Question

Find range of k such that range of y € R
y = x^2 - x - 2 / x^2 + kx​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \blue{y = \dfrac{ {x}^{2} - x - 2}{ {x}^{2} + kx}}}$

$\sf{ \implies \: y( {x}^{2} + kx) = {x}^{2} - x - 2}$

$\sf{ \implies \: y{x}^{2} + kyx = {x}^{2} - x - 2}$

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

$\sf{ \implies \: x = \dfrac{ - (ky + 1) \pm \sqrt{(ky + 1)^{2} - 8(y - 1) } }{2(y - 1)} }$

$\sf{ \implies \: x = \dfrac{ - (ky + 1) \pm \sqrt{k^{2} y^{2} + 2ky + 1- 8y + 8} }{2(y - 1)} }$

$\sf{ \implies \: x = \dfrac{ - (ky + 1) \pm \sqrt{k^{2} y^{2} + (2k - 8)y + 9} }{2(y - 1)} }$

Since, $y\in\mathbb{R}$ so,

$\sf(2k - 8)^{2} - 36 {k}^{2} \geqslant 0$

$\sf \implies4(k - 4)^{2} - 36 {k}^{2} \geqslant 0$

$\sf \implies(k - 4)^{2} - 9 {k}^{2} \geqslant 0$

$\sf \implies \: k ^{2} - 8k + 16 - 9 {k}^{2} \geqslant 0$

$\sf \implies \: - 8k ^{2} - 8k + 16 \geqslant 0$

$\sf \implies \: - 8(k ^{2} + k - 2) \geqslant 0$

$\sf \implies \: k ^{2} + k - 2 \leqslant 0$

$\sf \implies \: k ^{2} + 2k - k - 2 \leqslant 0$

$\sf \implies \:k( k + 2) -1( k + 2 )\leqslant 0$

$\sf \implies \:(k - 1)( k + 2) \leqslant 0$

$\sf \implies \:k \in [ - 2 , 1]$