ax)ⁿ = 1 + nax + n(n −⁠ 1)2a²x² + …
na = 12 and n(n −⁠ 1)2a² = 64.
Put a = 12/⁠n: n(n −⁠ 1)2 · 144n² = 64 ⇒ 72(n − 1)n = 64.
72n − 72 = 64n ⇒ n = 9 (and a = 4/⁠3).

Shortcut. Divide: (n − 1)a²n/⁠2(na)² = 64144 ⇒ n − 12n = 49 ⇒ n = 9.

Answer: (c) 6

Method. in (1 + t)ⁿ + (1 − t)ⁿ only even powers of t survive

Let t = 2⁠√3⁠ x. (1 + t)¹¹ + (1 − t)¹¹ = 2[¹¹C₀ + ¹¹C₂t² + ¹¹C₄t⁴ + … + ¹¹C₁₀t¹⁰].
The odd powers t, t³, …, t¹¹ cancel.
Surviving powers t⁰, t², …, t¹⁰: 6 terms, each with a non-zero coefficient.

Answer: (c) 51

Method. in (x + a)ⁿ + (x − a)ⁿ the odd-power-of-a terms cancel

(x + a)¹⁰⁰ + (x − a)¹⁰⁰ = 2[¹⁰⁰C₀x¹⁰⁰ + ¹⁰⁰C₂x⁹⁸a² + … + ¹⁰⁰C₁₀₀a¹⁰⁰].
The terms with odd powers of a cancel; those with a⁰, a², …, a¹⁰⁰ survive.
Number of even values 0, 2, …, 100 = 100/⁠2 + 1 = 51.

Answer: (b) n − 45

Method. general term T_r+1 = ⁿC_r x^n−⁠r(−⁠y)^r, then ratio of consecutive binomial coefficients

T₅ = ⁿC₄ x^n−⁠4(−⁠y)⁴ = ⁿC₄ x^n−⁠4y⁴.
T₆ = ⁿC₅ x^n−⁠5(−⁠y)⁵ = −⁠ⁿC₅ x^n−⁠5y⁵.
T₅ + T₆ = 0 ⇒ ⁿC₄ x^n−⁠4y⁴ = ⁿC₅ x^n−⁠5y⁵ ⇒ ⁿC₄ x = ⁿC₅ y.
xy = ⁿC₅ⁿC₄ = n − 45.

Shortcut. Use ⁿC_r/⁠ⁿC_r−⁠1 = (n − r + 1)/⁠r directly with r = 5.

Answer: (a) 1

Method. (1 + 1x)²ⁿ = (1 + x)²ⁿx²ⁿ

(1 + x)²ⁿ(1 + 1x)²ⁿ = (1 + x)²ⁿ · (1 + x)²ⁿx²ⁿ = (1 + x)⁴ⁿx²ⁿ.
Coefficient of x²ⁿ here = coefficient of x⁴ⁿ in (1 + x)⁴ⁿ = ⁴ⁿC_4n = 1.

Answer: (c) 4

Method. T_r+1 is rational when every surd exponent becomes an integer

T_r+1 = ¹²C_r(3^1/⁠2)^12−⁠r(5^1/⁠4)^r = ¹²C_r3^{12−⁠r/⁠2}5^r/⁠4, r = 0, 1, …, 12.
Rational ⇔ 12 − r2 and r4 are integers ⇔ r is a multiple of 4 (then 12 − r is even too).
r = 0, 4, 8, 12: 4 rational terms.

Answer: (c) 7

Method. (1 + x)^p(1 + x)^q = (1 + x)^p+q

(1 + x)^p(1 + x)^q = (1 + x)^p+q; coefficient of x³ = ^p+qC₃.
^p+qC₃ = 35.
⁷C₃ = 7 × 6 × 56 = 35, and ⁿC₃ increases with n(⁶C₃ = 20, ⁸C₃ = 56).
So p + q = 7.

Answer: (a) −⁠1

Method. 2 − x² − 1x² = −⁠(x − 1x)² = −(1 − x²)²x², which collapses the product to a polynomial

2 − x² − 1x² = −⁠(x² − 2 + 1x²) = −⁠(x − 1x)² = −(1 − x²)²x².
So (2 − x² − 1x²)^−⁠5 = (−⁠1)^−⁠5x¹⁰(1 − x²)¹⁰ = −x¹⁰(1 − x²)¹⁰.
Product = (1 − x²)²⁰ · (−x¹⁰(1 − x²)¹⁰) = −⁠x¹⁰(1 − x²)¹⁰.
Coefficient of x¹⁰ = −⁠1 × (constant term of (1 − x²)¹⁰) = −⁠1.

Answer: (a) p

Method. coefficient of x^r in (p + qx)⁹ is ⁹C_rp^9−⁠rq^r

Coefficient of x³ = ⁹C₃p⁶q³; coefficient of x⁶ = ⁹C₆p³q⁶.
⁹C₃ = ⁹C₆ = 84, so 84p⁶q³ = 84p³q⁶.
Divide by 84p³q³ (p, q ≠ 0): p³ = q³ ⇒ q = p.

Answer: (b) p/⁠q

Method. the two middle terms of a 9th power are T₅ and T₆

(p + qx)⁹ has 10 terms; the middle terms are T₅ and T₆.
T₅ = ⁹C₄p⁵q⁴x⁴; T₆ = ⁹C₅p⁴q⁵x⁵.
⁹C₄ = ⁹C₅ = 126, so the ratio of coefficients = 126p⁵q⁴126p⁴q⁵ = pq.

Answer: (b) p² : q² = 7 : 2

Method. coefficient of x^r in (p + qx)⁹ is ⁹C_rp^9−⁠rq^r

Coefficient of x² = ⁹C₂p⁷q² = 36p⁷q².
Coefficient of x⁴ = ⁹C₄p⁵q⁴ = 126p⁵q⁴.
Equal ⇒ 36p⁷q² = 126p⁵q⁴ ⇒ 36p² = 126q² (p, q ≠ 0).
p²q² = 12636 = 72, i.e. p² : q² = 7 : 2.

Answer: (a) 180

Method. general term; set the power of x to zero

T_r+1 = ¹⁰C_r(2x²)^10−⁠r(−⁠√x)^r⁠ = ¹⁰C_r2^10−⁠r(−⁠1)^rx^{−⁠2(10−⁠r) + r/⁠2}.
Power of x: −⁠20 + 2r + r2 = 0 ⇒ 5r2 = 20 ⇒ r = 8.
T₉ = ¹⁰C₈ · 2² · (−⁠1)⁸ = 45 × 4 = 180.

Answer: (b) ±3

Method. general term; set the power of x to zero

T_r+1 = ¹⁰C_r(⁠√x)^10−⁠r(−kx²)^r⁠ = ¹⁰C_r(−⁠k)^rx10−⁠r2 − 2r.
Constant term: 10 − r2 − 2r = 0 ⇒ 10 − r = 4r ⇒ r = 2.
T₃ = ¹⁰C₂k² = 45k².
45k² = 405 ⇒ k² = 9 ⇒ k = ±3.

Answer: (b) 14

Method. ⁿC_a = ⁿC_b ⇒ a = b or a + b = n

Coefficient of the (2r + 1)th term = ⁴³C_2r; of the (r + 2)th term = ⁴³C_r+1.
⁴³C_2r = ⁴³C_r+1 ⇒ 2r = r + 1 or 2r + (r + 1) = 43.
2r = r + 1 gives r = 1, which is excluded.
3r + 1 = 43 ⇒ r = 14.

Answer: (b) α = β

Method. ⁿC_r = ⁿC_n−⁠r

α = coefficient of a^m in (1 + a)^m+n = ^m+nC_m.
β = coefficient of aⁿ = ^m+nC_n = ^m+nC_(m+n)−⁠m = ^m+nC_m.
So α = β.

Answer: (d) The coefficient of x⁵⁰

Method. for an even power n the greatest binomial coefficient is ⁿC_n/⁠2

The coefficient of x^r in (1 + x)¹⁰⁰ is ¹⁰⁰C_r.
n = 100 is even, so the greatest coefficient is the middle one, ¹⁰⁰C₅₀, the coefficient of x⁵⁰.
(The coefficient of x⁵¹ is ¹⁰⁰C₅₁ = 5051 · ¹⁰⁰C₅₀, smaller; those of x⁹⁹ and x¹⁰⁰ are 100 and 1.)

Answer: (c) Fifth

Method. sum of binomial coefficients is 2ⁿ; the greatest coefficient of an even power is the middle one

Sum of binomial coefficients = 2ⁿ = 256 = 2⁸ ⇒ n = 8.
(x + y)⁸ has 9 terms; the greatest coefficient is ⁸C₄ = 70, in the middle term T₅.
So it occurs in the fifth term.

Answer: (a) 1

Method. (x + y)²ⁿ⁺¹(x − y)²ⁿ⁺¹ = (x² − y²)²ⁿ⁺¹; its two middle terms have equal binomial coefficients

The product is (x² − y²)²ⁿ⁺¹, with 2n + 2 terms; the middle terms are T_n+1 and T_n+2.
T_n+1 = ²ⁿ⁺¹C_n(x²)ⁿ⁺¹(−⁠y²)ⁿ; T_n+2 = ²ⁿ⁺¹C_n+1(x²)ⁿ(−⁠y²)ⁿ⁺¹.
²ⁿ⁺¹C_n = ²ⁿ⁺¹C_n+1, so T_n+1 + T_n+2 = (−⁠1)ⁿ·²ⁿ⁺¹C_nx²ⁿy²ⁿ(x² − y²).
Sum = 0 with x, y ≠ 0 ⇒ x² = y² ⇒ x²y² = 1.

Answer: (c) Both 1 and 2

Method. greatest binomial coefficient is the middle one; ⁿC_r = ⁿC_n−⁠r

(x + y)¹⁰ has 11 terms; the (r + 1)th term has coefficient ¹⁰C_r.
1. The greatest coefficient is ¹⁰C₅ = 252, in the 6th term (the middle term). Correct.
2. 3rd term: ¹⁰C₂ = 45; 9th term: ¹⁰C₈ = ¹⁰C₂ = 45. Correct.
Both statements are correct.

Answer: (a) 8064

Method. write the trinomial as a perfect square, then middle term of a binomial

1 + 4x + 4x² = (1 + 2x)², so the expression is (1 + 2x)¹⁰.
11 terms; the middle term is T₆ = ¹⁰C₅(2x)⁵.
Coefficient = 252 × 32 = 8064.

Answer: (a) 10

Method. middle term T_n+1 of a 2n-th power, then match the power of x

(x² + 1x)²ⁿ has 2n + 1 terms; the middle term is T_n+1.
T_n+1 = ²ⁿC_n(x²)ⁿ(1x)ⁿ = ²ⁿC_nxⁿ.
xⁿ = x¹⁰ ⇒ n = 10.
Check: ²⁰C₁₀ = 184756. (holds)

Answer: (d) 216

Method. middle term of (a + b)⁴ is T₃ = ⁴C₂a²b²

(2 + 3x)⁴ has 5 terms; the middle term is T₃.
T₃ = ⁴C₂(2)²(3x)² = 6 × 4 × 9x² = 216x².
Coefficient = 216.

Answer: (c) α = β + γ

Method. Pascal's rule ⁿC_r−⁠1 + ⁿC_r = ⁿ⁺¹C_r

(1 + x)²ⁿ has 2n + 1 terms; middle term T_n+1, so α = ²ⁿC_n.
(1 + x)^2n−⁠1 has 2n terms; middle terms T_n and T_n+1, so β = ^2n−⁠1C_n−⁠1 and γ = ^2n−⁠1C_n.
β + γ = ^2n−⁠1C_n−⁠1 + ^2n−⁠1C_n = ²ⁿC_n = α.

Answer: (d) 5

Method. subsets of size r and of size (2n + 1 − r) pair off, so at most n elements accounts for half of 2²ⁿ⁺¹

Number = ²ⁿ⁺¹C₀ + ²ⁿ⁺¹C₁ + … + ²ⁿ⁺¹C_n.
Since ²ⁿ⁺¹C_r = ²ⁿ⁺¹C_2n+1−⁠r, this equals the sum of the other half, so it is 2²ⁿ⁺¹2 = 2²ⁿ.
2²ⁿ = 1024 = 2¹⁰ ⇒ n = 5.

Answer: (b) 6

Method. subsets of size r and of size (2n + 1 − r) pair off, so each half of the 2²ⁿ⁺¹ subsets has 2²ⁿ

Subsets with at most n elements: ²ⁿ⁺¹C₀ + ²ⁿ⁺¹C₁ + … + ²ⁿ⁺¹C_n.
Since ²ⁿ⁺¹C_r = ²ⁿ⁺¹C_2n+1−⁠r, this sum equals the sum of the remaining coefficients, so it is half of 2²ⁿ⁺¹, i.e. 2²ⁿ.
2²ⁿ = 4096 = 2¹² ⇒ 2n = 12 ⇒ n = 6.

Answer: (d) n2^n-1

Method. r · ⁿC_r = n · ^n−⁠1C_r−⁠1, or differentiate (1 + x)ⁿ and put x = 1

Here T_r = ⁿC_r.
Differentiate: n(1 + x)^n−⁠1 = T₁ + 2T₂x + 3T₃x² + … + nT_nx^n−⁠1.
Put x = 1: T₁ + 2T₂ + … + nT_n = n2^n−⁠1.

Answer: (d) 1

Method. differentiate (1 + x)ⁿ and put x = −⁠1

Here T_r = ⁿC_r. Differentiate: n(1 + x)^n−⁠1 = T₁ + 2T₂x + 3T₃x² + … + nT_nx^n−⁠1.
Put x = −⁠1 (n ≥ 2): 0 = T₁ − 2T₂ + 3T₃ − … + (−⁠1)^n−⁠1nT_n.
So −⁠T₁ + 2T₂ − 3T₃ + … + (−⁠1)ⁿnT_n = 0.
The required value = 1 + 0 = 1.

Answer: (b) 2ⁿ - 1

Method. sum of binomial coefficients is 2ⁿ

Put x = 1: 1 + T₁ + T₂ + … + T_n = 2ⁿ.
So T₁ + T₂ + … + T_n = 2ⁿ − 1.

Answer: (b) 3ⁿ

Method. binomial theorem (1 + x)ⁿ = Σ ⁿC_rx^r with x = 2

(1 + x)ⁿ = nΣr=0 C(n, r)x^r.
Put x = 2: nΣr=0 2^rC(n, r) = (1 + 2)ⁿ = 3ⁿ.

Answer: (c) 1 + n²

Method. coefficients ⁿC₀, ⁿC₁, ⁿC_n−⁠1, ⁿC_n

The (r + 1)th term of (1 + x)ⁿ has coefficient ⁿC_r.
p = ⁿC₀ = 1, q = ⁿC₁ = n, r = ⁿC_n−⁠1 = n, s = ⁿC_n = 1.
ps + qr = 1 × 1 + n × n = 1 + n².

Answer: (b) 2ⁿ - 1

Method. sum of all binomial coefficients is 2ⁿ

Put x = 1 in (1 + x)ⁿ: C₀ + C₁ + … + C_n = 2ⁿ.
C₀ = 1, so C₁ + C₂ + … + C_n = 2ⁿ − 1.

Answer: (c) 1 + 2 + 2² + 2³ + … + 2^n-1

Method. sum of binomial coefficients and the geometric series 1 + 2 + … + 2^n−⁠1

C(n, 1) + C(n, 2) + … + C(n, n) = 2ⁿ − 1.
Geometric series: 1 + 2 + 2² + … + 2^n−⁠1 = 2ⁿ − 12 − 1 = 2ⁿ − 1.
So the sum equals option (c).
Check the others: (a) = 2ⁿ⁺¹ − 2, (b) = 2ⁿ⁺¹ − 1, (d) = 2ⁿ − 2.

Answer: (b) 2

Method. first and last binomial coefficients are both 1

First term coefficient = ²ⁿC₀ = 1; last term coefficient = ²ⁿC_2n = 1.
Sum = 1 + 1 = 2.

Answer: (b) 64

Method. alternating sum of coefficients: put x = −⁠1

Put x = −⁠1 on both sides: a₀ − a₁ + a₂ − … + a₁₂ = (1 − 2 − 1)⁶.
(−⁠2)⁶ = 64.

Answer: (a) 256

Method. symmetry ⁿC_r = ⁿC_n−⁠r splits the 10 coefficients into two equal halves

(1 + x)⁹ has 10 coefficients ⁹C₀, …, ⁹C₉, adding to 2⁹ = 512.
The last five, ⁹C₅ + … + ⁹C₉, equal the first five, ⁹C₄ + … + ⁹C₀, by symmetry.
So the last five add to 512/⁠2 = 256.

Answer: (b) 2⁴⁹

Method. sum of odd-placed and even-placed binomial coefficients are equal, each 2^n−⁠1

Put x = 1: C₀ + C₁ + … + C₅₀ = 2⁵⁰.
Put x = −⁠1: C₀ − C₁ + C₂ − … + C₅₀ = 0.
Subtract and halve: C₁ + C₃ + … + C₄₉ = 2⁵⁰/⁠2 = 2⁴⁹.

Answer: (a) 254

Method. Pascal's rule C(n, r) + C(n, r + 1) = C(n + 1, r + 1)

Each bracket: C(7, r) + C(7, r + 1) = C(8, r + 1), for r = 0, 1, …, 6.
Sum = C(8, 1) + C(8, 2) + … + C(8, 7).
All coefficients of (1 + x)⁸ add to 2⁸ = 256; remove C(8, 0) = 1 and C(8, 8) = 1.
Sum = 256 − 2 = 254.

Answer: (c) 8

Method. 5² = 25 = 26 − 1, and 26 is a multiple of 13

5⁹⁹ = 5 · (5²)⁴⁹ = 5(26 − 1)⁴⁹.
(26 − 1)⁴⁹ = 26⁴⁹ − ⁴⁹C₁26⁴⁸ + … + ⁴⁹C₄₈26 − 1 = 13m − 1 for some integer m.
5⁹⁹ = 5(13m − 1) = 65m − 5 = 13(5m − 1) + 8.
Remainder = 8.

Answer: (b) 1

Method. write 7 = 1 + 6 and expand; terms from the third onward contain 6² = 36

7ⁿ = (1 + 6)ⁿ = 1 + 6n + ⁿC₂6² + ⁿC₃6³ + … + 6ⁿ.
Every term after the second is a multiple of 36.
So 7ⁿ − 6n = 1 + 36k for some integer k.
Remainder = 1 (for n = 100, or any n).

Answer: (a) 1

Method. write 2³ = 8 = 7 + 1 and expand

2¹²⁰ = (2³)⁴⁰ = 8⁴⁰ = (7 + 1)⁴⁰.
(7 + 1)⁴⁰ = 7⁴⁰ + ⁴⁰C₁7³⁹ + … + ⁴⁰C₃₉7 + 1.
Every term except the last is a multiple of 7, so the remainder is 1.

Answer: (b) 2000

Method. aⁿ − bⁿ is divisible by (a − b); group the four terms in two ways

Grouping 1: (121ⁿ − 25ⁿ) + (1900ⁿ − (−⁠4)ⁿ) is divisible by 121 − 25 = 96 and by 1900 + 4 = 1904; both are multiples of 16, so the number is divisible by 16.
Grouping 2: (121ⁿ − (−⁠4)ⁿ) + (1900ⁿ − 25ⁿ) is divisible by 121 + 4 = 125 and by 1900 − 25 = 1875 = 15 × 125, so the number is divisible by 125.
16 and 125 are coprime, so the number is divisible by 16 × 125 = 2000.
Check n = 1: 121 − 25 + 1900 + 4 = 2000, which is not divisible by 1904, 2002 or 2006.

Shortcut. Put n = 1: the value is 2000, and only option (b) divides it.

Answer: (d) 12

Method. aⁿ + bⁿ is divisible by (a + b) when n is odd

5 is odd, so 5⁵ + 7⁵ is divisible by 5 + 7 = 12.
Hence it is divisible by 6 and 12; 12 is the greatest option that divides it.
Check: 5⁵ + 7⁵ = 3125 + 16807 = 19932 = 12 × 1661, and 19932/⁠8 is not an integer.

Answer: (c) Both I and II

Method. (⁠√2⁠ + 1)¹⁰ + (⁠√2⁠ − 1)¹⁰ is an integer, and 0 < v < 1

(⁠√2⁠ + 1)¹⁰ + (⁠√2⁠ − 1)¹⁰ = 2[¹⁰C₀(⁠√2)¹⁰⁠ + ¹⁰C₂(⁠√2)⁸⁠ + … + ¹⁰C₁₀]: odd powers of ⁠√2⁠ cancel, even powers are integers.
So u + f + v is an integer: I is correct.
u is an integer, so f + v = (u + f + v) − u is an integer: II is correct.
(Indeed 0 < f < 1 and 0 < v = (⁠√2⁠ − 1)¹⁰ < 1, so f + v = 1.)

Answer: (c) v²

Method. (⁠√2⁠ + 1)(⁠√2⁠ − 1) = 1, so (⁠√2⁠ + 1)^−⁠1 = ⁠√2⁠ − 1

(⁠√2⁠ + 1)(⁠√2⁠ − 1) = 2 − 1 = 1 ⇒ 1⁠√2⁠ + 1 = ⁠√2⁠ − 1.
1(⁠√2⁠ + 1)²⁰ = (⁠√2⁠ − 1)²⁰ = [(⁠√2⁠ − 1)¹⁰]² = v².

Answer: (b) 1

Method. f + v is an integer lying strictly between 0 and 2

From the stem, (⁠√2⁠ + 1)¹⁰ + (⁠√2⁠ − 1)¹⁰ = u + f + v is an integer (odd powers of ⁠√2⁠ cancel), so f + v is an integer.
0 < f < 1 and 0 < v < 1 (since 0 < ⁠√2⁠ − 1 < 1), so 0 < f + v < 2.
Hence v + f = 1.

Answer: (d) 6725

Method. u = (⁠√2⁠ + 1)¹⁰ + (⁠√2⁠ − 1)¹⁰ − 1, computed via (⁠√2⁠ ± 1)² = 3 ± 2⁠√2⁠

u + f + v = (⁠√2⁠ + 1)¹⁰ + (⁠√2⁠ − 1)¹⁰ and f + v = 1, so u = (⁠√2⁠ + 1)¹⁰ + (⁠√2⁠ − 1)¹⁰ − 1.
(⁠√2⁠ ± 1)² = 3 ± 2⁠√2,⁠ so the sum is (3 + 2⁠√2)⁵⁠ + (3 − 2⁠√2)⁵⁠ = 2[3⁵ + ⁵C₂3³(2⁠√2)²⁠ + ⁵C₄3(2⁠√2)⁴].⁠
= 2[243 + 10 × 27 × 8 + 5 × 3 × 64] = 2[243 + 2160 + 960] = 2 × 3363 = 6726.
u = 6726 − 1 = 6725.

Answer: (c) 0 < uv < 1

Method. u + f = 1/⁠v, so uv = 1 − fv

(⁠√2⁠ + 1)¹⁰(⁠√2⁠ − 1)¹⁰ = 1 ⇒ (u + f)v = 1 ⇒ uv = 1 − fv.
0 < f < 1 and 0 < v < 1 ⇒ 0 < fv < 1 ⇒ 0 < uv < 1.
(Numerically v ≈ 0.000149 and uv ≈ 0.99985.)

Answer: (b) 13

Method. expand (⁠√2⁠ + 1)³ exactly, then bound ⁠√2⁠

(⁠√2⁠ + 1)³ = (⁠√2)³⁠ + 3(⁠√2)²⁠ + 3⁠√2⁠ + 1 = 2⁠√2⁠ + 6 + 3⁠√2⁠ + 1 = 7 + 5⁠√2.⁠
1.414 < ⁠√2⁠ < 1.415 ⇒ 14.07 < 7 + 5⁠√2⁠ < 14.08.
Working: '… so k < 14.08 and k > 12.07, so k = 13 or 14.' Add to note: 'The condition alone does not fix k (13 and 14 both satisfy it); of the two, only 13 is offered, so (b).'

Answer: (d) 1

Method. (8 + 3⁠√7)²⁰⁠ + (8 − 3⁠√7)²⁰⁠ is an integer, and 0 < W < 1

(8 + 3⁠√7)²⁰⁠ + (8 − 3⁠√7)²⁰⁠ = 2[²⁰C₀8²⁰ + ²⁰C₂8¹⁸(3⁠√7)²⁠ + …]: odd powers of 3⁠√7⁠ cancel, even powers are integers. So U + V + W is an integer.
U is an integer, so V + W is an integer.
8 − 3⁠√7⁠ = 64 − 638 + 3⁠√7⁠ = 18 + 3⁠√7,⁠ which lies between 0 and 1, so 0 < W < 1.
0 < V < 1 and 0 < W < 1 ⇒ 0 < V + W < 2, so V + W = 1.

Answer: (b) 1

Method. conjugate surds: (8 + 3⁠√7)(8⁠ − 3⁠√7)⁠ = 1

(U + V)W = (8 + 3⁠√7)²⁰(8⁠ − 3⁠√7)²⁰⁠ = [(8 + 3⁠√7)(8⁠ − 3⁠√7)]²⁰.⁠
(8 + 3⁠√7)(8⁠ − 3⁠√7)⁠ = 64 − 63 = 1.
So (U + V)W = 1²⁰ = 1.