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#### Let a be a positive real number such that where [x] is the greatest integer less than or equal to x. Then a is equal to  Option: 1Option: 2Option: 3Option: 4

\begin{aligned} & \int_{0}^{a} e^{x-[x]} d x \\ =& \int_{0}^{a} e^{\{x\}} d x . \end{aligned}

Let,  $a=[a]+\{a\}$

$=\int_{0}^{[a]} e^{\{x\}} d x+\int_{[a]}^{[a]+\{a\}} e^{\{x\}} d x .$

Now {x} is periodic with period $1 \Rightarrow e^{x}$ is periodic with period 1.

\begin{aligned} &=[a] \int_{0}^{1} e^{\{ x\}} d x+\int_{0}^{\{a\}} e^{\{x\}} dx\,\, \text{(Properties of Periodic functions)}\\ &\text { In } 0 \leq x<1,\{x\}=x \\ &=[a] \int_{0}^{1} e^{x} d x+\int_{0}^{\{a\}} e^{x} d x \\ &=\left.[a]\left(e^{x}\right)\right|_{0} ^{1}+\left.e^{x}\right|_{0} ^{\{a\}} \end{aligned}

\begin{aligned} &=[a](e-1)+e^{\{a\}}-1 \\ &=[a] e+\left(-[a]+e^{\{a\}}-1\right) \end{aligned}

Given that this equals $10 e-9$

$\Rightarrow[a] e+\left(e^{\{a\}}-[a]-1\right)=10 e-9$

Comparing  $\Rightarrow[a]=10 \text { and } e^{\{a\}}-10-1=-9 \Rightarrow\{a\}=\ln 2$

$\Rightarrow a= [a]+\{a\}=10+\ln (2)$

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#### Using integration find the area of the region bounded between the two circles and

x - cordinate of A = 3/2

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