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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
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
    $\log _{e}\left(\frac{4}{3}\right)$
  • ✓
    $\log _{e} 2$
  • D
    $\log _{e} \sqrt{2}$
✓

Answer

Correct option: C.
$\log _{e} 2$
c
$4 \alpha\left[\int_{-1}^{0} e^{\alpha x} d x+\int_{0}^{2} e^{-\alpha x} d x\right]=5$

$\Rightarrow 4 \alpha\left(\left[\frac{\mathrm{e}^{\alpha \mathrm{x}}}{\alpha}\right]_{-1}^{0}+\left[\frac{\mathrm{e}^{-\alpha \mathrm{x}}}{-\alpha}\right]_{0}^{2}\right)=5$

$\Rightarrow 4 \mathrm{e}^{-2 \alpha}+4 \mathrm{e}^{-\alpha}-3=0$

Let $\mathrm{e}^{-\alpha}=\mathrm{t}, 4 \mathrm{t}^{2}+4 \mathrm{t}-3=0,$$ \mathrm{t}=\frac{1}{2}, \frac{-3}{2}$ (Rejected)

$\mathrm{e}^{-\alpha}=\frac{1}{2} ; \quad \alpha=\ln 2$

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