JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 18)
Let $${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$$, where n$$\in$$N. If (20)I10 = $$\alpha$$I9 + $$\beta$$I8, for natural numbers $$\alpha$$ and $$\beta$$, then $$\alpha$$ $$-$$ $$\beta$$ equals to ___________.
Answer
1
Explanation
$${I_n} = 2\int\limits_1^e {{x^{19}}{{(\ln x)}^n}\,.\,dx} $$
$$ = {(\ln x)^n}\,.\,\left. {{{{x^{20}}} \over {20}}} \right|_1^e -\int\limits_1^e {n{{{{(\ln x)}^{n - 1}}} \over x}{{{x^{20}}} \over {20}}dx} $$
$${I_n} = {{{e^{20}}} \over {20}} - {n \over {20}}({I_{n - 1}})$$
$$20{I_n} = {e^{20}} - n\,{I_{n - 1}}$$
Putting n = 10, we get
$$20{I_{10}} = ({e^{20}} - 10{I_9})$$ ...... (1)
Putting n = 9, we get
$$20{I_9} = {e^{20}} - 9{I_8}$$ ....... (2)
Subtracting (2) from (1), we get
$$20{I_{10}} = 10{I_9} + 9{I_8}$$
By comparing with (20)I10 = $$\alpha$$I9 + $$\beta$$I8, we get
$$\alpha$$ = 10, $$\beta$$ = 9 $$ \Rightarrow $$ $$\alpha$$ $$-$$ $$\beta$$ = 1
$$ = {(\ln x)^n}\,.\,\left. {{{{x^{20}}} \over {20}}} \right|_1^e -\int\limits_1^e {n{{{{(\ln x)}^{n - 1}}} \over x}{{{x^{20}}} \over {20}}dx} $$
$${I_n} = {{{e^{20}}} \over {20}} - {n \over {20}}({I_{n - 1}})$$
$$20{I_n} = {e^{20}} - n\,{I_{n - 1}}$$
Putting n = 10, we get
$$20{I_{10}} = ({e^{20}} - 10{I_9})$$ ...... (1)
Putting n = 9, we get
$$20{I_9} = {e^{20}} - 9{I_8}$$ ....... (2)
Subtracting (2) from (1), we get
$$20{I_{10}} = 10{I_9} + 9{I_8}$$
By comparing with (20)I10 = $$\alpha$$I9 + $$\beta$$I8, we get
$$\alpha$$ = 10, $$\beta$$ = 9 $$ \Rightarrow $$ $$\alpha$$ $$-$$ $$\beta$$ = 1
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