Solve the following inequality :Also show the answer on number line. C

1.	Solve the following inequality :Also show the answer on number line. Class-11 inequality
Answer» $\large\underline{\sf{Solution-}}$ Given inequality is $\rm :\longmapsto\:\tt \dfrac{x(x + 6)(x + 2)^{2} (x - 3)}{(x - 4) ^{3}(x + 1) ^{4} } > <klux>0</klux>$ Let first find the critical points. So, critical points are x = 0, - 6, - 2, 3, 4 and - 1 Let rearranged ACCORDING to NUMBER line - 6, - 2, - 1, 0, 3, 4 So, Let PLOT these points on number line and let check the interval and check the sign. $\begin{gathered}\boxed{\begin{array}{c\|c} \bf Interval & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( - \infty, - 6) & \sf + \\ \\ \sf ( - 6, - 2) & \sf - \\ \\ \sf ( - 2, - 1) & \sf - \\ \\ \sf ( - 1,0) & \sf - \\ \\ \sf (0,3) & \sf + \\ \\ \sf (3,4) & \sf - \\ \\ \sf (4, \infty ) & \sf + \end{array}} \\ \end{gathered}$ SINCE, it is given that, $\rm :\longmapsto\:\tt \dfrac{x(x + 6)(x + 2)^{2} (x - 3)}{(x - 4) ^{3}(x + 1) ^{4} } > 0$ So, required solution set is $\bf\implies \:x \: \in \: ( - \infty ,6) \: \cup \:(0,3) \: \cup \:(4, \infty )$ Additional Information :- $\boxed{ \rm{ x > y\rm \: \implies\: - x < - y}}$ $\boxed{ \rm{ x < y\rm \: \implies\: - x > - y}}$ $\boxed{ \rm{ x \geqslant y\rm \: \implies\: - x \leqslant - y}}$ $\boxed{ \rm{ x \leqslant y\rm \: \implies\: - x \geqslant - y}}$ $\boxed{ \rm{ x > - y \: \rm\implies \: - x < y}}$ $\boxed{ \rm{ x < - y \: \rm\implies \: - x > y}}$ $\boxed{ \rm{ x \leqslant - y \: \rm\implies \: - x \geqslant y}}$

Solve the following inequality :Also show the answer on number line. Class-11 inequality

Answer»

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

Given inequality is

$\rm :\longmapsto\:\tt \dfrac{x(x + 6)(x + 2)^{2} (x - 3)}{(x - 4) ^{3}(x + 1) ^{4} } > <klux>0</klux>$

Let first find the critical points.

So, critical points are x = 0, - 6, - 2, 3, 4 and - 1

Let rearranged ACCORDING to NUMBER line - 6, - 2, - 1, 0, 3, 4

So, Let PLOT these points on number line and let check the interval and check the sign.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( - \infty, - 6) & \sf + \\ \\ \sf ( - 6, - 2) & \sf - \\ \\ \sf ( - 2, - 1) & \sf - \\ \\ \sf ( - 1,0) & \sf - \\ \\ \sf (0,3) & \sf + \\ \\ \sf (3,4) & \sf - \\ \\ \sf (4, \infty ) & \sf + \end{array}} \\ \end{gathered}$