Question:

Step 1: Expand the matrix equation. $[f(2)\ f(1)\ f(0)]\begin{bmatrix}x\\y\\1\end{bmatrix} = f(2)\cdot x + f(1)\cdot y + f(0)\cdot 1 = 2x + y + 2$. Step 2: Compare coefficients: $f(2) = 2$, $f(1) = 1$, $f(0) = 2$. Step 3: Let $f(x) = ax^2 + bx + c$. From $f(0)=2$: $c=2$. From $f(1)=1$: $a+b+2=1 \Rightarrow a+b=-1$. From $f(2)=2$: $4a+2b+2=2 \Rightarrow 4a+2b=0 \Rightarrow 2a+b=0$. Step 4: Solve: $2a+b=0$ and $a+b=-1$ give $a=1$, $b=-2$. So $f(x) = x^2 - 2x + 2 = (x-1)^2 + 1$. Step 5: Range of $f(x)$: minimum value is $1$ at $x=1$, so range is $[1,\infty)$. ✓ Option (1) is correct. Option (2) is incorrect. Step 6: Find intersection of $y=f(x)=(x-1)^2+1$ and $y=2-x$: $$(x-1)^2+1 = 2-x \Rightarrow x^2-2x+2 = 2-x \Rightarrow x^2-x=0 \Rightarrow x(x-1)=0$$ So $x=0$ and $x=1$. Step 7: Compute area: $$\text{Area} = \int_0^1 [(2-x)-(x^2-2x+2)]\,dx = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \frac{1}{6}$$ Option (4) is correct. ✓ Option (3) is incorrect.