Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Let $$P(x)=ax^3+bx^2+cx+d$$ be a cubic polynomial such that $$P(2)=7$$, $$P(3)=13$$ and $$P(5)=7$$. If the sum of the three roots of $$P(x)=0$$ is $$40$$, the value of $$P(35)$$ is
Correct Answer: 205
The root-sum condition gives $$-\frac{b}{a}=40$$, so $$b=-40a$$. Solving this relation together with the three given function values yields $$P(x)=\frac{1}{10}x^3-4x^2+\frac{241}{10}x-26$$. Substitution of $$x=35$$ gives $$P(35)=205$$.
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation