Mathematics - Quadratic Equation Question with Solution | TestHub

The given expression can be rewritten as a sum of squares.

$$E={\left(x-(1+\cos t)\right)}^2+{\left(\frac{K}{x}-(1+\sin t)\right)}^2$$

For minimum value, we need $x=1+\cos t$ and $\frac{K}{x}=1+\sin t$.

This implies $K=(1+\cos t)(1+\sin t)$.

We need to find the maximum value of $(1+\cos t)(1+\sin t)$.

Let $1+\cos t=A$ and $1+\sin t=B$.

$A^2+B^2=(1+\cos t{)}^2+(1+\sin t{)}^2=1+2\cos t+{\cos }^{2}t+1+2\sin t+{\sin }^{2}t=3+2(\cos t+\sin t)$.

Let $\cos t+\sin t=\sqrt{2}\sin (t+\frac{\pi }{4})$. Max value is $\sqrt{2}$.

So, max value of $A^2+B^2=3+2\sqrt{2}=(\sqrt{2}+1{)}^2$.

Since $K=AB$, we want to maximize $AB$ subject to $A^2+B^2=(\sqrt{2}+1{)}^2$.

This occurs when $A=B$, so $1+\cos t=1+\sin t⟹\cos t=\sin t$.

For $0\le t\le 2\pi$, this means $t=\frac{\pi }{4}$ or $t=\frac{5\pi }{4}$.

At $t=\frac{\pi }{4}$, $1+\cos t=1+\frac{1}{\sqrt{2}}$ and $1+\sin t=1+\frac{1}{\sqrt{2}}$.

So, $K=(1+\frac{1}{\sqrt{2}}{)}^2$.

The minimum value of the expression is $2{\left\{\sqrt{K}-(1+\frac{1}{\sqrt{2}})\right\}}^2$.