Question-46225

Question Number 46225 by rahul 19 last updated on 22/Oct/18

Answered by MrW3 last updated on 23/Oct/18

let f(x)=a(x−b)^2 +c (b)⇒b=1, c=(1/2) (a)⇒a(0−1)^2 +(1/2)=0⇒a=−(1/2) ⇒f(x)=−(1/2)(x−1)^2 +(1/2) A=((1.5×1.5)/2)−2×((0.5×0.5)/2)−2×(2/3)×1×0.5 =(9/8)−(1/4)−(2/3) =(7/8)−(2/3) =(5/(24)) ⇒24A=5

$${let}\:{f}\left({x}\right)={a}\left({x}−{b}\right)^{\mathrm{2}} +{c} \\ $$$$\left({b}\right)\Rightarrow{b}=\mathrm{1},\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({a}\right)\Rightarrow{a}\left(\mathrm{0}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}=\frac{\mathrm{1}.\mathrm{5}×\mathrm{1}.\mathrm{5}}{\mathrm{2}}−\mathrm{2}×\frac{\mathrm{0}.\mathrm{5}×\mathrm{0}.\mathrm{5}}{\mathrm{2}}−\mathrm{2}×\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{1}×\mathrm{0}.\mathrm{5} \\ $$$$=\frac{\mathrm{9}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}}−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{24}} \\ $$$$\Rightarrow\mathrm{24}{A}=\mathrm{5} \\ $$

Commented by MrW3 last updated on 23/Oct/18