Q36 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
The value of $\lim _{x \rightarrow 0} \frac{\log _{e}\left(\sec (e x) \cdot \sec \left(e^{2} x\right) \cdot \ldots \cdot \sec \left(e^{10} x\right)\right)}{e^{2}-e^{2 \cos x}}$ is equal to
- Correct answerA$\frac{\left(e^{20}-1\right)}{2\left(e^{2}-1\right)}$
- B$\frac{\left(e^{10}-1\right)}{2\left(e^{2}-1\right)}$
- C$\frac{\left(e^{10}-1\right)}{2 e^{2}\left(e^{2}-1\right)}$
- D$\frac{\left(e^{20}-1\right)}{2 e^{2}\left(e^{2}-1\right)}$
Solution
Numerator: $\log_e(\sec(ex)\sec(e^2x)\cdots\sec(e^{10}x)) = \sum_{k=1}^{10} \log_e(\sec(e^kx))$.
Using $\sec(u) \approx 1 + \frac{u^2}{2}$ for small $u$, we get $\log_e(\sec(e^kx)) \approx \frac{e^{2k}x^2}{2}$.
Thus the numerator is $\approx \frac{x^2}{2}\sum_{k=1}^{10} e^{2k} = \frac{x^2 e^2(e^{20}-1)}{2(e^2-1)}$.
Denominator: Using $\cos x \approx 1 - \frac{x^2}{2}$, we have $e^{2\cos x} \approx e^2(1-x^2)$, so $e^2 - e^{2\cos x} \approx e^2 x^2$.
Therefore: $\lim_{x \to 0} = \frac{\frac{x^2 e^2(e^{20}-1)}{2(e^2-1)}}{e^2 x^2} = \frac{e^{20}-1}{2(e^2-1)}$.
Using $\sec(u) \approx 1 + \frac{u^2}{2}$ for small $u$, we get $\log_e(\sec(e^kx)) \approx \frac{e^{2k}x^2}{2}$.
Thus the numerator is $\approx \frac{x^2}{2}\sum_{k=1}^{10} e^{2k} = \frac{x^2 e^2(e^{20}-1)}{2(e^2-1)}$.
Denominator: Using $\cos x \approx 1 - \frac{x^2}{2}$, we have $e^{2\cos x} \approx e^2(1-x^2)$, so $e^2 - e^{2\cos x} \approx e^2 x^2$.
Therefore: $\lim_{x \to 0} = \frac{\frac{x^2 e^2(e^{20}-1)}{2(e^2-1)}}{e^2 x^2} = \frac{e^{20}-1}{2(e^2-1)}$.
Hence, the correct answer is option: 'A'.