JEE Mains · Maths · STD 12 - 7.2 definite integral
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:
- A \(10+\log _{e} 3\)
- B \(10-\log _{e}(1+e)\)
- C \(10+\log _{e} 2\)
- D \(10+\log _{e}(1+e)\)
Answer & Solution
Correct Answer
(C) \(10+\log _{e} 2\)
Step-by-step Solution
Detailed explanation
\(a\,>\,0\) Let \(\geq a\,<\,n+1, n \in W\) \(\therefore a=[a]+\{a\}\) \(\quad\,\,\,\,\,\,\,\,\,\)G.I.F \(\,\,\,\,\,\,\,\) Fractional part Here \([a]=n\) Now, \(\int_{0}^{a} e^{x-[x]} d x=10 e-9\) \(\Rightarrow \int_{0}^{a} e^{x} d x+\int_{n}^{a} e^{x-[x]} d x=10 e-9\)…
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