Suppose that the temperature T at every point $(x,y)$ in the cartesian plane is given by the formula $T=1-x^2+2y^2$. The correct statement about the maximum and minimum temperature along the line $x+y=1$ is

It is given that $T=1-x^2+2y^2$, where $x+y=1$

Replacing $y=1-x$ in $T$ we get,

$$\begin{gathered} \Rightarrow \mathrm{T}=1-{\mathrm{x}}^2+2{(1-\mathrm{x})}^2 \\ \Rightarrow \mathrm{T}=1-{\mathrm{x}}^2+2(1+{\mathrm{x}}^2-2\mathrm{x}) \end{gathered}$$

$$\begin{gathered} \Rightarrow T=x^2-4x+3 \\ \Rightarrow T=(x-2{)}^2-1 \end{gathered}$$

As we know ${\left(x-2\right)}^2\ge 0$

$\Rightarrow {\left(x-2\right)}^2-1\ge -1$

But the maximum value does not exist

$\Rightarrow T_{\max }=$Does not exist and $T_{min}=-1$

Hence option-$1$ is the correct answer.