JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
Let Iₙ(x)=∫₀ˣ 1/(t²+5)ⁿ d t, n=1,2,3, … . Then :
- (a) 50 I₆-9 I₅=x I₅′
- (b) 50 I₆-11 I₅=x I₅′
- (c) 50 I₆-9 I₅=I₅′
- (d) 50 I₆-11 I₅=I₅′
Answer and solution
Answer (a) 50 I₆-9 I₅=x I₅′
Solution
Maths sir · 4 steps
Integrate I₅ by parts with 1 as the second function, and I₆ appears by itself.
- Integrate by parts. Take u = (t²+5)⁻⁵ and dv = dt, so du = -10t (t²+5)⁻⁶ dt and v = t. The boundary term is x/(x²+5)⁵, because the lower limit gives 0.
I₅ = x/(x²+5)⁵ + 10∫₀ˣ t²/(t²+5)⁶ dt
- Split the numerator. Write t² = (t²+5) - 5. The remaining integral then becomes I₅ - 5I₆, so I₅ = x/(x²+5)⁵ + 10I₅ - 50I₆.
- Recognise the derivative. By the fundamental theorem of calculus, I₅'(x) = 1/(x²+5)⁵. The boundary term is therefore x I₅', and the relation reads I₅ = xI₅' + 10I₅ - 50I₆.
- So. Rearranging gives 50I₆ - 9I₅ = xI₅', which is option (a).
The idea to keep. For ∫ (t²+a)⁻ⁿdt, integrate by parts with 1 as the second function, then write t² = (t²+a) - a. This gives the reduction formula connecting Iₙ and Iₙ₊₁.
Answer from the NTA final answer key. Solution by PrepWiser.