Question

I need proper Explanation, I'd.k why but I'm not that clear how to deal with questions like this ~

​

Thanks for your effort ~

Answer 1

Answer:

Hey There!

Let's solve....

$f(x) = \frac{(x + b)(x + c)}{(b - a)(c - a)} + \frac{(x + c)(x + a)}{(c - b)(a - b)} + \frac{(x + a)(x + b)}{(a - c)(b - c)} - 1$

$f( - a) = \frac{( - a + \cancel{b})( \cancel{ - a} + c)}{ (\cancel{b} - a)(c - \cancel{a})} + 0 + 0 - 1 = 1 - = 0$

$f( - b) = 0 + \frac{( \cancel{ - b} + c)( \cancel{ - b} + a)}{ (c - \cancel{b})(a - \cancel{b})} + 0 - 1 = 1 - 1 = 0$

$f( - c) = 0 + 0 + \\ \frac{ (\cancel{ - c} + a)( \cancel{ - c} + b)}{(a - \cancel{c})(b - \cancel{c})} - 1 \\ \\ = 1 - 1 = 0$

$f(x) \to {max}^{m} \: power = 2$

Maximum power where quadratic equation is 2

$\therefore \: f(x) = 0$

x belongs r

Now let's identify

$\to \: \frac{(x + b)(x + c)}{(b - a)(c - a)} + \frac{(x + c)(x + a)}{(c - b)(a - b)} + \frac{(x + a)(x + b)}{(a - c)(b - c)} - 1 = 0$

So x belongs r is identified

Hence finally solved...

I hope it is helpful to you..

cheers!__________