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MCQMediumJEE Main 2026 · 28 January, Shift 1Limits

Mathematics Question from JEE Main 2026 · 28 January, Shift 1

The value of lim⁡x→0log⁡e ⁣(sec⁡(ex)⋅sec⁡(e2x)⋯sec⁡(e10x))e2−e2cos⁡x\lim_{x\to 0}\frac{\log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big)} {e^2-e^{2\cos x}} is equal to:

  • A

    e10−12e2(e2−1)\dfrac{e^{10}-1}{2e^2(e^2-1)}

  • B

    e20−12e2(e2−1)\dfrac{e^{20}-1}{2e^2(e^2-1)}

  • C

    e10−12(e2−1)\dfrac{e^{10}-1}{2(e^2-1)}

  • D

    e20−12(e2−1)\dfrac{e^{20}-1}{2(e^2-1)}

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