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Question

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


A
g(1)=377
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B
g(1)=277
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C
g(1)=17
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D
g(1)=3714
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Solution

The correct options are
A $$\displaystyle g(1)=\frac{3^{7}}{7}$$
C $$\displaystyle g(-1)=\frac{1}{7}$$
$$\displaystyle g(x)=\int x^{27}(1+x+x^{2})^{6}(6x^{2}+5x+4)dx$$
$$=\displaystyle \int (x^{4}+x^{5}+x^{6})^{6}(6x^{5}+5x^{4}+4x^{3})dx$$
Let $$\displaystyle x^{6}+x^{5}+x^{4}=t  \Rightarrow  (6x^{5}+5x^{4}+4x^{3})dx=dt$$
$$\therefore \displaystyle g(x)=\int t^{6}dt$$
             $$\displaystyle =\frac{t^{7}}{7}+C$$
$$\displaystyle \Rightarrow g(x)=\frac{1}{7}(x^{4}+x^{5}+x^{6})^{7}+C$$
$$\displaystyle g(0)=0   \Rightarrow C=0 $$
$$\displaystyle g(x)=\frac{1}{7}(x^{4}+x^{5}+x^{6})^{7}$$
 $$\displaystyle\Rightarrow g(1)=\frac{3^{7}}{7}  $$ and  $$g(-1)=\displaystyle \frac{1}{7}$$

Mathematics

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