Maths

14. If a and b are distinct integers, prove that a – b is a factor of aⁿ– bⁿ, whenever n is a positive integer.

[Hint: write aⁿ= (a – b + b)ⁿ and expand]

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14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ = $(a - b)$ k where k = [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ] is a natural number.

Therefore (a – b) is a factor o

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14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ = $(a - b)$ k where k = [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ] is a natural number.

Therefore (a – b) is a factor of aⁿ – bⁿwhere n is positive integer.

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13. Find a if the coefficients of x²and x³in the expansion of (3 + ax)⁹are equal.

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13. The general term of the expansion ${(3 + a x)}^{9}$ is

T_r₊₁ = ⁹C_r $3^{(9 - r)}$ ${(a x)}^{r}$

= ⁹C_r $3^{(9 - r)} a^{r} x^{r}$

At r = 2,

T₂₊₁ = ⁹C₂ $3^{(9 - 2)} a^{2} x^{2}$

= $\frac{9!}{2! (9 - 2)!}$ 3⁷a²x²

= $9′$ $8′$ $7!$ / $2′$ $1′$ $7!$ 3⁷a²x²

= 36 *3⁷a²x²

At r = 3,

T₃₊₁ = ⁹C₃ $3^{(9 - 3)} a^{3} x^{3}$

= $\frac{9!}{3! (9 - 3)!}$ 3⁶a³x³

= 9'8'7'6!/3'2'1'6! 3⁶a³x³

= 84 *3⁶a³x³

Given that,

Co-efficient of $x^{2}$ = co-efficient of $x^{3}$

=> 36 * $3^{7} a^{2}$ = 84 * $3^{6} a^{3}$

=> $\frac{a^{3}}{a^{2}}$ = 36' 3⁷^/84'3⁶

=> $a$ = $3′3/7$

= $\frac{9}{7}$

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133. Kindly consider the following

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133. Given, $c o s y = x c o s (a + y) .$

$x = \frac{c o s y}{c o s (a + y)}$

Differentiating w r t 'y' we get,

$\frac{d x}{d y} = \frac{d}{d y} (\frac{c o s y}{c o s (a + y)}) .$

$= \frac{c o s (a + y) \frac{d}{d y} c o s y - c o s y \frac{d}{d y} c o s (a + y)}{c o s^{2} (a + y) .}$

$= \frac{c o s (a + y) (- s i n y) - c o s y (- s i n (a + y))}{c o s^{2} (a + y) .}$

$= \frac{- c o s (a + y) s i n y + s i n (a + y) c o s y}{c o s^{2} (a + y)}$

$= \frac{s i n (a + y) c o s y - c o s (a + y) s i n y .}{c o s^{2} (a + y)}$

$\frac{d x}{d y} = \frac{s i n (a + y - y)}{c o s^{2} (a + y)} {\begin{array}{r} ? s i n (A - B) = s i n A c o s B & - c o s A s i n B \end{array}}$

So, $\frac{d y}{d x} = \frac{c o s^{2} (a + y)}{s i n a}$

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132. Kindly consider the following

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132. Given, ${(x - a)}^{2} + {(y - b)}^{2} = c^{2} .$

Differentiating w r t 'x' we get

$\frac{d}{d x} {(x - a)}^{2} + \frac{d}{d x} {(y - b)}^{2} = \frac{d}{d x} c^{2}$

$\Rightarrow 2 (x - a) + 2 (y - b) \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{2 (x - a)}{2 (y - b)} = - \frac{(x - a)}{(y - b)}$

Again, $\frac{d^{2} y}{d x^{2}} = - {\frac{(y - b) \frac{d}{d x} (x - a) - (x - a) \frac{d}{d x} (y - b)}{{(y - b)}^{2}}}$

$= - {\frac{(y - b) - (x - a) \frac{d y}{d x}}{{(y - b)}^{2}}}$

$= - {\frac{(y - b) + (x - a) \frac{(x - a)}{(y - b)}}{{(y - b)}^{2}}}$

$= - {\frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{3}}}$

$= - \frac{c^{2}}{{(y - b)}^{3}} {? {(x - a)}^{2} + {(y - b)}^{2} = c^{2}}$

Then, L.H.S = $\frac{{1 + {(\frac{d y}{d x})}^{2}}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = \frac{{1 + \frac{{(x - a)}^{2}}{{(y - b)}^{2}}}^{3 / 2}}{\frac{- c^{2}}{{(y - b)}^{2}}}$

$= \frac{{{(y - b)}^{2} + {(x - a)}^{2}}^{3 / 2}}{{(y - b)}^{3}} * \frac{{(y - b)}^{3}}{- c^{2}}$

$= \frac{c^{2 * 3 / 2}}{- c^{2}} = \frac{c^{3}}{- c^{2}} = - c$ Where c is a constant and is independent of a and b.

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131. Kindly consider the following

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131. Kindly go through the solution

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12. 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.

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1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n * (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n * (n - 1) * (n - 2)!}{2 * 1 * (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²* $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ * $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ * 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ * 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ = $\frac{25}{3}$

=> 10 – $\frac{b}{a}$ = $\frac{25}{3}$ [since, $\frac{n b}{a}$ &

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1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n * (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n * (n - 1) * (n - 2)!}{2 * 1 * (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²* $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ * $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ * 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ * 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ = $\frac{25}{3}$

=> 10 – $\frac{b}{a}$ = $\frac{25}{3}$ [since, $\frac{n b}{a}$ = 10]

=> $\frac{b}{a}$ = 10 – $\frac{25}{3}$

= $\frac{30 - 25}{3}$

= $\frac{5}{3}$ --------------- (5)

Putting equation (5) in (4) we get,

n* $\frac{5}{3}$ = 10

=>n = 10 * $\frac{3}{5}$

=>n = 6

So putting the value of n in equation (1) we get,

a6 = 729

=>a⁶= 36

=>a = 3

And putting a = 3 in equation (5) we get,

$\frac{b}{a}$ = $\frac{5}{3}$

=>b = $\frac{5}{3}$ *a

= $\frac{5}{3}$ * 3

= 5

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130. Kindly consider the following

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130. Kindly go through the solution

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11. Find a positive value of m for which the coefficient of x²in the expansion (1 + x)^mis 6.

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11. The general term of the expansion ${(1 + x)}^{m}$ is given by,

T_r₊₁ = ^mC_r $1^{m - r} x^{r}$

= ^mC_rx^r

At r = 2,

T₂₊₁ = ^mC₂x²

Given that, co-efficient of x² = 6

=>^mC₂ = 6

=> $\frac{m!}{2! (m - 2)!}$ = 6

=>m² – m = 12

=>m² – m – 12 = 0

=>m² + 3m – 4m – 12 = 0

=>m (m + 3) – 4 (m+ 3) = 0

=> (m – 4) (m + 3) = 0

=>m = 4 and m = –3

Since, we need a positive value of m we have, m = 4

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129. Kindly consider the following

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129. Given, $y = 12 (1 - c o s t) . x = 10 (t - s i n t) .$

Differentiating w r t 't' we get,

$\frac{d y}{d t} = 12 \frac{d}{d t} (1 - c o s t) = 12 (0 - (- s i n t)) = 12 s i n t .$

$\frac{d x}{d t} = 10 \frac{d}{d t} (t - s i n t) = 10 (1 - c o s t) .$

$∴ \frac{d y}{d x} = \frac{d y / d t}{d x / d t}$

$= \frac{12 s i n t}{10 (1 - c o s t)} = \frac{12 s i n t}{10 [1 - c o s t]}$

$= \frac{12 * 2 s i n t / t c o s t / 2}{10 * 2 s i n^{2} t / 2}$

${\begin{matrix} ? s i n 2 θ = 2 s i n θ c o s θ & c o s 2 θ = 1 - 2 s i n^{2} θ \end{matrix}}$ $= \frac{6}{5} \frac{c o s t / 2}{s i n t / 2} = \frac{6}{5} c o t t / 2$

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128. Kindly consider the following

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