JEE Mathematics 2025 Question with Solution

Given:

and

Find: $\alpha + 2\beta$

Replace $x$ by $\frac{1}{x}$ in the given functional equation:

Now use the two linear equations

Multiply the second equation by $6$ and add to the first:

So,

Hence,

Substitute this into the limit expression:

For the limit as $x \to 0$ to exist, the coefficient of $\frac{1}{x}$ must be zero:

Therefore,

Then the limit becomes

So,

Finally,

Therefore, the correct option is C.

Given: the function satisfies a pair of linear equations in the two unknowns $f(x)$ and $f\left(\frac{1}{x}\right)$.

Find: the values of $\alpha$ and $\beta$ needed to compute $\alpha + 2\beta$.

Write the two equations explicitly:

Multiply the second equation by $6$:

Now add it to the first equation:

Divide by $-35$:

Now evaluate the limit condition:

Substitute $f(x)$:

Group the singular terms:

For this limit to be finite, the coefficient of $\frac{1}{x}$ must vanish.

Thus,

which gives

Then

Hence,

Therefore, the answer is $4$.