Question

the maximum value of x^2+4x+1/x^2+x+1 are​

Answer 1

Step-by-step explanation:

Let,

$y = \frac{ {x}^{2} + 4x + 1}{ {x}^{2} + x + 1 }$

$\implies \: y( {x}^{2} + x + 1)= {x}^{2} + 4x + 1$

$\implies \:( y - 1) {x}^{2} +(y - 4) x + (y - 1)= 0$

$\forall \: x \in \:\real, \: D \geqslant 0$

Also, y ≠ 1

So,

$(y - 4)^{2} - 4(y - 1) ^{2} \geqslant 0$

$\implies(y - 4)^{2} - \{ 2(y - 1) \}^{2} \geqslant 0$

$\implies\{y - 4 - 2(y - 1) \} \{y - 4 + 2(y - 1) \} \geqslant 0$

$\implies\{y - 4 - 2y + 2 \} \{y - 4 + 2y -2 \} \geqslant 0$

$\implies( - y -2 ) (3y - 6) \geqslant 0$

$\implies( y + 2 ) (y - 2) \leqslant 0$

$\implies \: y \in [ - 2,2 ]$