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Question Number 156584 by Ghaniy last updated on 12/Oct/21

$$\mathrm{find}\mathrm{p}\mathrm{if}\mathrm{y}=1-\mathrm{px}-{3\mathrm{x}}^2$$$$\mathrm{if}\mathrm{the}\mathrm{maximum}\mathrm{is}13\left(\mathrm{help}\mathrm{pls}\right)$$

Commented by john_santu last updated on 14/Oct/21

$$maxf\left(x\right)=-3x^2-px+1$$$$whenx=-\left(\frac{-p}{2(-3)}\right)=-\frac{p}{6}$$$$so\Rightarrow 13=1-p(-\frac{p}{6})-3\left(\frac{p^2}{36}\right)$$$$\Rightarrow 12=\frac{p^2}{6}-\frac{p^2}{12}$$$$\Rightarrow 12=\frac{p^2}{12}\Rightarrow p=\pm 12$$

Answered by mr W last updated on 13/Oct/21

$$y=1-px-3x^2$$$$y=1-3(\frac{p}{3}x+x^2)$$$$y=1-3(2\frac{p}{6}x+x^2)$$$$y=1+\frac{p^2}{12}-3(\frac{p^2}{36}+2\frac{p}{6}x+x^2)$$$$y=1+\frac{p^2}{12}-3{(x+\frac{p}{6})}^2⩽1+\frac{p^2}{12}$$$$y_{max}=1+\frac{p^2}{12}=13$$$$p^2={12}^2$$$$\Rightarrow p=\pm 12$$

Commented by otchereabdullai@gmail.com last updated on 13/Oct/21

$$\mathrm{nice}\mathrm{question}+\mathrm{nice}\mathrm{solution}$$

Commented by Ghaniy last updated on 14/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}\mathrm{Mr}\mathrm{W}$$

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