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Sol:- Let P_{3}(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=p(x)where a_{0}, a_{1}, a_{2} and a_{3} are to be determined.since P(x) passes through the points (0,0),(0,5, y),(1,3) and (2,2).And coefficient of x^{3} in p_{3}(x) is 6\Rightarrow a_{3}=6so, P(x)=a_{0}+a_{1} x+a_{2} x^{2}+6 x^{3}P(x) Passes +hrough (0,0)\begin{array}{c}\Rightarrow \quad 0=a_{0}+0+0+0 \\\Rightarrow a_{0}=0\end{array}P(x) Passes through (0.5, y)\begin{array}{c}\Rightarrow \quad y=a_{0}+(0.5) a_{1}+(0.5)^{2} a_{2}+6(0.5)^{3} \\\Rightarrow y=a_{0}+0.5 a_{1}+0.25 a_{2}+0.125 \times 6 \\\because a_{0}=0 \\\Rightarrow \quad y=0.5 a_{1}+0.25 a_{2}+0.750\end{array}p(x) passes through (1,3)\begin{array}{l}\Rightarrow \quad 3=a_{0}+a_{1}+a_{2}+6 \\\because a_{0}=0 \\\Rightarrow \quad 3=a_{1}+a_{2}+6 \Rightarrow a_{1}+a_{2}=-3 \\\end{array}P(x) passes through (2,2)\begin{array}{c}\Rightarrow \quad 2=a_{0}+2 a_{1}+4 a_{2}+6 \times(2)^{3} \\\Rightarrow \quad 2=a_{0}+2 a_{1}+4 a_{2}+48 \\\because a_{0}=0 \\\Rightarrow \quad 2=2 a_{1}+4 a_{2}+48 \\\Rightarrow \quad a_{1}+2 a_{2}=-23-3\end{array}Now, equation (2) and (3)\begin{array}{r}a_{1}+a_{2}=-3 \\a_{1}+2 a_{2}=-23 \\\hline-a_{2}=20 \\a_{2}=-20\end{array}p_{4}+a_{2}=-20 in eqution (2)\text { a) } \begin{array}{r}-20=-3 \\a_{1}=17\end{array}p_{4}+a_{0}=0, a_{1}=17, a_{2}=-20 in equation (1). we get.\begin{aligned}y & =0+(0.5) a_{1}+0.25 a_{2}+0.750 \\\Rightarrow y & =0.5 \times 17+0.25 \times(-20)+0.750 \\y & =8.5-5+7.50 \\y & =4.250 \text { or }=y=\frac{17}{4}\end{aligned}so y=4.250 ...