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Let $\mathrm{I}=\int_0^1 \frac{\sin \mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}$ and $\mathrm{J}=\int_0^1 \frac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}$. Then which one of the following is true?
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1807 Upvotes
Verified Answer
The correct answer is:
$I < \frac{2}{3}$ and $J < 2$
$I < \frac{2}{3}$ and $J < 2$
$$
\begin{aligned}
& \mathrm{I}=\int_0^1 \frac{\sin x}{\sqrt{x}} \mathrm{dx} < \int_0^1 \frac{\mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}=\int_0^1 \sqrt{\mathrm{x}} \mathrm{dx}=\left.\frac{2}{3} \mathrm{x}^{3 / 2}\right|_0 ^1=\frac{2}{3} \\
& \Rightarrow \mathrm{I} < \frac{2}{3} \\
& \mathrm{~J}=\int_0^1 \frac{\cos x}{\sqrt{\mathrm{x}}} \mathrm{dx} < \int_0^1 \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}=\left.2 \sqrt{\mathrm{x}}\right|_0 ^1=2 \\
& \therefore \mathrm{J} \leq 2 .
\end{aligned}
$$
\begin{aligned}
& \mathrm{I}=\int_0^1 \frac{\sin x}{\sqrt{x}} \mathrm{dx} < \int_0^1 \frac{\mathrm{x}}{\sqrt{\mathrm{x}}} \mathrm{dx}=\int_0^1 \sqrt{\mathrm{x}} \mathrm{dx}=\left.\frac{2}{3} \mathrm{x}^{3 / 2}\right|_0 ^1=\frac{2}{3} \\
& \Rightarrow \mathrm{I} < \frac{2}{3} \\
& \mathrm{~J}=\int_0^1 \frac{\cos x}{\sqrt{\mathrm{x}}} \mathrm{dx} < \int_0^1 \frac{1}{\sqrt{\mathrm{x}}} \mathrm{dx}=\left.2 \sqrt{\mathrm{x}}\right|_0 ^1=2 \\
& \therefore \mathrm{J} \leq 2 .
\end{aligned}
$$
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