JEE Main 2025 Jan 24 Shift 1, Mathematics Q3: Introduction To Limits

Start with the given relation and form a second equation by replacing $x$ with $1/x$:

(i) $f(x)-6f(1/x)=\frac{35}{3x}-\frac{5}{2}$

(ii) $f(1/x)-6f(x)=\frac{35x}{3}-\frac{5}{2}$

Multiply (ii) by 6 and add to (i) to eliminate $f(1/x)$:

$-35f(x)=\frac{35}{3x}-\frac{5}{2}+6\left(\frac{35x}{3}-\frac{5}{2}\right)=\frac{35}{3x}+70x-\frac{35}{2}$

$f(x)=-\frac{1}{3x}-2x+\frac{1}{2}$

Now evaluate the limit. The divergence at $x\to 0$ comes from $-\frac{1}{3x}$, so $\frac{1}{\alpha x}$ must cancel it:

$\frac{1}{\alpha x}+f(x)=\left(\frac{1}{\alpha x}-\frac{1}{3x}\right)-2x+\frac{1}{2}$

The bracket vanishes only when $\alpha =3$. The remaining limit then gives $\beta =\frac{1}{2}$.

$\alpha +2\beta =3+1=4$.

$\therefore$ Option (3).