A curve `g(x)=intx^(27)(1+x+x^2)^6(6x^2+5x+4)dx` is passing through origin. Then `g(1)=(3^7)/7` (b) `g(1)=(2^7)/7` `g(-1)=1/7` (d) `g(-1)=(3^7)/(14)`

Question

A curve `g(x)=intx^(27)(1+x+x^2)^6(6x^2+5x+4)dx` is passing through origin. Then `g(1)=(3^7)/7` (b) `g(1)=(2^7)/7` `g(-1)=1/7` (d) `g(-1)=(3^7)/(14)`

A curve `g(x)=intx^(27)(1+x+x^2)^6(6x^2+5x+4)dx` is passing through origin. Then `g(1)=(3^7)/7` (b) `g(1)=(2^7)/7` `g(-1)=1/7` (d) `g(-1)=(3^7)/(14)`
A. `g(1)=(3^(7))/(7)`
B. `g(1)=(2^(7))/(7)`
C. `g(-1)=(1)/(7)`
D. `g(-1)=(3^(7))/(14)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A::C
`g(x)=int x^(27)(1+x+x^(2))^(6)(6x^(2)+5x+4)dx`
`=int(x^(4)+x^(5)+x^(6))^(6)(6x^(5)+5x^(4)+4x^(3))dx `
`"Let " x^(6)+x^(5)+x^(4)=t " or "(6x^(5)+5x^(4)+4x^(3))dx=dt `
` :. g(x)=int t^(6)dt=(t^(7))/(7)+C=(1)/(7)(x^(4)+x^(5)+x^(6))^(7)+C `
`g(0)=0 impliesC=0 impliesg(1)=(3^(7))/(7) " and " g(-1)=(1)/(7)`