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1

### AIEEE 2006

If $${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}},$$ then $${{{dy} \over {dx}}}$$ is
A
$${y \over x}$$
B
$${{x + y} \over {xy}}$$
C
$$xy$$
D
$${x \over y}$$

## Explanation

$${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$

$$\Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$

Differentiating both sides.

$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$

$$\Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$$

$$\Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$$

$$\Rightarrow {{dy} \over {dx}} = {y \over x}$$
2

### AIEEE 2006

The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is
A
$$\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$$
B
$$\left( { - \infty ,1} \right) \cup \left( { - 1,\infty } \right)$$
C
$$\left( { - \infty ,\infty } \right)$$
D
$$\left( {0,\infty } \right)$$

## Explanation

$$f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right.$$

$$\Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$$

$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.
3

### AIEEE 2005

If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals
A
$$-2$$
B
$$3$$
C
$$2$$
D
$$1$$

## Explanation

Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots

Then $$\alpha + \alpha + 1 = b =$$ sum of -

roots $$\alpha \left( {\alpha + 1} \right) = c$$

$$=$$ product of roots

$$\therefore$$ $${b^2} - 4c$$

$$= {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)$$

$$= 1.$$
4

### AIEEE 2005

The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is
A
$$1$$
B
$$0$$
C
$$3$$
D
$$2$$

## Explanation

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$

$$\Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$

$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta$$

$$= {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$

For min. value of $${\alpha ^2} + {\beta ^2}$$ where $$\alpha$$ is an integer

$$\Rightarrow \,\,a = 1.$$

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