TS EAMCET 5 May 2018 Shift 2 Solved Paper

The given equation of curves are, $y=8x-x^{2}$ It can be written as, $y=-(x^{2}-8x+4^{2}-4^{2})$ $y=-((x-4)^{2}-16)$ $y-16=-(x-4)^{2}$ The above equation represents a parabola having vertex (4,16) . Consider the equation, $8x-4y+11=0$ It represent a straight line passing through the points $\left(0,\frac{11}{4}\right)$ and$\left(-\frac{11}{8},0\right)$. The intersection is calculated as, $8x-4(8x-x^{2})+11=0$ $4x^{2}-24x+11=0$ $4x^{2}-22x-2x+11=0$ $(2x-1)(2x-11)=0$ Thus, $x=\frac{1}{2},\frac{11}{2}$ Consider the figure shown below. The required area is the shaded region that is calculated as, $A=\int\limits_{\frac{1}{2}}^{\frac{11}{2}} \left[8x-x^{2}-\left(\frac{8x+11}{4}\right)\right]dx$ $=-\frac{1}{4}\int\limits_{\frac{1}{2}}^{\frac{11}{2}}\left[4x^{2}-24x+11\right]dx$ $=-\frac{1}{4}\left[\frac{4x^{3}}{3}-24\frac{x^{2}}{2}+11x\right]_{\frac{1}{2}}^{\frac{11}{2}}$ $=-\frac{1}{4}[\frac{1331}{6}-363+\frac{121}{2}$ $-\frac{1}{6}+3-\frac{11}{2}]$ Solve further, $A=-\frac{1}{4}\left\{\frac{1330}{6}-360+55\right\}$ $=-\frac{1}{4}\left\{\frac{665-915}{3}\right\}$ $=\frac{250}{12}$ $=\frac{125}{6}$ sq units