JEE Mathematics 2026 Question with Solution

Given: $f(3)=18$, $f'(3)=0$, $f''(3)=4$.

Find: $\lim_{x \to 1} \log_e \left[ \frac{f(2+x)}{f(3)} \right]^{\frac{18}{(x-1)^2}}$.

Use the logarithm property first:

$$L = \lim_{x \to 1} \frac{18}{(x-1)^2} \log_e \left[ \frac{f(2+x)}{f(3)} \right]$$

Then

$$L = \lim_{x \to 1} \frac{18\log_e(f(2+x)) - 18\log_e(f(3))}{(x-1)^2}$$

As $x \to 1$, this becomes the indeterminate form $\frac{0}{0}$, so apply L'Hopital's Rule.

After the first differentiation,

$$L = \lim_{x \to 1} \frac{\frac{18f'(2+x)}{f(2+x)}}{2(x-1)}$$

Again, substituting $x=1$ gives $\frac{0}{0}$ because $f'(3)=0$. Apply L'Hopital's Rule once more.

After the second differentiation,

$$L = \lim_{x \to 1} \frac{18 \frac{f''(2+x)f(2+x) - (f'(2+x))^2}{(f(2+x))^2}}{2}$$

Now substitute $x=1$:

$$L = \frac{18}{2} \cdot \frac{f''(3)f(3) - (f'(3))^2}{(f(3))^2}$$

Using the given values,

$$L = 9 \cdot \frac{4 \times 18 - 0^2}{18^2}$$ $$L = 9 \cdot \frac{72}{324} = 9 \cdot \frac{2}{9} = 2$$

Therefore, the correct option is A.

Given: $f(3)=18$, $f'(3)=0$, $f''(3)=4$.

Find: the given limit using expansion near $x=1$.

Let $h = x-1$ so that $h \to 0$ and $2+x = 3+h$. By Taylor expansion around $3$,

$$f(3+h) \approx f(3) + f'(3)h + \frac{f''(3)}{2}h^2$$

Substituting the given values,

$$f(3+h) \approx 18 + 0 \cdot h + \frac{4}{2}h^2 = 18 + 2h^2$$

Hence,

$$\frac{f(3+h)}{f(3)} = \frac{18+2h^2}{18} = 1 + \frac{h^2}{9}$$

So the limit becomes

$$L = \lim_{h \to 0} \frac{18}{h^2} \log_e\left(1 + \frac{h^2}{9}\right)$$

Now use the standard limit $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$ with $u = \frac{h^2}{9}$:

$$L = \lim_{h \to 0} \frac{18}{h^2} \left(\frac{\log_e(1+h^2/9)}{h^2/9}\right) \cdot \frac{h^2}{9}$$ $$L = \frac{18}{9} = 2$$

Therefore, the value of the limit is $2$, so the correct option is A.