Question

If $f(x)=\left(x^3+b x^2+c x+d\right)\left|(x-1)(x-2)(x-3)^2(x-4)^3\right|$ is derivable for all $x \in R$ and$f ^{\prime}(3)+ f ^{\prime \prime \prime}(4)=0$ then $| b + c + d |$ equals

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$ \because f ^{\prime \prime \prime}(4)=0 \Rightarrow( x -4)$ must be a factor of $\left( x ^3+ bx ^2+ cx + d \right)$ and two other factors are $(x-1)$ and $(x-2)$
$\therefore x ^3+ bx ^2+ cx + d =( x -1)( x -2)( x -4) $
$1+ b + c + d =0 \Rightarrow| b + c + d |=1$