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# Gate 2016 EE Set -2 - Question 33

Using the formula, $\int_{-\infty }^{\infty }f_{x}(x)=1$ and applying it to the equation given in the question, we now have, $\int_{-\infty }^{0}ae^{4x}dx+\int_{0}^{\infty }\frac{3}{2}e^{-3x}dx=1$

$\left [ \frac{ae^{4x}}{4} \right ]_{-\infty }^{0}+\left [ \frac{3e^{-3x}}{2(-3)} \right ]_{0}^{\infty }=1$

$\Rightarrow \frac{a}{4}+\frac{3}{6}=1\Rightarrow a=2$

$P\left [ X\leq 0 \right ]=\int_{-\infty }^{0}2e^{4x}dx=\left [ \frac{2e^{4x}}{4} \right ]_{-\infty }^{0}=\frac{1}{2}$

$\Rightarrow 2,\frac{1}{2}$

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