Question

If the curve $ y=2{{x}^{3}}+a{{x}^{2}}+bx+c $ passes through the origin and the tangents drawn to it at $ x=-1 $ and $ x=2 $ are parallel to the $ x- $ axis, then the values of a, b and c are respectively

• A. $ 12,-3 $ and 0
• B. $ -3,-12 $ and 0
• C. $ -3,12 $ and 0
• D. $ 3,-12 $ and 0

Answer 1

Answer: (B)

Given equation of curve is $ y=2{{x}^{3}}+a{{x}^{2}}+bx+c $ ...(i) Since, it is passes through (0, 0) $ \Rightarrow $ $ 0=2(0)+a(0)+b(0)+c $ $ \Rightarrow $ $ c=0 $ ...(ii) On differentiating Eq. (i), we get $ \frac{dy}{dx}=6{{x}^{2}}+2ax+b $ Since, the tangents at $ x=-1 $ and $ x=2 $ are parallel to $ x- $ axis. $ \therefore $ $ \frac{dy}{dx}=0 $ At $ x=-1 $ $ 6{{(-1)}^{2}}+2a(-1)+b=0 $ $ \Rightarrow $ $ 6-2a+b=0 $ ?.(iii) At $ x=2 $ $ 6{{(2)}^{2}}+2a(2)+b=0 $ On solving Eqs. (iii) and (iv), we get $ a=-3,\text{ }b=-12 $ and $ c=0 $