Question

(a) Let $y=\sqrt{\left(\frac{(x+1)(x-3)}{x-2}\right)}$.
Find all the real values of x for which y takes real values.
(b) Determine all real values of x for which the expression ${\{\frac{2}{x^2-x+1}-\frac{1}{x+1}-\frac{2x+1}{x^1+1}\}}^{1/2}$ takes real values.

Answer 1

(a) $y=\sqrt{\frac{(x+1)(x-3)}{x-2}}$
If y is real then $y^2$ is $+ive$ or zero :
$\therefore$ $\frac{(x+1)(x-3)}{x-2}\ge 0$
or $\frac{(x+1)(x-3)(x-2)}{{(x-2)}^2}\ge 0$
Now $D^r$ is $+ive$ $\therefore$ $N^r$ is to be $+ive$
$N^0=0$ if $x=-1,3,2$
Marking these values in ascending order on real line, write $+ive$ in extreme right and move towards left with alternately changing the sign. From the fig. It is clear that $N^r$ is $+ive$ when $x\ge 3,-1\le x<2$
(x=2 is exluded as it will make $D^r$ infinite)
(b) $y=\sqrt{E}$ is real if $E\ge 0$
or $\frac{-x^2+x}{(x+1)(x^2-x+1)}\ge 0$
Now $x^2-x+1$ is a quadratic expression whose Discriminant $\Delta =-ive$ and hence its sign is same as that of Ist term i.e.$+ive$.
Hence we must have $\frac{-x^2+x}{x+1}\ge 0$
Multplying by '_' sign and changing the sign of inequality, we must have
$\frac{x(x-1)(x+1)}{{(x+1)}^2}\le 0$
The change points are $-1,0,1$ in ascending order.
As in part (a) the inequality is satisfied for
$xϵ[0,1]$ and $]-\infty ,-1[,$
$xϵ]-\infty -1[\cup [0,1]$
or $0\le x\le 1$ and $x<-1$