JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\alpha\) for which \(4 \alpha \int\limits_{-1}^{2} \mathrm{e}^{-\alpha \mathrm{|x|} } \mathrm{d} \mathrm{x}=5,\) is
- A \(\log _{e}\left(\frac{3}{2}\right)\)
- B \(\log _{e}\left(\frac{4}{3}\right)\)
- C \(\log _{e} 2\)
- D \(\log _{e} \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(\log _{e} 2\)
Step-by-step Solution
Detailed explanation
\(4 \alpha\left[\int_{-1}^{0} e^{\alpha x} d x+\int_{0}^{2} e^{-\alpha x} d x\right]=5\) \(\Rightarrow 4 \alpha\left(\left[\frac{\mathrm{e}^{\alpha \mathrm{x}}}{\alpha}\right]_{-1}^{0}+\left[\frac{\mathrm{e}^{-\alpha \mathrm{x}}}{-\alpha}\right]_{0}^{2}\right)=5\)…
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