Question

Let $f\left(x\right)=a{x}^2+b{x}^2+cx+d$ and $g\left(x\right)=x^2+x-2.$
If $\underset{x\to 1}{lim}\frac{f\left(x\right)}{g\left(x\right)}=1$ and $\underset{x\to 2}{lim}\frac{f\left(x\right)}{g\left(x\right)}=4,$ then find the value of $\frac{c^2+d^2}{a^2+b^2}$

Answer 1

$f\left(x\right)=a{x}^3+b{x}^2+cx+d,g\left(x\right)=x^2+x-2=(x-1)(x-2)$

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{g\left(x\right)}=1$

For limit to exist, $f\left(1\right)=0⟹a+b+c+d=0\cdots \left(1\right)$

$\underset{x\to 1}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\underset{x\to 1}{lim}\frac{3a{x}^2+2bx+c}{2x+1}=1⟹3a+2b+c=3\cdots \left(2\right)$

$\underset{x\to 2}{lim}\frac{f\left(x\right)}{g\left(x\right)}=2$

For limit to exist, $f\left(2\right)=0⟹8a+4b+2c+d=0\cdots \left(3\right)$

$\underset{x\to 2}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\underset{x\to 2}{lim}\frac{3a{x}^2+2bx+c}{2x+1}=1⟹12a+4b+c=20\cdots \left(4\right)$

Solving $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$

$a=23,b=-95,c=124,d=-52$

$\frac{c^2+d^2}{a^2+b^2}=\frac{18080}{9554}=1.89$