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(A) Given points are $A(1,1), B(3,2), C(2,3)$.The slope of line $l_2$ passing through $A(1,1)$ and $B(3,2)$ is $m = \frac{2-1}{3-1} = \frac{1}{2}$.Sin

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$2$

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(A) Given points are $A(1,1), B(3,2), C(2,3)$.The slope of line $l_2$ passing through $A(1,1)$ and $B(3,2)$ is $m = \frac{2-1}{3-1} = \frac{1}{2}$.Since $l_1$ is parallel to $l_2$ and tangent to the curve at $C(2,3)$,the equation of line $l_1$ is $y - 3 = \frac{1}{2}(x - 2)$,which simplifies to $y = \frac{x}{2} + 2$.The area of the shaded region is the area between the line $l_1$ and the curve $f(x)$ from $x=1$ to $x=3$,minus the area of the trapezoid formed by the line $l_2$ and the $x$-axis,or more simply,the area between $l_1$ and $l_2$ minus the area between $f(x)$ and $l_2$.Alternatively,the shaded area is $\int_{1}^{3} (l_1(x) - f(x)) dx$.Area $= \int_{1}^{3} (\frac{x}{2} + 2) dx - \int_{1}^{3} f(x) dx$.Given $\int_{1}^{3} f(x) dx = 4$.Area $= [\frac{x^2}{4} + 2x]_{1}^{3} - 4$.Area $= (\frac{9}{4} + 6) - (\frac{1}{4} + 2) - 4 = (2 + 4) - 4 = 2$ sq units.

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