Question

(a) Find all integral values of $x$ for which
$\begin{array}{ccc}(5x-1)<& {(x+1)}^2<& (7x-3)\\ I& II& III\end{array}$.
(b) $Solve\left|\frac{12x}{4x^2+9}\right|\le 1$.

Answer 1

(a) Both the following inequalities have to hold good simultaneously.
$x^2-3x+2=(x-1)(x-2)>0$ from I and II
$x^2-5x+4=(x-1)(x-4)<0$ from II and III
$\therefore$ $x<1$ or $x>2$ and $1<x<4$
$\therefore$ From above we conclude that $2<x<4$ satisfy both. Since x is integral therefore $x=3$.
(b) From the given relation we have
$-1\le \frac{12x}{4x^2+9}\le 1$ by definition of mod.
Since $4x^2+9$ is $+ive$ for all real x, we can multiply the inequality by $4x^2+9$.
$\therefore$ $-(4x^2+9)\le 12x\le (4x^2+9)$
${(2x+3)}^2\ge 0$ and ${(2x-3)}^2\ge 0$
Above is true for all real values of x being perfect squares.