JEE MAIN - Mathematics (2007 - No. 9)
Let $$I = \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx} $$ and $$J = \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx} .$$ Then which one of the following is true?
$$1 > {2 \over 3}$$ and $$J > 2$$
$$1 < {2 \over 3}$$ and $$J < 2$$
$$1 < {2 \over 3}$$ and $$J > 2$$
$$1 > {2 \over 3}$$ and $$J < 2$$
Explanation
We know that $${{\sin x} \over x} < 1,$$ for $$x \in \left( {0,1} \right)$$
$$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $$ on $$x \in \left( {0,1} \right)$$
$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1$$
$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}$$
Also $${{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}$$ for $$x \in \left( {0,1} \right)$$
$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} } $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2$$
$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2} $$
$$ \Rightarrow J < 2$$
$$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $$ on $$x \in \left( {0,1} \right)$$
$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1$$
$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}$$
Also $${{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}$$ for $$x \in \left( {0,1} \right)$$
$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} } $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2$$
$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2} $$
$$ \Rightarrow J < 2$$
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