Question bank/JEE Main/Mathematics

If two sides of a square
are$3x - 4y + n = 0,(n > 0),4x + 3y + m = 0,(m > 0)$, having (1,1) as
center and 16 as area then the smallest value of k for which both the
roots of equation
$x^{2} - (n - m)kx + (m + n + 2)\left( k^{2} - k + 1 \right) = 0\$are
real, distinct and have values at least 4 is

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[IMAGE] As area =16 ⇒side =4
Line passing through centre and parallel to a side is 3x - 4 y + 1= 0
$\Rightarrow \frac{4}{2} = \frac{|n - 1|}{\sqrt{9 + 16}} \Rightarrow n = 11, - 9\ and\ \frac{4}{2}\frac{|m - 7|}{\sqrt{9 + 16}} \Rightarrow m = 3, - 17$
taking n = 11, m =3
$f(x) \equiv x^{2} - 8kx + 16\left( k^{2} - k + 1 \right) = 0$
$\frac{8k}{2 \times 1} > 4$
$8^{2}k^{2} - 4.1 \cdot 16\left( k^{2} - k + 1 \right) > 0$
$\Rightarrow k \in \lbrack 2,\infty)$
smallest value of k is 2

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx