NIMCET 2014 — Mathematics PYQ

Concept:
For the least value of the function, the sum of the square of the roots of the equation x² - (a - 2)x - (a + 1) = 0 is zero

Calculations:

Consider, the roots of the equation ${\mathrm{x}}^2-($a-2)x-(a+1)=0 are $\alpha$ and $\beta$
$\Rightarrow \alpha +\beta =$a-2 and $\alpha .\beta =-($a+1)
Given, the sum of the square of the roots of the equation ${\mathbf{x}}^2-(\mathbf{a}-2)\mathbf{x}-(\mathbf{a}+1)=0$, is the least value
$$\begin{array}{rl}& amp;\text{Olven, ule sum ol une squa}\\ & amp;\Rightarrow {\alpha }^2+{\beta }^2=0\\ & amp;\Rightarrow (\alpha +\beta {)}^2-2\alpha \beta =0\\ & amp;\Rightarrow (\mathrm{a}-2{)}^2+2(\mathrm{a}+1)=0\\ & amp;\Rightarrow {\mathbf{a}}^2-2\mathbf{a}+6=0\\ & amp;\Rightarrow {\mathbf{a}}^2-2\mathbf{a}+1+5=0\\ & amp;\Rightarrow (\mathbf{a}-1{)}^2+5=0\\ & amp;\Rightarrow \mathbf{a}=1\end{array}$$

Hence, the value of a, for which the sum of the square of the roots of the equation $x^2$- ( a- 2) x- ( a+ 1) = 0, assumes the least value is 1.