Question

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$\;\;\;\;\;\;\;\;\;\bullet\;\;\;\;\;\sf{\dfrac{x^2-x-6}{x^2+6x}\geqslant0}$
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Find the integrals of 'x' by using wavy curve method.​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - x - 6}{ {x}^{2} + 6x} \geqslant 0$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 3x + 2x - 6 }{x(x + 6)} \geqslant 0$

$\rm :\longmapsto\:\dfrac{x(x - 3) + 2(x - 3)}{x(x + 6)} \geqslant 0$

$\rm :\longmapsto\:\dfrac{(x - 3)(x + 2)}{x(x + 6)} \geqslant 0 \: \: and \: x \ne \: 0, \: - 6$

$\begin{gathered}\boxed{\begin{array}{c|c} \sf interval & \bf sign \: of \: \dfrac{(x - 3)(x + 2)}{x(x + 6)} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x < - 6 & \sf + ve \\ \\ \sf - 6 < x \leqslant - 2 & \sf - ve \\ \\ \sf - 2 \leqslant x < 0 & \sf + ve\\ \\ \sf 0 < x \leqslant 3 & \sf - ve\\ \\ \sf x \geqslant 3 & \sf + ve \end{array}} \\ \end{gathered}$

$\bf\implies \:x \in \: ( - \infty , - 6) \: \cup \: [ - 2,0) \: \cup \: [3, \infty )$

Basic Concept :-

• Factorize the given expression.

• Get the values of x, by equating every factor to 0.

• Plot these points on the number line and assign the sign.

• The requured interval is the solution set.