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The solution of the inequality

$\frac{\left ( x+1 \right )\left ( x-1 \right )^{4}}{\left ( x-3 \right )^{2}\left ( x-2 \right )}\geq 0$ is
Option: 1
$\left ( -\infty,-1 \right )\cup \left ( 2,\infty\right )$

Option: 2
$\left ( -\infty ,-1\right ]\cup \left ( 2,\infty \right )$

Option: 3
$\left ( -\infty,-1 \right ] \cup \left ( 2,\infty \right )$ $\cup \{1\}$

Option: 4
none of these

Answers (1)

best_answer

1. 0 is on one side

2. All linear factors

3. Coefficients of x are positive

4. Critical points

$x+1=0\Rightarrow x=-1\\$

$x-2=0\Rightarrow x=2\\$

(Do not include $x-1=0\Rightarrow x=-1\\$ and $x-3=0\Rightarrow x=3$ , as they have even powers)

5. Draw number line and mark critical points on it

6. Mark (+) sign on RHS and write alternate signs towards left

7. We need x for which expression is (+)ve, so all numbers from $-\infty\: to\: -1\\$ and from $2\: to\: \infty$ are to be included in the answer

8. Now check at all critical points (including ones with even powers)

At $x=-1$ expression is $0$ , so included in the answer

At $x=1$ expression $=0$ , so included

At $x=3,$ expression is not defined (as denominator $= 0$ ), so excluded

At $x=2,$ expression is not defined , so excluded.

9. Answer is $\left ( -\infty,-1 \right )\cup \left ( 2,\infty \right )\cup \{-1,1\}\\$

$= \left ( -\infty,-1 \right ]\cup \left ( 2,\infty \right )\cup \{1\}$

Posted by

Shailly goel

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