JEE MAIN - Mathematics (2005 - No. 9)
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 :
$$1$$
$$0$$
$$3$$
$$2$$
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
$$ \Rightarrow \,\,a = 1.$$
$$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$
$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$
$$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$
For min. value of $${\alpha ^2} + {\beta ^2}$$ where $$\alpha $$ is an integer
$$ \Rightarrow \,\,a = 1.$$
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