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Solution of inequality $\frac{x\left ( 2-x \right )}{x^{2}-4x+3}> 0$ is
Option: 1
$\left ( 0,1 \right )\cup \left ( 2,3 \right )$

Option: 2
$\left [ 0,1 \right )\cup \left [ 2,3 \right )$

Option: 3
$\left ( -\infty, 0 \right )\cup \left ( 1,2 \right )\cup \left ( 3,\infty \right )$

Option: 4
$\left ( -\infty,0 \right ]\cup \left ( 1,2 \right]\cup \left ( 3,\infty \right )$

Answers (1)

best_answer

1. 0 is on one side

2. we do not have all linear factor, so

$\frac{x\left ( 2-x \right )}{\left ( x-1 \right )\left ( x-3 \right )}> 0$

3. Coefficient of x is negative in $\left ( 2-x \right )$ so

$\frac{x\cdot \left ( -1 \right )\left ( x-2 \right )}{\left ( x-1 \right )\left ( x-3 \right )}> 0$

This $\left ( -1 \right )$ has to be cross multiplied to remove it,but it is negative, so direction of inequality will change

$\frac{x\left ( x-2 \right )}{\left ( x-1 \right )\left ( x-3 \right )}< 0$

4. Critical point

$x=0\Rightarrow x=0\\$

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

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

$x-3=0\Rightarrow x=3$

(All powers are odd)

5. Number line

6. Rightmost internal is (+)ve, but we need negative values, so we do not need rightmost interval and we put (-) sign on it and then put alternative signs

In answer all (+)ve intervals are taken. So $\left ( 0,1 \right )$ and $\left ( 2,3\right )$ are included in the answer

7. Check all critical points

At $x=1$ and $x=3$ expression is not defined, so excluded.

At $x=0,2$ , expression $= 0$ . BUt these are also excluded as we need $< 0$ and not $\leq 0$

8. So,answer is $\left ( 0,1 \right )\cup \left ( 2,3 \right )$

Posted by

Pankaj Sanodiya

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