MCQ
$\int_{ - \pi /2}^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}\,dx = } $
- ✓$1$
- B$0$
- C$ - 1$
- Dએકપણ નહીં.
Putting $x = - t$ in $\int_{ - \pi /2}^0 {\frac{{\cos x}}{{1 + {e^x}}}dx} $, we get
$I = \int_{ - \pi /2}^0 {\frac{{\cos x}}{{1 + {e^x}}}dx = \int_0^{\pi /2} {\frac{{{e^x}\cos x}}{{1 + {e^x}}}dx} } $
$I = \int_0^{\pi /2} {\frac{{{e^x}\cos x}}{{1 + {e^x}}}dx + \int_0^{\pi /2} {\frac{{\cos x}}{{1 + {e^x}}}dx} } $
$ = \int_{\,0}^{\,\pi /2} {\frac{{(1 + {e^x})\cos x\,dx}}{{(1 + {e^x})}}} $
$ = \int_0^{\pi /2} {\cos x\,dx = [\sin x]_0^{\pi /2} = 1} $.
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ધારો કે $a \in S$ અને $A =\left[\begin{array}{ccc}1 & 0 & a \\ -1 & 1 & 0 \\ - a & 0 & 1\end{array}\right]$ છે.
જો $\sum_{ a \in S } \operatorname{det}(\operatorname{adj} A )=100 \lambda$ હોય, તો $\lambda$ .........