JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
Let f be a continuous function satisfying ∫0^(t²)(f(x)+x²) d x=4/3 t³, ∀ t > 0. Then f(π²/4) is equal to :
- (a) -π(1+π³/16)
- (b) π(1-π³/16)
- (c) -π²(1+π²/16)
- (d) π²(1-π²/16)
Answer and solution
Answer (b) π(1-π³/16)
Solution
Maths sir · 4 steps
Differentiate both sides with respect to t, and do not forget the chain rule on the upper limit t².
- Differentiate both sides. By the Fundamental Theorem with the chain rule, the derivative of the left side is the integrand at x = t², times the derivative of t².
(f(t²) + t⁴) · 2t = 4t²
- Find f. Since t > 0, divide by 2t to get f(t²) + t⁴ = 2t. Put x = t², so t = √x. Then f(x) = 2√x - x².
- Put in the value. At x = π²/4 we have √x = π/2 and x² = π⁴/16. So f(π²/4) = π - π⁴/16.
- So. Taking π common gives π(1 - π³/16), which is option (b).
The idea to keep. When the unknown function sits inside an integral with a variable limit, differentiate to free it: d/dt∫₀ᵍ⁽ᵗ⁾ h(x) dx = h(g(t)) g'(t).
Answer from the NTA final answer key. Solution by PrepWiser.