Question

I have a question...
Give me solution please...​

Answer 1

Answer:

Step-by-step explanation:

Let us suppose

$\tt{y=\dfrac{9\cdot3^{2x}+6\cdot3^{x}+4}{9\cdot3^{2x}-6\cdot3^{x}+4}}$

$\tt{\implies\,y(9\cdot3^{2x}-6\cdot3^{x}+4)=9\cdot3^{2x}+6\cdot3^{x}+4}$

$\tt{\implies\,9y\cdot3^{2x}-6y\cdot3^{x}+4y=9\cdot3^{2x}+6\cdot3^{x}+4}$

$\tt{\implies\,9y\cdot3^{2x}-6y\cdot3^{x}+4y-9\cdot3^{2x}-6\cdot3^{x}-4=0}$

$\tt{\implies\,9(y-1)\cdot3^{2x}-6(y+1)\cdot3^{x}+4(y-1)=0}$

$\tt{\implies\,9(y-1)\cdot\left(3^{x}\right)^2-6(y+1)\cdot3^{x}+4(y-1)=0}$

This is a quadratic equation 3ˣ, so, its discriminant will be positive

$\sf{36(y+1)^2-36\cdot4\cdot(y-1)^2>0}$

$\sf{\implies(y+1)^2-4(y-1)^2>0}$

$\sf{\implies\,y^2+2y+1-4y^2+8y-4>0}$

$\sf{\implies\,-3y^2+10y-3>0}$

$\sf{\implies\,3y^2-10y+3<0}$

$\sf{\implies\,3y^2-9y-y+3<0}$

$\sf{\implies\,3y(y-3)-1(y-3)<0}$

$\sf{\implies\,(3y-1)(y-3)<0}$

$\sf{\implies\,3\left(y-\dfrac{1}{3}\right)(y-3)<0}$

$\sf{\implies\,\left(y-\dfrac{1}{3}\right)(y-3)<0}$

$\sf{\implies\,y\in\left(\dfrac{1}{3},3\right)}$