JEE 2023 — Mathematics PYQ

Is equation ko solve karne ke liye hum substitution method ka use karenge.

Step 1: Equation ko simplify karna

Maana $e^x=t$. Kyunki $e^x$ hamesha positive hota hai, isliye t > 0.

Ab equation aise dikhegi:

Kyunki $t\ne 0$, hum poori equation ko $t^2$ se divide kar sakte hain:

Ab terms ko rearrange karte hain:

Step 2: Substitution ka use

Maana $y=t-\frac{1}{t}$.

Dono sides square karne par:

Iska matlab, $t^2+\frac{1}{t^2}=y^2+2$.

Ab in values ko original equation mein rakhte hain:

Step 3: Quadratic equation solve karna

Is quadratic equation ko factorize karte hain:

Toh $y=-3$ ya $y=-5$.

Step 4: $x$ ki value nikalna

Ab wapas $t-\frac{1}{t}=y$ mein values rakhte hain:

Case 1: $y=-3$

Quadratic formula se: $t=\frac{-3\pm \sqrt{3^2-4(1)(-1)}}{2}=\frac{-3\pm \sqrt{13}}{2}$

Kyunki t = e^x > 0, hum sirf positive value lenge:

Yahan 0 < t < 1, isliye $x=\ln (t)$ ek negative value hogi.

Case 2: $y=-5$

Quadratic formula se: $t=\frac{-5\pm \sqrt{5^2-4(1)(-1)}}{2}=\frac{-5\pm \sqrt{29}}{2}$

Sirf positive value lene par:

Yahan bhi 0 < t < 1, isliye $x=\ln (t)$ ek dusri negative value hogi.

Conclusion

Hume $x$ ki do (2) solutions mili hain aur dono hi negative hain.

Sahi Option: (3) two solutions and both are negative.