If x is real, the function frac{{(x - a)(x - | vedclass

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Question

(d) Let $y = \frac{{(x - a)(x - b)}}{{(x - c)}}$

or $y(x - c) = {x^2} - (a + b)x + ab$

or ${x^2} - (a + b + y)x + ab + cy = 0$

$\Delta = {(a

Vedclass · Accepted Answer

$a < c < b$

Vedclass · Answer

(d) Let $y = \frac{{(x - a)(x - b)}}{{(x - c)}}$ 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 $ herefore$ $c$ 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.

If $x$ is real, the function $\frac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided

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