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If $(2, a)$ and $(b, 19)$ are two stationary points of the curve $y=2 x^3-15 x^2+36 x+c$, then $a+b+c=$
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The correct answer is:
$15$
Given the function $y=2 x^3-15 x^2+36 x+C$
Now, $\frac{d y}{d x}=6 x^2-30 x+36$
For stationary points, $\frac{d y}{d x}=0$
$6 x^2-3 x+36 \Rightarrow(x-2)(x-3)=0 \Rightarrow x=2,3$
So $b=3$
$\Rightarrow 19=2 \times 27-15 \times 9+36 \times 3+c \Rightarrow c=-8$
and $a=2 \times 8-15 \times 4+36 \times 2-8$
$\Rightarrow a=16-60+72-8=20$
so $a+b+c=20+3-8=15$
Now, $\frac{d y}{d x}=6 x^2-30 x+36$
For stationary points, $\frac{d y}{d x}=0$
$6 x^2-3 x+36 \Rightarrow(x-2)(x-3)=0 \Rightarrow x=2,3$
So $b=3$
$\Rightarrow 19=2 \times 27-15 \times 9+36 \times 3+c \Rightarrow c=-8$
and $a=2 \times 8-15 \times 4+36 \times 2-8$
$\Rightarrow a=16-60+72-8=20$
so $a+b+c=20+3-8=15$
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