$I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x<\int_{0}^{1} \frac{x}{\sqrt{x}} d x$$=\int_{0}^{1} \sqrt{x} d x=\left[\frac{2}{3} x^{3 / 2}\right]_{0}^{1}=\frac{2}{3}$
$I<\frac{2}{3}$
Also $J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x<\int_{0}^{1} \frac{1}{\sqrt{x}} d x$$=[2 \sqrt{x}]_{0}^{1}=2$
$J<2$
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અહી શ્રેણિક $A=\left[a_{i j}\right]_{3 \times 3}$ કે જ્યાં
$a_{i j}=J_{6+i, 3}-J_{i+3,3}, \quad i \leq j$
$\quad\quad\quad\quad\quad\quad0 , \quad\quad\quad i>j$.
તો $\left|\operatorname{adj} A^{-1}\right|$ મેળવો.
વિધાન $1$:કોઇક $c\; \in R$ માટે, $f\left( c \right) = \frac{1}{3}$
વિધાન $2$:$0 < f\left( x \right) < \frac{1}{{2\sqrt 2 }}\;,\forall x\; \in R$
વિધાન $-2$ : સમીકરણ કે જે $\alpha $ સ્વરૂપ માં છે
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
નું એક માત્ર બીજ અંતરાલ $\left( {0\,,\,\frac{\pi }{2}} \right)$ માં છે .