# 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, $$\int_{-\infty }^{0}ae^{4x}dx+\int_{0}^{\infty }\frac{3}{2}e^{-3x}dx=1$$

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

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

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

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