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

Answered by MrW3 last updated on 23/Oct/18

$$letf\left(x\right)=a{(x-b)}^2+c$$$$\left(b\right)\Rightarrow b=1,c=\frac{1}{2}$$$$\left(a\right)\Rightarrow a{(0-1)}^2+\frac{1}{2}=0\Rightarrow a=-\frac{1}{2}$$$$\Rightarrow f\left(x\right)=-\frac{1}{2}{(x-1)}^2+\frac{1}{2}$$$$A=\frac{1.5\times 1.5}{2}-2\times \frac{0.5\times 0.5}{2}-2\times \frac{2}{3}\times 1\times 0.5$$$$=\frac{9}{8}-\frac{1}{4}-\frac{2}{3}$$$$=\frac{7}{8}-\frac{2}{3}$$$$=\frac{5}{24}$$$$\Rightarrow 24A=5$$

Commented by MrW3 last updated on 23/Oct/18

Commented by rahul 19 last updated on 23/Oct/18

thank you sir !����

Answered by ajfour last updated on 22/Oct/18

$$24A=5.$$

Commented by rahul 19 last updated on 23/Oct/18

$$Yessir!$$

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