First slide

Theory of expressions

Question

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

Moderate

A

$a > b > c$
B

$a < b < c$
C

$a > c < b$
D

$a < c < b$

Solution

Let $y = \frac{(x - a) (x - b)}{(x - c)}$
or $y (x - c) = x^{2} - (a + b) x + ab$
or $x^{2} - (a + b + y) x + ab + cy = 0$
$∆ = {(a + b + y)}^{2} - 4 (ab + cy)$
$= y^{2} + 2 y (a + b - 2 c) + {(a - b)}^{2}$
Since x is real and y assumes all real values, we must have $∆ \geq 0$ for all real values of y. The sign of a quadratic in y is same as of first term provided its discriminant $B^{2} - 4 AC < 0$
This will be so if $4 {(a + b - 2 c)}^{2} - 4 {(a - b)}^{2} < 0$
or $4 (a + b - 2 c + a - b) (a + b - 2 c - a + b) < 0$
or $16 (a - c) (b - c) < 0 or 16 (c - a) (c - b) = - ve$
$∴ c lies between a and b i . e ., a < c < b . . . . (i)$
$where a < b, but if b < a$ then the above condition will be
$b < c < a or a > c > b . . . (ii)$
Hence from (i) and (ii) we observe that Option 4 is correct answer.

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