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Solution of inequality  \frac{x^{2}}{x-1}\geq 1  is

Option: 1

\left [ 1,\infty \right )


Option: 2

\left ( 1,\infty \right )


Option: 3

\left ( -\infty,\infty \right )


Option: 4

\left ( -\infty,1 \right )


Answers (1)

best_answer

1.     0 is not on one side, so we shift 1 to LHS

        \frac{x^{2}}{x-1}\geq 1\Rightarrow \frac{x^{2}}{x-1}-1\geq 0

        \Rightarrow \frac{x^{2}-\left ( x-1 \right )}{\left ( x-1 \right )}\geq 0

        \Rightarrow \frac{x^{2}-x+1}{x-1}\geq 0

        Now we have 0 on the one side
        (Note : we have not cross multiplied \left ( x-1 \right ) as it can be (+) or (-) depending on the value of x)

2.     All linear factors: not present. So we will try to factorize numerator

        x^{2}-x+1

        It's discriminant = b^{2}-4ac=1-4=-3

        So, D< 0, so, it does not have real roots and so it cannot be factorized.

        Such quadratic factors are always (+)ve or always (-)ve.

        Now D< 0 and a=1\left ( a> 0 \right ), so x^{2}-x+1 is always positive and thus it can be cross multiplied

        \frac{x^{2}-x+1}{x-1}\geq 0\\\\ \Rightarrow \frac{1}{\left ( x-1 \right )}\geq \frac{0}{x^{2}-x+1}\\ \Rightarrow \frac{1}{\left ( x-1 \right )}\geq 0

3.    Coefficient of x are positive

4.     Critical points

        x-1=0\Rightarrow x=1

5.  

6.    Rightmost interval is positive and we need  \frac{1}{x-1}\geq 0 , means we need positive part, So we put + sign in this interval.
        We alternate signs in other intervals towards left.

        

        So \left ( 1,\infty \right ) included in the answer

7.    Check at all critical points x=1\Rightarrow expression is not defined so excluded.

8.    Answer is \left ( 1,\infty \right )

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