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1

### AIEEE 2006

For natural numbers $$m$$ , $$n$$, if $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + ..........$$ and $${a_1} = {a_2} = 10,$$ then $$\left( {m,\,n} \right)$$ is
A
$$\left( {20,\,45} \right)$$
B
$$\left( {35,\,20} \right)$$
C
$$\left( {45,\,35} \right)$$
D
$$\left( {35,\,45} \right)$$

## Explanation

$${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,$$

= $$\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)$$ -
$$\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)$$

$${a_1}$$ = Coefficient of y = $${{}^n{C_1}}$$ - $${{}^m{C_1}}$$ = 10

$$\Rightarrow$$ n - m = 10

$${a_2}$$ = Coefficient of y2

= $${}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10$$

$$\Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10$$

$$\Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20$$

$$\Rightarrow$$ (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20

$$\Rightarrow$$ 90 + 19m + m2 - 2m2 - 20m + m2 - m - 20 = 0

$$\Rightarrow$$ 70 - 2m = 0

$$\Rightarrow$$ m = 35

$$\therefore$$ n = 10 + 35 = 45
2

### AIEEE 2006

If the expansion in powers of $$x$$ of the function $${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$ is $${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....$$ then $${a_n}$$ is
A
$${{{b^n} - {a^n}} \over {b - a}}$$
B
$${{{a^n} - {b^n}} \over {b - a}}$$
C
$${{{a^{n + 1}} - {b^{n + 1}}} \over {b - a}}$$
D
$${{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}$$

## Explanation

$${1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}$$

= $${\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}}$$

= $$\left[ {1 + \left( { - 1} \right)\left( { - ax} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - ax} \right)}^2} + ...} \right]$$ -
$$\left[ {1 + \left( { - 1} \right)\left( { - bx} \right) + {{\left( { - 1} \right)\left( { - 2} \right)} \over {1.2}}{{\left( { - bx} \right)}^2} + ...} \right]$$

= $$\left[ {1 + ax + {a^2}{x^2} + ... + {a^{n - 1}}{x^{n - 1}} + {a^n}{x^n}}+.... \right]$$ -
$$\left[ {1 + bx + {b^2}{x^2} + ... + {b^{n - 1}}{x^{n - 1}} + {b^n}{x^n}}+.... \right]$$

Coefficient of xn =

$${a^n} + {a^{n - 1}}b + {a^{n - 2}}{b^2} + .... + {b^n}$$

= $${a^n}\left[ {1 + {b \over a} + {{{b^2}} \over {{a^2}}} + ..... + {{{b^n}} \over {{a^n}}}} \right]$$

= $${a^n}\left[ {{{{{\left( {{b \over a}} \right)}^{n + 1}} - 1} \over {{b \over a} - 1}}} \right]$$

= $${a^n}\left[ {{{{b^{n + 1}} - {a^{n + 1}}} \over {{a^{n + 1}}\left( {{{b - a} \over a}} \right)}}} \right]$$

= $${{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}}$$
3

### AIEEE 2005

If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation
A
$${m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0$$
B
$${m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0$$
C
$${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$
D
$${m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0$$

## Explanation

Let r = 2

$$\therefore$$ 2nd, 3rd and 4th terms are in AP.

2nd term = T2 = $${}^m{C_1}.y$$

Coefficient of T2 = $${}^m{C_1}$$

3rd term = T3 = $${}^m{C_2}.{y^2}$$

Coefficient of T3 = $${}^m{C_2}$$

4th term = T4 = $${}^m{C_3}.{y^3}$$

Coefficient of T2 = $${}^m{C_3}$$

$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$

$$\Rightarrow$$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$

$$\Rightarrow$$ 6m2 - 6m = 6m +m(m2 - 3m + 2)

$$\Rightarrow$$ 6m2 - 6m = 6m + m3 - 3m2 + 2m

$$\Rightarrow$$ 6m - 6 = 6 + m2 - 3m + 2

$$\Rightarrow$$ m2 - 9m + 14 = 0

Now put r = 2 at each option and find answer.

In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get

m2 - 9m + 14 = 0. So Option C is correct.
4

### AIEEE 2005

If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as
A
$$1 - {3 \over 8}{x^2}$$
B
$$3x + {3 \over 8}{x^2}$$
C
$$- {3 \over 8}{x^2}$$
D
$${x \over 2} - {3 \over 8}{x^2}$$

## Explanation

$${\left( {1 + x} \right)^{{3 \over 2}}}$$ = 1 + $${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$$

= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)

$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$

= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$

= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$

= $$- {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$

[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]

= $$- {3 \over 8}{x^2}$$

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