4.2 Quadratic equations and inequations — MATHS STD 11 Science — Question

or $y(x - c) = {x^2} - (a + b)x + ab$

or ${x^2} - (a + b + y)x + ab + cy = 0$

$\Delta = {(a + b + y)^2} - 4(ab + cy)$

$ = {y^2} + 2y(a + b - 2c) + {(a - b)^2}$

Since $x$ is real and $y$ assumes all real values, we must have $\Delta \ge 0$ for all real values of $y$.The sign of a quadratic in $y$ is same as of first term provided its discriminant ${B^2} - 4AC < 0$

This will be so if $4{(a + b - 2c)^2} - 4{(a - b)^2} < 0$

or $4(a + b - 2c + a - b)(a + b - 2c - a + b) < 0$

or $16(a - c)(b - c) < 0$ or $16(c - a)(c - b) = - $ve

$\therefore$ $c$ lies between $a$ and $b$ i.e., $a < c < b$.....$(i)$

Where $a < b$, but if $b < a$then the above condition will be $b < c < a$or $a > c >b$.....$(ii)$

Hence from $(i)$ and $(ii)$ we observe that $(d)$ is correct answer.