Question

(a) For real values of x, prove that the value of the expression $\frac{11x^2+12x+6}{x^2+4x+2}$ cannot lie between $-5$ and $3$
(b) $x^2+(a-b)x+(1-a-b)=0,a,bϵR$. Find the condition on a, for which both roots of the equation are real and unequal.
(c) Determine the values of x which satisfy the inequalities $x^2-3x+2>0$ and $x^2-3x-4\le 0$.

Answer 1

(a) Let $\frac{11x^2+12x+6}{x^2+4x+2}=y$.

$\therefore$ $x^2(11-y)+4x(3-y)+2(3-y)=0$........(1)

For real values of x, $B^2-4AC$ of (1) should be $\ge 0$

or $16{(3-y)}^2-8(11-y)(3-y)\ge 0$ i.e., $+ive$

$8(3-y)[6-2y-11+y]\ge 0$

or $8(3-y)(-5-y)\ge 0$

or $8(y-3)(y+5)\ge 0$ i.e. $+ive$

or $8[y-(-5\left)\right](y-3)$ is $+ive$

Hence arguing as in part (a), y should not lie between $-5$ and $3$.

(b) For roots to be real and unequal $D>0$ i.e., $+ive$

$\therefore$ ${(a-b)}^2-4(1-a-b)>0\forall bϵR$

Arranging as a quadratic in b,

$b^2-2(a-2)b+(a^2+4a-4)$ is $+ive\forall bϵR$.

Its sign will be same as of its first term i.e., $+ive$ provided its $\Delta$ is $-ive$.

$\therefore$ $4(a^2-4a+4)-4(a^2+4a-4)<0$

or $-8a+8<0$

or $a-1>0$ $\therefore$ $a>1$.

(c) $(x-1)(x-2)>0\Rightarrow x<1,x>2$

$(x+1)(x-4)\le 0\Rightarrow -1\le x\le 4$.

$\therefore$ $-1\le x<1$ and $2<x\le 4$ for both

$\therefore$ $xϵ[-1,1]\cup [2,4]$. Both are semi-open; 1 and 2 are not included.