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# The inequality ${\log _{{x^2}}}\left| {x - 1} \right| > 0$ is not defined for some integral values of x, find the sum of their magnitudes.

Last updated date: 25th Feb 2024
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Hint: The above inequality can be solved by solving all the values of x. Hence calculate their magnitudes according to the inequality given and sum up their magnitudes.

Here, it is given ${\log _{{x^2}}}\left| {x - 1} \right| > 0$
From the given inequality equation, it is observed that ${x^2} \ne 1$that means x is neither equal to 1 nor equal to -1.
$\Rightarrow x \ne \pm 1$ Which means ${x^2} > 1$
As, $\left| {x - 1} \right| > 0$ where modulus will open two times: one for positive and other for negative.
So, for positive it will open as $x - 1 > 0$ which implies $x > 1$
And other for negative $- \left( {x - 1} \right) > 0$
On further solving inequality we get, $- x + 1 > 0$ which implies $x < - 1$
By concluding from all the positive and negative modulus we can say that
$\left| {x - 1} \right| > 1$
And hence, by opening this we get, $- 1 > x - 1 > 1$
On further simplifying this equality, $0 > x > 2$ which means x is neither equal to 1 nor equal to 2.
$\Rightarrow x \ne 1,2$
Hence, the magnitude of x comes out to be -1, 1, and 2.
So, the sum of their magnitudes = 1-1+2 = 2.

The inequality is used for comparing numbers with help of different signs. Example: $> , < , \le {\rm{and}} \ge$.These signs help us in finding the numbers which are less than or greater than the variables given in the question. The simple inequalities can be solved by just adding and subtracting the values by taking variables at one side and finding the result on the other side.