If $x$ is real, the function $\frac{{(x - a)(x - b)}}{{(x - c)}}$ will assume all real values, provided
$a > b > c$
$a < b < c$
$a > c < b$
$a < c < b$
Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. Then the value of $\mathrm{m}$ is..........
Number of solutions of equation $|x^2 -2|x||$ = $2^x$ , is
If $\log _{(3 x-1)}(x-2)=\log _{\left(9 x^2-6 x+1\right)}\left(2 x^2-10 x-2\right)$, then $x$ equals
If $a,b,c$ are real and ${x^3} - 3{b^2}x + 2{c^3}$ is divisible by $x - a$ and$x - b$, then
Suppose $a, b, c$ are three distinct real numbers, let $P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$ When simplified, $P(x)$ becomes