Question

For real x, the expression $\frac{(x-a)(x-b)}{x-c}$ will assume real values provided

• A. a>b>c
• B. a < b < c
• C. c >a > b
• D. a < c < b

Answer 1

The correct option is B a < b < c

Put $y=\frac{(x-a)(x-b)}{(x-c)}$

So, $x^2-(a+c+y)x+(ac+by)=0$

Now for roots to be real, its discriminant be positive

$O>o$ or $(a+c+y{)}^2-4(ac+by)>0$

$(a+c{)}^2+y^2+2y(a+c)-4ax-4by=0$

$y^2+2y(a+c-2b)+(a-c{)}^2>0-----(1)$

Here for any value of $y\left(belowtoR\right)$ we have, $y^2$ and $(a-c{)}^2$ always positive so for eq $\left(1\right)$ to be positive for all $y(-ve$ or $+ve)$,

$2(a+c,-2b)$ should be zero

i.e., $a+c-2b=0$ or $a,b,c$ in $AP$ service four where we can say that either $a<B<c$ or $c,b<a$