Hi Guys,

The range of inequality can also be found out by using the

'Wavy Line' method. It's a powerful method of solving inequalities especially when you have trouble in finding the range algebraically. Let me illustrate briefly how to use this method.

Finding Range of \(x^3 - x^2 > 0\) through 'Wavy Line' method. We can simplify the expression to \(x^2 (x -1) > 0\). Now, the LHS of the inequality becomes 0 at 0 and 1. So,

{0, 1} are the zero points of the inequality.

Plot these points on the number line and start from the top right corner and draw a Wavy Line. Be ready to alternate the region of the wave based on how many times a a point is root to the equation. If the power of a term is odd, the wave simply passes through the point into the other region. If the power of a term is even, the wave bounces back to the same region. In this case, power of the term was 2 for the zero point 0, hence the wave bounced back. The range of the inequality will be +ve for region above the number line and -ve for region below the number line.

We observe from the graph that the range is +ve when \(x > 1\). So, all we need is to find from the statements- I & II is if \(x >1\) or not.

Statement-ISt-I tells us that \(x > 0\). This does not tell us for sure if \(x >1\). Hence, the statement is insufficient.

Statement-II- Finding Range of \(x^2 -x >0\) through Wavy Line methodThe expression in st-II can be simplified to \(x(x-1) > 0\). Let's find the range by the Wavy Line method. The

zero points of the expression are { 0,1}.

The Wavy Line graph would be represented in the manner shown below:

We see that the value of the inequality is +ve for \(x < 0\) and \(x > 1\). But the inequality in the question statement is +ve only when \(x > 1\). Hence, Statement-II is insufficient.

Combining Statement-I & IIIf we combine both the statements, we can infer that \(x > 1\), which is what we needed. Hence, combining both the statements is sufficient to answer the question.

You can practice a similar question at is-x-196185.html. Try solving it by Wavy Line method.

Also, an article for you on the Wavy Line method inequalities-trick-91482-80.html#p1465609Hope it helps!

Regards

Harsh

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