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:

T_{1} of (a + b)^{n} = a^{n} = 729 ……………….. (1)

T_{2} of (a + b)^{n} = ^{n}C_{1}a^{n-1}b = 7290 ……………… (2)

T_{3} of (a + b)^{n} = ^{n}C_{2}a^{n-2}b^{2} = 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 a^{6} = 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 x^{2} and x^{3} in the expan sion of (3 + ax)^{9} are equal.

Solution:

General Term = T_{r+1} = ^{9}C_{r}.3^{9-r}a^{r}x^{r}.

Putting r = 2, we get

Coefficient of x^{2} = ^{9}C_{2}3^{9-2}.a^{2} = \(\frac{9×8}{2}\).3^{7}a^{2} = 4.3^{9}a^{2} …………………. (1)

Putting r = 3, we get

Coefficient of x^{3} = ^{9}C_{3}3^{9-3}a^{3} = \(\frac{9×8×7}{6}\) × 3^{6}.a^{3}

= 4 × 7 × 3^{7}.a^{3}

Equating (1) and (2), we get

4.3^{9}.a^{2} = 4 × 7 × 3^{7}a^{3}, we get

or 3^{2} = 7a ⇒ a = \(\frac{9}{7}\).

Question 3.

Find the coefficient of x^{5} in the expansion of the product (1 + 2x)^{6}(1 – x)^{7}.

Solution:

(1 + 2x)^{6} = ^{6}C_{0}.1 + ^{6}C_{1}(2x) + ^{6}C_{2}(2x)^{2} +

^{6}C_{3}(2x)^{3} + ^{6}C_{4}(2x)^{4} + ^{6}C_{5}(2x)^{5} + ^{6}C_{6}(2x)^{6}

= 1 + 12x + 60x^{2} + 20 × 8x^{3}

+ 15 × 16x^{4} + 6 × 32x^{6} + 64x^{5}

= 1 + 12x + 60x^{2} + 160x^{3}

+ 240x^{4} + 192x^{5} + 64x^{5} …………………. (1)

(1 – x)^{7} = 1 – ^{7}C_{1}x + ^{7}C_{2}x^{2}

– ^{7}C_{3}x^{3} + ^{7}C_{4}x^{4} – ^{7}C_{5}x^{5} + ^{7}C_{6}x^{6} – ^{7}C_{7}x^{7}

= 1 – 7x + 21x^{2} – 35x^{3} + 35x^{4} – 21x^{5} + 7x^{6} – x^{7} …………………… (2)

Multiplying (1) and (2) and collecting the coefficients of x^{5}, we get

Coefficient of x^{5} 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 a^{n} – b^{n}, whenever n is a positive integer?

Solution:

Now a = a + b – b = b + (a – b)

a^{n} = b + (a – b)^{n}

= b^{n} + ^{n}C_{1}a^{n-1}(a – b) + ^{n}C_{2}b^{n-2}(a – b)^{2} + ………………… + (a – b)^{n}

or a^{n} – b^{n} = ^{n}C_{1}b^{n-1}(a – b) + ^{n}C_{2}(a – b)^{2 }+ ……………. + (a – b)^{n}

= (a – b) = (a – b)[^{n}C_{1}b^{n-1} + ^{n}C_{2}b^{n-2}(a – b) + ……………. + (a – b)^{x-1}]

Thus, (a – b) is a factor of (a^{n} – b^{n}).

Question 5.

Evaluate (\(\sqrt{3}\) + \(\sqrt{2}\))^{6} – (\(\sqrt{3}\) – \(\sqrt{2}\))^{6}.

Solution:

(\(\sqrt{3}\) + \(\sqrt{2}\))^{6} = (\(\sqrt{3}\))^{6} + ^{6}C_{1}(\(\sqrt{3}\))^{5}(\(\sqrt{2}\))

+ ^{6}C_{2}(\(\sqrt{3}\))^{4}(\(\sqrt{2}\))^{2} + ^{6}C_{5}(\(\sqrt{3}\))(\(\sqrt{2}\))^{5} + (\(\sqrt{2}\))^{5} ………………. (1)

(\(\sqrt{3}\) – \(\sqrt{2}\))^{6} = (\(\sqrt{3}\))^{6} – ^{6}C_{1}(\(\sqrt{3}\))^{5}(\(\sqrt{2}\)) + ^{6}C_{2}(\(\sqrt{3}\))^{4}(\(\sqrt{2}\))^{4}

– ^{6}C_{5}(\(\sqrt{3}\))(\(\sqrt{2}\))^{5} + (\(\sqrt{2}\))^{5} ………………. (2)

Subtracting (2) from (1), we get

(\(\sqrt{3}\) + \(\sqrt{2}\))^{6} – (\(\sqrt{3}\) – \(\sqrt{2}\))^{6}

= 2[^{6}C_{1}(\(\sqrt{3}\))^{5}(\(\sqrt{2}\)) + ^{6}C_{3}(\(\sqrt{3}\))^{3} + ^{6}C_{5}(\(\sqrt{3}\))(\(\sqrt{2}\))^{5}]

= 2[6.3^{5/2}.2^{1/2} = 20.3^{3/2}2^{3/2} + 6.3^{1/2}.2^{5/2}]

= 2.3^{1/2}.2^{1/2}[6.3^{2} + 20.3^{3/2}2^{3/2} + 6.3^{1/2}.2^{5/2}]

= 2.3^{1/2}.2^{1/2}[6.3^{2} + 20.3.2 + 6.2^{2}]

= 2\(\sqrt{6}\) × 198 = 396\(\sqrt{6}\).

Question 6.

Find the value of (a^{2} + \(\sqrt{a^{2}-1}\))^{4} + (a^{2} – \(\sqrt{a^{2}-1}\))^{4}.

Solution:

Put a^{2} = x, \(\sqrt{a^{2}-1}\) = y.

∴ (x + y)^{4} = x^{4} + ^{4}C_{1}x^{3}y + ^{4}C_{2}x^{2}y^{2} + ^{4}C_{3}xy^{3} + ^{4}C_{4}y^{4} ………………… (1)

(x – y)^{4} = x^{4} – ^{4}C_{1}x^{3}y + ^{4}C_{2}x^{2}y^{2} – ^{4}C_{3}xy^{3} + ^{4}C_{4}y^{4} ………………. (2)

Adding (1) and (2), we get

(x + y)^{4} + (x – y)^{4} = 2[x^{4} + ^{4}C_{2}x^{2}y^{2} + ^{4}C_{4}y^{4}] = 2[x^{4} + 6x^{2}y^{2} + y^{4}].

∴ (a^{2} + \(\sqrt{a^{2}-1}\)^{4} + (a^{2} – \(\sqrt{a^{2}-1}\))^{4}

= 2[a^{2})^{4} + 6(a^{2})^{2}(\(\sqrt{a^{2}-1}\))^{2} + (\(\sqrt{a^{2}-1}\))^{4}].

= 2[a^{8} + 6a^{4}(a^{2} – 1) + (a^{2} – 1)^{2}]

= 2[a^{8} + 6a^{4}(a^{2} – 1) + a^{4} – 2a^{2} + 1]

= 2[a^{8} + 6a^{6} – 5a^{4} – 2a^{2} + 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 – 5C_{1} × (0.01) + ^{5}C_{2} × (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 (3x^{2} – 2ax + 3a^{2})^{3}, using Binomial theorem?

Solution:

[3x^{2} – a(2x – 3a)]^{3}

= (3x^{2})^{3} – ^{3}C_{1}(a^{2})^{2}.a(2x – 3a) + ^{3}C_{2}(3x^{2})a^{2} – a^{3}(2x – 3a)^{3}.

= 27x^{6} – 27x^{4}ax – 3a) + 9x^{2}a^{2}(4x^{2} – 12ax + 9a^{2})

– a^{3}(8x^{3} – 3.4x^{2}.3a + 3.2x9a^{2} – 27a^{3}).

= 27x^{6} – 54x^{5}a + 81a^{2}x^{4} + 36a^{2}x^{4} – 108a^{3}x^{3}

+ 81a^{4}x^{2} – 8a^{3}x^{3} + 36a^{4}x^{2} – 54a^{5}x + 27a^{6}.

= 27x^{6} – 54ax^{5} + 117a^{2}x^{4} – 116a^{3}x^{3} + 117a^{4}x^{2} – 54a^{5}x + 27a^{6}.