- ✓$21$
- B$0$
- C$19$
- D$-21$
$x \rightarrow 1-x$
$I=\int \limits_0^1 \frac{e^{2-4 x} d x}{\left(5+2 x-2 x^2\right)\left(1+ e ^{2-4 x}\right)}$
Add $(i)$ and $(ii)$
$2 I=\int \limits_0^1 \frac{d x}{5+2 x-2 x^2}=\int \limits_0^1 \frac{d x}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)}$
$I=\frac{1}{\sqrt{11}} \ln \left(\frac{\sqrt{11}+1}{\sqrt{10}}\right) \quad \begin{array}{l}\alpha=\sqrt{11} \\\beta=\sqrt{10}\end{array}$
$\alpha^4-\beta^4=121-100=21$
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| $X$ | $0$ | $2$ | $4$ | $6$ | $8$ |
| $P(X)$ | $a$ | $2a$ | $a+b$ | $2b$ | $3b$ |
નું મધ્યક જો $\frac{46}{9}$ હોય, તો વિતરણ નું વિચરણ ............ છે.
$\begin{array}{|l|l|l|l|l|l|} \hline X=x & 0 & 1 & 2 & 3 & 4 \\ \hline P(X=x) & \frac{1}{3} & \frac{1}{2} & 0 & \frac{1}{6} & 0 \\ \hline \end{array}$
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