Let I=∫01sinxxdx and J=∫01cosxxdx, then

I=∫01sinxxdxU.K.T sinx<x ∀x∈(0,1)sinxx<x∫01sinxxdx<∫01xdxI=<x323201I<23J=∫01cosxxdx0<cosx<1⇒cosxx<1xJ<∫011xdx⇒J<2

Let I=∫01sinxxdx and J=∫01cosxxdx, then

I=∫01sinxxdxU.K.T sinx<x ∀x∈(0,1)sinxx<x∫01sinxxdx<∫01xdxI=<x323201I<23J=∫01cosxxdx0<cosx<1⇒cosxx<1xJ<∫011xdx⇒J<2

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Let $I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x a n d J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x,$ then

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$I > \frac{2}{3}$

$I < \frac{2}{3}$

$J > 2$

$J < 2$

answer is B, C.

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Detailed Solution

$I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x$

U.K.T $\sin x < x \forall x \in (0, 1)$

$\frac{\sin x}{\sqrt{x}} < \sqrt{x}$

$\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x < \int_{0}^{1} \sqrt{x} d x$

$I = < {(\frac{x^{\frac{3}{2}}}{\frac{3}{2}})}_{0}^{1}$

$I < \frac{2}{3}$

$J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x$

$0 < \cos x < 1$

$\Rightarrow \frac{\cos x}{\sqrt{x}} < \frac{1}{\sqrt{x}}$

$J < \int_{0}^{1} \frac{1}{\sqrt{x}} d x$

$\Rightarrow J < 2$

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Let I=∫01sinxxdx and J=∫01cosxxdx, then

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Let $I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x a n d J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x,$ then