JEE 2023 — Mathematics PYQ

Step 1: Denominator ko simplify karna

Sabse pehle denominator waale integral ko dekhte hain:

$D={\int }_0^{\frac{\pi }{4}}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx$

Hum jaante hain ki $1+{\tan }^{2}x={\sec }^{2}x$, isliye:

$D={\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{49}x(1+{\tan }^{2}x)dx$

$D={\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{49}x{\sec }^{2}xdx$

Step 2: Integration by Parts (IBP) apply karna

Ab hum integral $D$ par IBP ka istemal karenge:

Maana $u=e^{-x}$ aur $dv={\tan }^{49}x{\sec }^{2}xdx$.

• Tab $du=-e^{-x}dx$

• Aur $v=\int {\tan }^{49}x{\sec }^{2}xdx=\frac{{\tan }^{50}x}{50}$

Formula $\int udv=uv-\int vdu$ ka use karne par:

$D={\left[e^{-x}\cdot \frac{{\tan }^{50}x}{50}\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{50}x}{50}(-e^{-x})dx$

Step 3: Limits put karna aur simplify karna

$D=\left(\frac{e^{-\pi /4}{\tan }^{50}(\pi /4)}{50}-\frac{e^0{\tan }^{50}(0)}{50}\right)+\frac{1}{50}{\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx$

Kyunki $\tan (\pi /4)=1$ aur $\tan (0)=0$:

$D=\frac{e^{-\pi /4}}{50}+\frac{1}{50}{\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx$

Puri equation ko $50$ se multiply karne par:

$50\cdot D=e^{-\pi /4}+{\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx$

Step 4: Final Ratio nikalna

Humein sawal mein pucha gaya hai:

$\text{Value}=\frac{e^{-\frac{\pi }{4}}+{\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx}{D}$

Upar waali equation se value substitute karne par:

$\text{Value}=\frac{50\cdot D}{D}=50$

Correct Option: (3) 50