Question

A curve $g\left(x\right)=\int x^{27}(1+x+x^2{)}^6(6x^2+5x+4)dx$ is passing through origin. Then

• A. $g\left(1\right)=\frac{3^7}{7}$
• B. $g\left(1\right)=\frac{2^7}{7}$
• C. $g(-1)=\frac{1}{7}$
• D. $g(-1)=\frac{3^7}{14}$

Answer 1

Answer: (A), (C)

The correct options are
A $g\left(1\right)=\frac{3^7}{7}$
C $g(-1)=\frac{1}{7}$

$g\left(x\right)=\int x^{27}(1+x+x^2{)}^6(6x^2+5x+4)dx$

$=\int (x^4+x^5+x^6{)}^6(6x^5+5x^4+4x^3)dx$

Let $x^6+x^5+x^4=t\Rightarrow (6x^5+5x^4+4x^3)dx=dt$

$\therefore g\left(x\right)=\int t^{6}dt$
$=\frac{t^7}{7}+C$

$\Rightarrow g\left(x\right)=\frac{1}{7}(x^4+x^5+x^6{)}^7+C$
$g\left(0\right)=0\Rightarrow C=0$

$g\left(x\right)=\frac{1}{7}(x^4+x^5+x^6{)}^7$

$\Rightarrow g\left(1\right)=\frac{3^7}{7}$ and $g(-1)=\frac{1}{7}$