If graph of y = ax^3 + bx^2 + cx + d is symmetric about line x = k then

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4-2.Quadratic Equations and Inequations

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If the graph of $y = ax^3 + bx^2 + cx + d$ is symmetric about the line $x = k$ then

A

$k=c$

B

$k = -\frac{c}{b}$

C

$a + \frac{c}{{2b}} + k = 0$

D

none of these

Solution

$\therefore a=0$ and $y=b x^{2}+c x+d$ is symmetric

about $x=-\frac{c}{2 b}$

$\therefore \mathrm{x}=\mathrm{k}=-\frac{c}{2 b} \Rightarrow k+\frac{c}{2 b}=0$

$\Rightarrow a+\frac{c}{2 b}+k=0$

Standard 11

Mathematics

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