(A) 33/2
(B) 5/3
(C) 27/2
(D) 41/3
Correct Answer: 33/2
Given,
f(x2 + 1) = x4 + 5x2 + 2
Let,
x2 + 1 = t ⇒ x2 = t − 1
Then,
f(t) = (t − 1)2 + 5(t − 1) + 2
f(t) = t2 − 2t + 1 + 5t − 5 + 2
f(t) = t2 + 3t − 2
Hence,
f(x) = x2 + 3x − 2
Now evaluate the integral:
∫03 (x2 + 3x − 2) dx
= [ x3/3 + 3x2/2 − 2x ]03
= (27/3 + 27/2 − 6)
= 9 + 13.5 − 6
= 33/2
Hence, the required value is 33/2.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.