If the line $y = 1 - 4 x$ touches the curve $y = x^{3} + a x^{2} - b x + c$ at $(0, 1)$ and also touches the curve $y = \frac{x^{2}}{2} - p x + 3$ at $(α, β)$ then

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sum of all possible values of $α$ is $0$

$b + c = 5$

sum of all possible value of $p$ is $8$

sum of all possible values of $β$ is $2 .$

answer is A, B, C, D.

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Detailed Solution

$(0, 1)$ lies on first curve $\Rightarrow c = 1$

Also, $\frac{dy}{dx} = - 4 \Rightarrow - b = - 4 \Rightarrow b = 4$

$∴ b + c = 5$

Also $1 - 4 x = \frac{x^{2}}{2} - p x + 3 \Rightarrow x^{2} + (8 - 2 p) x + 4 = 0$

$D = 4 (4 - p)^{2} - 16 = 0$

$(4 - p) = \pm 2$

if $p = 6 \Rightarrow x = 2$

$p = 2 \Rightarrow x = - 2$

$\Rightarrow$ point of contact with second curve $(2, - 7)$ on $(- 2, 9)$

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