Question

If $I_1=\underset{0}{\overset{1}{\int }}\frac{1+x^8}{1+x^4}dx$ and $I_2=\underset{0}{\overset{1}{\int }}\frac{1+x^9}{1+x^3}dx$, then :

• A. $I_1<I_2$
• B. $I_1>I_2$
• C. $I_1>I_2>1$
• D. $I_2<I_1<1$

Answer 1

Answer: (B), (D)

$x\in (0,1)$
so $1+x^8>1+x^9$
and $1+x^3>1+x^4$
$\Rightarrow \left(1+x^8\right)\left(1+x^3\right)>\left(1+x^9\right)\left(x+x^4\right)$
$\Rightarrow \frac{1+x^8}{1+x^4}>\frac{1+x^9}{1+x^3}$
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{1+x^8}{1+x^4}dx>\underset{0}{\overset{1}{\int }}\frac{1+x^9}{1+x^3}dx\Rightarrow I_1>I_2$
Now again $\frac{1+x^8}{1+x^4}<1\because 1+x^4>1+x^8$
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{1+x^8}{1+x^4}dx<\underset{0}{\overset{1}{\int }}dx\Rightarrow I_1<1$