JEE 2023 — Mathematics PYQ

1. Left Hand Limit (LHL) nikalna ($x\to {\frac{\pi }{2}}^{-}$):

Jab $x\to {\frac{\pi }{2}}^{-}$, tab $\cos x\to 0^{+}$. Yeh limit $1^{\infty }$ form ki hai.

Hume pata hai ${\lim }_{z\to 0}(1+z{)}^{1/z}=e$, isliye:

2. Right Hand Limit (RHL) nikalna ($x\to {\frac{\pi }{2}}^{+}$):

Power ko simplify karte hain: ${\lim }_{x\to {\frac{\pi }{2}}^{+}}\frac{\cot 6x}{\cot 4x}$. Yeh $\frac{0}{0}$ form hai kyunki $\cot (3\pi )=0$ aur $\cot (2\pi )$ undefined hota hai? Nahi, dhyan dein: $x=\frac{\pi }{2}+h$ rakhne par:

Small angle approximation ($\tan \theta \approx \theta$) se:

Isliye, $RHL=e^{2/3}$.

3. Continuity ki condition apply karna:

Continuity ke liye $LHL=RHL=f(\frac{\pi }{2})$ hona chahiye.

• $e^{\lambda }=e^{2/3}⟹\lambda =\frac{2}{3}$

• $\mu =e^{2/3}$

4. Final expression ki value nikalna:

Hume $9\lambda +6{\log }_e\mu +{\mu }^6-e^{6\lambda }$ find karna hai:

• $9\lambda =9\times \frac{2}{3}=6$

• $6{\log }_e\mu =6{\log }_e(e^{2/3})=6\times \frac{2}{3}=4$

• ${\mu }^6=(e^{2/3}{)}^6=e^4$

• $e^{6\lambda }=e^{6\times \frac{2}{3}}=e^4$

Ab values rakhte hain:

Correct Option: (1)