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The curve $\mathrm{y(x)=a x^{3}+b x^{2}+c x+5}$ touches the x-axis at the point $\mathrm{P(-2,0)}$ and cuts the y-axis at the point $\mathrm{Q}$ where $\mathrm{y'}$ is equal to $\mathrm{3}$ . Then the local maximum value of $\mathrm{y(x)}$ is :
Option: 1
$\frac{27}{4}$

Option: 2
$\frac{29}{4}$

Option: 3
$\frac{37}{4}$

Option: 4
$\frac{9}{2}$

Answers (1)

best_answer

$\mathrm{y(x)=a x^{3}+b x^{2}+c x+5}$ is passing through $\mathrm{(-2,0)}$

then $\mathrm{8 a-4 b+2 c=5}$ ...(1)

$\mathrm{y^{\prime}(x)=3 a x^{2}+2 b x+c}$ touches x-axis at $\mathrm{(-3,0)}$

$\mathrm{12 a-4 b+c=0}$ ...(2)

$\mathrm{again, for\; x=0, y^{\prime}(x)=3}\\$

$\mathrm{c=3}\\$ ...(3)

$\mathrm{Solving \;eq.(1), (2) \& (3)\;\; a=\frac{-1}{2}, b=-\frac{3}{4}}\\$

$\mathrm{y^{\prime}(x)=-\frac{3}{2} x^{2}-\frac{3}{2} x+3}\\$

$\mathrm{y(x) has\; local \;maxima \;at\; x=1}$

$\mathrm{y(1)=\frac{27}{4}}$

Hence correct option is 1

Posted by

Shailly goel

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