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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

Let a be a positive real number such that \int_{0}^{a} e^{x-[x]} d x=10 e-9 where [x] is the greatest integer less than or equal to x. Then a is equal to  
Option: 1 \\10-\log _{\mathrm{e}}(1+\mathrm{e})\\
Option: 2 \\10+\log _{\mathrm{e}} 2\\
Option: 3 \\10+\log _{\mathrm{e}} 3\\
Option: 4 \\10+\log _{e}(1+e)

Answers (1)

best_answer

\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)

Posted by

Kuldeep Maurya

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