JEE Main 2021 — Mathematics Calculus

Given $\displaystyle {I}_{n}={\int }_{1}^{e}{x}^{19}(\mathrm{log}|x|{)}^{n}dx,n\in N$In​=∫1e​x19(log∣x∣)ndx,n∈N

$\displaystyle \Rightarrow {I}_{n}={\int }_{1}^{e}{x}^{19}{(\mathrm{log}x)}^{n}dx,\because$⇒In​=∫1e​x19(logx)ndx,∵ for $1<x<e,x>0$1<x<e,x>0

Integrating by using by parts rule, $\displaystyle {\int }_{a}^{b}(u\cdot v)dx={[u\int vdx]}_{a}^{b}-{\int }_{a}^{b}[\frac{d}{dx}(u)\int vdx]dx,$∫ab​(u⋅v)dx=[u∫vdx]ab​−∫ab​[dxd​(u)∫vdx]dx, taking u=\mathrm{log}x&v={x}^{19}, we get

$${I}_{n}={[{(\mathrm{log}x)}^{n}\int {x}^{19}dx]}_{1}^{e}-{\int }_{1}^{e}[\frac{d}{dx}{(\mathrm{log}x)}^{n}\int {x}^{19}dx]dx$$In​=[(logx)n∫x19dx]1e​−∫1e​[dxd​(logx)n∫x19dx]dx

$$\Rightarrow {I}_{n}={[{(\mathrm{log}x)}^{n}(\frac{{x}^{20}}{20})]}_{1}^{e}-{\int }_{1}^{e}[n{(\mathrm{log}x)}^{n-1}\times \frac{1}{x}\times (\frac{{x}^{20}}{20})]dx$$⇒In​=[(logx)n(20x20​)]1e​−∫1e​[n(logx)n−1×x1​×(20x20​)]dx

$$\Rightarrow {I}_{n}=[{(\mathrm{log}e)}^{n}(\frac{{e}^{20}}{20})-0]-{\int }_{1}^{e}[n{(\mathrm{log}x)}^{n-1}\times (\frac{{x}^{19}}{20})]dx$$⇒In​=[(loge)n(20e20​)−0]−∫1e​[n(logx)n−1×(20x19​)]dx

$$\Rightarrow {I}_{n}=(\frac{{e}^{20}}{20})-\frac{n}{20}{\int }_{1}^{e}{x}^{19}{(\mathrm{log}x)}^{n-1}dx$$⇒In​=(20e20​)−20n​∫1e​x19(logx)n−1dx

$$\Rightarrow 20{I}_{n}={e}^{20}-n{I}_{n-1}$$⇒20In​=e20−nIn−1​

Hence, we get $20{I}_{10}={e}^{20}-10{I}_{9}...(i)$20I10​=e20−10I9​...(i) and

$$20{I}_{9}={e}^{20}-9{I}_{8}...(ii)$$20I9​=e20−9I8​...(ii)

On subtracting the above two equations, we get

$$20{I}_{10}-20{I}_{9}=-10{I}_{9}+9{I}_{8}$$20I10​−20I9​=−10I9​+9I8​

$$\Rightarrow 20{I}_{10}=10{I}_{9}+9{I}_{8}$$⇒20I10​=10I9​+9I8​

Given $20{I}_{10}=\alpha {I}_{9}+\beta {I}_{8}$20I10​=αI9​+βI8​

$\Rightarrow \alpha =10,\beta =9$⇒α=10,β=9 and hence $\alpha -\beta =1.$α−β=1.