Daily Question 110

Daily Question 110

$\textbf{Q110.}$ If $I_{1}=\displaystyle\int\limits_{0}^{1}\dfrac{1+x^8}{1+x^{4}}\mathrm{d}x$ and $I_2=\displaystyle\int\limits_{0}^{1}\dfrac{1+x^{9}}{1+x^{3}}\mathrm{d}x$, then

(A) $I_1 > 1, I_2 < 1$

(B) $I_1 < 1, I_2 > 1$

(C) $1 < I_1 < I_2$

(D) $I_2 < I_1 < 1$

$\textbf{Ans.} (D)$

$\textbf{Sol.}$ As $x\in(0, 1)$

$\implies 1+x^8 < 1+x^4$

$\therefore I_{1}< 1$

Similarly, $I_2 < 1$

Now, $1+x^8 > 1+x^9$ and $1+x^4< 1+x^3$

Thus $\dfrac{1+x^8}{1+x^{4}}>\dfrac{1+x^{9}}{1+x^{3}}$

$\implies I_1 > I_2$

Hence $I_2 < I_1 < 1$

Compiled Daily Questions 1-100