JEE Mathematics 2026 Question with Solution

Given:

$$\lim_{x\to 0}\frac{\log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big)} {e^2-e^{2\cos x}}$$

Find: The value of the limit and the correct option.

Use logarithm to convert the product into a sum and then apply small-angle expansions.

$$\log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big)=\sum_{k=1}^{10}\log(\sec(e^k x))$$

For small $t$,

$$\sec t = 1+\frac{t^2}{2}+o(t^2), \qquad \log(\sec t)=\frac{t^2}{2}+o(t^2)$$

Therefore,

$$\log(\sec(e^k x)) \sim \frac{(e^k x)^2}{2}$$

So the numerator becomes

$$\text{Numerator} \sim \frac{x^2}{2}\sum_{k=1}^{10} e^{2k}$$

Now evaluate the geometric sum:

$$\sum_{k=1}^{10} e^{2k}=e^2\frac{e^{20}-1}{e^2-1}$$

Hence,

$$\text{Numerator} \sim \frac{x^2}{2}\, e^2\frac{e^{20}-1}{e^2-1}$$

For the denominator, using

$$\cos x = 1-\frac{x^2}{2}+o(x^2)$$

we get

$$2\cos x = 2-x^2+o(x^2)$$

Therefore,

$$e^{2\cos x} = e^{2-x^2+o(x^2)} = e^2(1-x^2+o(x^2))$$

So,

$$e^2-e^{2\cos x} \sim e^2x^2$$

Now substitute into the limit:

$$\lim_{x\to 0} \frac{\frac{x^2}{2}\, e^2\frac{e^{20}-1}{e^2-1}}{e^2x^2} = \frac{e^{20}-1}{2(e^2-1)}$$

Therefore, the value of the limit is $\dfrac{e^{20}-1}{2(e^2-1)}$, so the correct option is D.

Given: The numerator contains a product of secant terms inside logarithm.

Find: The asymptotic forms of numerator and denominator up to order $x^2$.

Because logarithm converts products into sums, write

$$\log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big) = \log(\sec(ex)) + \log(\sec(e^2x)) + \cdots + \log(\sec(e^{10}x))$$

Each term contributes only up to order $x^2$:

$$\log(\sec(e^k x)) = \frac{e^{2k}x^2}{2}+o(x^2)$$

Adding all terms,

$$\log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big) = \frac{x^2}{2}\sum_{k=1}^{10} e^{2k}+o(x^2)$$

Now,

$$\sum_{k=1}^{10} e^{2k}=e^2\frac{e^{20}-1}{e^2-1}$$

Thus the numerator is

$$\frac{x^2}{2}\,e^2\frac{e^{20}-1}{e^2-1}+o(x^2)$$

For the denominator,

$$\cos x = 1-\frac{x^2}{2}+o(x^2)$$

so

$$2\cos x = 2-x^2+o(x^2)$$

and then

$$e^{2\cos x}=e^{2-x^2+o(x^2)}=e^2\big(1-x^2+o(x^2)\big)$$

Hence,

$$e^2-e^{2\cos x}=e^2x^2+o(x^2)$$

Dividing the leading terms gives

$$\frac{\frac{x^2}{2}\,e^2\frac{e^{20}-1}{e^2-1}}{e^2x^2} = \frac{e^{20}-1}{2(e^2-1)}$$

Therefore, the correct option is D.