MCQ
The value of $\frac{e^{-\frac{\pi}{4}}+\int \limits_0^{\frac{\pi}{4}} e^{-x} \tan ^{50} x d x}{\int \limits_0^{\frac{\pi}{4}} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}$ is
- ✓$50$
- B$49$
- C$51$
- D$25$
$\left[-e^{-y}(\tan x)^{50}\right]_0^{\pi / 4}+\int \limits_0^{\pi / 4} e^{-x}(50)(\tan x)^{49} \sec ^2 x$
$=-e^{-\pi / 4}+0+50 \int \limits_0^{\pi / 4} e^{-x}(\tan x)^{49}\left(\tan ^2 x+1\right)$
$=-e^{-\pi / 4}+50\left(\int \limits_0^{\pi / 4} e^{-x}(\tan x)^{51}+(\tan x)^{49}\right) d x$
Now,$\frac{-e^{-\pi / 4}+\int \limits_0^{\pi / 4} e^{-x}(\tan x)^{50} d x}{\int \limits_0^{\pi / 4} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}$
$\frac{50 \int \limits_0^{\pi / 4} e^{-x}\left((\tan x)^{51}+(\tan x)^{49}\right) d x}{\int \limits_0^{\pi / 4} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}=50$
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($1$) $P(X>Y)$ is
($A$) $\frac{1}{4}$ ($B$) $\frac{5}{12}$ ($C$) $\frac{1}{2}$ ($D$) $\frac{7}{12}$
($2$) $P(X=Y)$ is
($A$). $\frac{11}{36}$ ($B$) $\frac{1}{3}$ ($C$) $\frac{13}{36}$ ($D$) $\frac{1}{2}$
Given the answer quetion ($1$) and ($2$)