If $\frac{(x - 5) (x^{2} - 2 x + 1)}{(x - 7) (x^{2} + 2 x + 3)}$ ${(x-5)(x^2-2x+1)}/{(x-7)(x^2+2x+3)}$ is positive for all real value of x,show that x has no value between 5 and 7?

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Answer 1 · 2018-02-18T17:15:45

Hence proved.

Let

$y = \frac{(x - 5) (x^{2} - 2 x + 1)}{(x - 7) (x^{2} + 2 x + 3)}$ $y=((x-5)(x^2-2x+1))/((x-7)(x^2+2x+3))$

Simplifying,

$y = \frac{(x - 5) {(x - 1)}^{2}}{(x - 7) ({(x + 1)}^{2} + 1)}$ $y=((x-5)(x-1)^2)/((x-7)((x+1)^2+1))$

Now, note that the ${(x - 1)}^{2}$ $(x-1)^2$ and ${(x + 1)}^{2} + 1$ $(x+1)^2+1$ are positive for all values of x

Thus, the sign of $y$ $y$ depends only on the $\frac{x - 5}{x - 7}$ $(x-5)/(x-7)$ term

For values between 5 and 7, The above term is -ve.

Thus,
For $y$ $y$ to be +ve, $x$ $x$ must not lie between 5 and 7.

Note: My answer is in no way be fit to be presented in an exam paper, it is only to help you understand the logic

If (x−5)(x2−2x+1)(x−7)(x2+2x+3){(x-5)(x^2-2x+1)}/{(x-7)(x^2+2x+3)} is positive for all real value of x,show that x has no value between 5 and 7?