If the area enclosed by curve y=f(x) and y=x2+2 between the abscissa x=2 and x=α,α>2, is α3−4α2+8 sq. unit. It is known that curve y=f(x) lies below the parabola y=x2+2
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
1
b
2
c
3
d
4
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
According to the question,α3−4α2+8=∫2α x2+2−f(x)dx Differentiating w.r.t. α on both sides, we get 3α2−8α=α2+2−f(α)∴ f(x)=−2x2+8x+2