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The curve $y=ax^{3}+bx^{2}+cx+d$ has a point of inflexion at $x=1$ then

\[\begin{array}{1 1}(1)a+b=0&(2)a+3b=0\\(3)3a+b=0&(4)3a+b=1\end{array}\]

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Let $f(x)=ax^3+bx^2+cx+d$
$f'(x)=3ax^2+2bx+c$
$f''(x)=6ax+2b$
Given that the curve has stationary point at $ x=1$
$f''(1) =0$
ie $6a \times 1 +2=0$
$3a+b=0$
Hence 3 is the correct answer.
answered May 20, 2014 by meena.p
 
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