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motikmotik
4 years ago
14

What are the restrictions for 4/x+1/x^2=1/5x^2

Mathematics
1 answer:
dmitriy555 [2]4 years ago
8 0

Answer:

The restrictions will be:

x ≠ 0;  

x² ≠ 0 ⇒  x ≠ 0 ;  

5x² ≠ 0 ⇒ x² ≠ 0 ⇒  x ≠ 0;

Step-by-step explanation:

As we know that

There are two genuine reasons why domains are restricted.

We cannot divide the expression by 0 as it would be undefined then.

We cannot take the square root of a negative number, as the result will not be a defined.

Considering the expression

\frac{4}{x}+\frac{1}{x^2}=\frac{1}{5x^2}

As we can not put x = 0 in the denominator as it would make the expression undefined.

In other words, the certain values which make the denominator equal to zero for a rational expression will be restricted values.

Hence, the restrictions will be:

x ≠ 0;  

x² ≠ 0 ⇒  x ≠ 0 ;  

5x² ≠ 0 ⇒ x² ≠ 0 ⇒  x ≠ 0;

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given  cos x - sin x = √2 cos (3 x)

Dividing '√2' on both sides , we get

\frac{1}{\sqrt{2} } cos (x) - \frac{1}{\sqrt{2} } sin (x) = \frac{\sqrt{2} cos (3 x)}{\sqrt{2} }

we will use trigonometry formulas

a) Cos ( A + B) = Cos A Cos B - sin A sin B

b)  cos \frac{\pi }{4} = \frac{1}{\sqrt{2} }

<u><em>Step(ii):-</em></u>

<u><em></em></u>\frac{1}{\sqrt{2} } cos (x) - \frac{1}{\sqrt{2} } sin (x) = \frac{\sqrt{2} cos (3 x)}{\sqrt{2} }<u><em></em></u>

cos (\frac{\pi }{4} ) cos x - sin(\frac{\pi }{4} ) sin x = cos 3x

cos (\frac{\pi }{4}+x ) = cos 3 x

<u><em>Step(iii):-</em></u>

<u><em>General solution of  cos x = cos ∝  is  x = 2 nπ+∝</em></u>

<u><em>we have </em></u> cos (\frac{\pi }{4}+x ) = cos 3 x

The general solution of  cos (\frac{\pi }{4}+x ) = cos 3 x is

⇒  3 x   = 2 n \pi  + (\frac{\pi }{4}+x )

⇒ 3 x- x = 2 n \pi + \frac{\pi }{4}

 2x = 2 n \pi + \frac{\pi }{4}

<em><u>final answer</u></em>:-

General solution is

 x = n \pi + \frac{\pi }{8}

             

8 0
3 years ago
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