JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q69

MathematicsSingle correct+4 / -1Question 69 of 90

Let $I_n(x)=\displaystyle\int_0^x\frac1{\left(t^2+5\right)^n}dt$ , $n=1,2,3,\ldots$ . Then :

50I_6-9I_5=xI_5'

50I_6-11I_5=xI_5'

50I_6-9I_5=I_5'

50I_6-11I_5=I_5'

Answer and worked solution

Official answer: A

Reduction. $\frac d{dt}\frac t{(t^2+5)^n}=\frac{1-2n}{(t^2+5)^n}+\frac{10n}{(t^2+5)^{n+1}}$ , so $\frac x{(x^2+5)^n}=(1-2n)I_n+10nI_{n+1}$ .

n = 5. $xI_5'=-9I_5+50I_6$ . (A)

Topics:Properties of definite integrals Integration by parts

JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 69