GSEB Solutions Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 1.
Find a, b and n in the expansion of (a + b)ⁿ, if the first three terms of the expansion are 729, 7290 and 30375 respectively.
Solution:
T₁ of (a + b)ⁿ = aⁿ = 729 ……………….. (1)
T₂ of (a + b)ⁿ = ⁿC₁a^n-1b = 7290 ……………… (2)
T₃ of (a + b)ⁿ = ⁿC₂a^n-2b² = 30375 …………………. (3)
Dividing (1) by (2), we get

Dividing (2) by (3), we get

Dividing (4) by 5, we get

or 2n = 12 ⇒ n = 6.
Also, putting n = 6 in (1), we get a⁶ = 729 ⇒ a = 3
Putting n = 6 and a = 3 in (4), we get
\(\frac{3}{6b}\) = \(\frac{1}{10}\).
∴ b = \(\frac{3×10}{6}\) = 5.
Thus, a = 3, b = 5, n = 6.

Question 2.
Find a, if the coefficients of x² and x³ in the expan sion of (3 + ax)⁹ are equal.
Solution:
General Term = T_r+1 = ⁹C_r.3^9-ra^rx^r.
Putting r = 2, we get
Coefficient of x² = ⁹C₂3^9-2.a² = \(\frac{9×8}{2}\).3⁷a² = 4.3⁹a² …………………. (1)
Putting r = 3, we get
Coefficient of x³ = ⁹C₃3^9-3a³ = \(\frac{9×8×7}{6}\) × 3⁶.a³
= 4 × 7 × 3⁷.a³
Equating (1) and (2), we get
4.3⁹.a² = 4 × 7 × 3⁷a³, we get
or 3² = 7a ⇒ a = \(\frac{9}{7}\).

Question 3.
Find the coefficient of x⁵ in the expansion of the product (1 + 2x)⁶(1 – x)⁷.
Solution:
(1 + 2x)⁶ = ⁶C₀.1 + ⁶C₁(2x) + ⁶C₂(2x)² +
⁶C₃(2x)³ + ⁶C₄(2x)⁴ + ⁶C₅(2x)⁵ + ⁶C₆(2x)⁶
= 1 + 12x + 60x² + 20 × 8x³
+ 15 × 16x⁴ + 6 × 32x⁶ + 64x⁵
= 1 + 12x + 60x² + 160x³
+ 240x⁴ + 192x⁵ + 64x⁵ …………………. (1)
(1 – x)⁷ = 1 – ⁷C₁x + ⁷C₂x²
– ⁷C₃x³ + ⁷C₄x⁴ – ⁷C₅x⁵ + ⁷C₆x⁶ – ⁷C₇x⁷
= 1 – 7x + 21x² – 35x³ + 35x⁴ – 21x⁵ + 7x⁶ – x⁷ …………………… (2)
Multiplying (1) and (2) and collecting the coefficients of x⁵, we get
Coefficient of x⁵ in the product (1 + 2x)⁰(1 – y)⁷
= 1 × (- 21) + 12 × 35 + 60 × (- 35) + 160 × 21 + 240 × (- 7) + 192 × 1
= – 21 + 420 – 2100 + 3360 – 1680 + 192 = 171.

Question 4.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer?
Solution:
Now a = a + b – b = b + (a – b)
aⁿ = b + (a – b)ⁿ
= bⁿ + ⁿC₁a^n-1(a – b) + ⁿC₂b^n-2(a – b)² + ………………… + (a – b)ⁿ
or aⁿ – bⁿ = ⁿC₁b^n-1(a – b) + ⁿC₂(a – b)² + ……………. + (a – b)ⁿ
= (a – b) = (a – b)[ⁿC₁b^n-1 + ⁿC₂b^n-2(a – b) + ……………. + (a – b)^x-1]
Thus, (a – b) is a factor of (aⁿ – bⁿ).

Question 5.
Evaluate (\(\sqrt{3}\) + \(\sqrt{2}\))⁶ – (\(\sqrt{3}\) – \(\sqrt{2}\))⁶.
Solution:
(\(\sqrt{3}\) + \(\sqrt{2}\))⁶ = (\(\sqrt{3}\))⁶ + ⁶C₁(\(\sqrt{3}\))⁵(\(\sqrt{2}\))
+ ⁶C₂(\(\sqrt{3}\))⁴(\(\sqrt{2}\))² + ⁶C₅(\(\sqrt{3}\))(\(\sqrt{2}\))⁵ + (\(\sqrt{2}\))⁵ ………………. (1)
(\(\sqrt{3}\) – \(\sqrt{2}\))⁶ = (\(\sqrt{3}\))⁶ – ⁶C₁(\(\sqrt{3}\))⁵(\(\sqrt{2}\)) + ⁶C₂(\(\sqrt{3}\))⁴(\(\sqrt{2}\))⁴
– ⁶C₅(\(\sqrt{3}\))(\(\sqrt{2}\))⁵ + (\(\sqrt{2}\))⁵ ………………. (2)
Subtracting (2) from (1), we get
(\(\sqrt{3}\) + \(\sqrt{2}\))⁶ – (\(\sqrt{3}\) – \(\sqrt{2}\))⁶
= 2[⁶C₁(\(\sqrt{3}\))⁵(\(\sqrt{2}\)) + ⁶C₃(\(\sqrt{3}\))³ + ⁶C₅(\(\sqrt{3}\))(\(\sqrt{2}\))⁵]
= 2[6.3^5/2.2^1/2 = 20.3^3/22^3/2 + 6.3^1/2.2^5/2]
= 2.3^1/2.2^1/2[6.3² + 20.3^3/22^3/2 + 6.3^1/2.2^5/2]
= 2.3^1/2.2^1/2[6.3² + 20.3.2 + 6.2²]
= 2\(\sqrt{6}\) × 198 = 396\(\sqrt{6}\).

Question 6.
Find the value of (a² + \(\sqrt{a^{2}-1}\))⁴ + (a² – \(\sqrt{a^{2}-1}\))⁴.
Solution:
Put a² = x, \(\sqrt{a^{2}-1}\) = y.
∴ (x + y)⁴ = x⁴ + ⁴C₁x³y + ⁴C₂x²y² + ⁴C₃xy³ + ⁴C₄y⁴ ………………… (1)
(x – y)⁴ = x⁴ – ⁴C₁x³y + ⁴C₂x²y² – ⁴C₃xy³ + ⁴C₄y⁴ ………………. (2)
Adding (1) and (2), we get
(x + y)⁴ + (x – y)⁴ = 2[x⁴ + ⁴C₂x²y² + ⁴C₄y⁴] = 2[x⁴ + 6x²y² + y⁴].
∴ (a² + \(\sqrt{a^{2}-1}\)⁴ + (a² – \(\sqrt{a^{2}-1}\))⁴
= 2[a²)⁴ + 6(a²)²(\(\sqrt{a^{2}-1}\))² + (\(\sqrt{a^{2}-1}\))⁴].
= 2[a⁸ + 6a⁴(a² – 1) + (a² – 1)²]
= 2[a⁸ + 6a⁴(a² – 1) + a⁴ – 2a² + 1]
= 2[a⁸ + 6a⁶ – 5a⁴ – 2a² + 1].

Question 7.
Find an approximation of (0.99)⁵, using the first three terms of its expansion?
Solution:
(0.99)⁵ = (1 – 0.01)⁵
= 1 – 5C₁ × (0.01) + ⁵C₂ × (0.01)² ……
= 1 – 0.05 + 10 × 0.0001 ……
= 1.001 – 0.05
= 0.951.

Question 8.
Find x, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (\(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\))ⁿ is \(\sqrt{6}\) : 1.
Solution:

Total number of terms = n + 1.
Fifth term from the end
= [(n + 1) – 5 + 1]^th term from the beginning
= (n – 3)^th term from the beginning

Dividing (1) by (2), we get

Question 9.
Expand using Binomial theorem (1 + \(\frac{x}{2}\) – \(\frac{2}{x}\))⁴, x ≠ 0.
Solution:

Question 10.
Find the expansion of (3x² – 2ax + 3a²)³, using Binomial theorem?
Solution:
[3x² – a(2x – 3a)]³
= (3x²)³ – ³C₁(a²)².a(2x – 3a) + ³C₂(3x²)a² – a³(2x – 3a)³.
= 27x⁶ – 27x⁴ax – 3a) + 9x²a²(4x² – 12ax + 9a²)
– a³(8x³ – 3.4x².3a + 3.2x9a² – 27a³).
= 27x⁶ – 54x⁵a + 81a²x⁴ + 36a²x⁴ – 108a³x³
+ 81a⁴x² – 8a³x³ + 36a⁴x² – 54a⁵x + 27a⁶.
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶.