If f(x) + 6f(1/x) = 35/(3x) − 7/2 for x ∈ R − {0}, and the Limit of (1/(αx) + f(x)) as x → 0 Exists Finitely and Is Equal to β, Then α − 2β Is

JEE Main 2025 — Previous Year Question.

Problem

If $f(x)$ satisfies the functional equation $f(x)+6f\left(\dfrac1x\right)=\dfrac{35}{3x}-\dfrac72$, $x\in\mathbb R-\{0\}$, and if $\displaystyle\lim_{x\to0}\left(\frac1{\alpha x}+f(x)\right)$ exists finitely and is equal to $\beta$, then $\alpha-2\beta$ is:

Key insight. A single functional equation with both $f(x)$ and $f(1/x)$ in it can’t be solved on its own — but substituting $x\to\frac1x$ produces a second equation, and the two together eliminate $f(1/x)$ and pin down $f(x)$ explicitly.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Substitute $x\to\dfrac1x$ into the given equation to get a second relation between $f(x)$ and $f\left(\dfrac1x\right)$. Eliminate $f\left(\dfrac1x\right)$ between the two equations to solve for $f(x)$ explicitly. With $f(x)$ known, examine the limit expression term by term: since it involves a $\dfrac1x$ singularity, the limit can only be finite if that singular part cancels — this condition fixes $\alpha$, after which the finite remainder gives $\beta$ directly.

Solution

Step 1 — Write the given equation

$$f(x)+6f\left(\frac1x\right) = \frac{35}{3x}-\frac72 \quad \text{…(1)}$$

Step 2 — Substitute x → 1/x to get a second equation

Replacing every $x$ with $\dfrac1x$:

$$f\left(\frac1x\right)+6f(x) = \frac{35}{3\cdot\frac1x}-\frac72 = \frac{35x}3-\frac72 \quad \text{…(2)}$$

Step 3 — Eliminate f(1/x)

To isolate $f(x)$, multiply equation (2) by 6 — so its $f\left(\frac1x\right)$ term becomes $6f\left(\frac1x\right)$, matching equation (1)’s own $f\left(\frac1x\right)$ term — then subtract equation (1):

$$6\times\text{(2)}: \quad 6f\left(\frac1x\right)+36f(x) = 6\left(\frac{35x}3-\frac72\right) = 70x-21$$

$$\text{(1)}: \quad f(x)+6f\left(\frac1x\right) = \frac{35}{3x}-\frac72$$

Subtracting equation (1) from $6\times$equation (2) cancels the matching $6f\left(\frac1x\right)$ terms on each side:

$$36f(x)-f(x) = (70x-21)-\left(\frac{35}{3x}-\frac72\right)$$

$$35f(x) = 70x-21-\frac{35}{3x}+\frac72$$

Step 4 — Simplify to find f(x)

$$35f(x) = 70x-\frac{35}{3x}-21+\frac72 = 70x-\frac{35}{3x}-\frac{42}2+\frac72 = 70x-\frac{35}{3x}-\frac{35}2$$

Dividing through by 35:

$$f(x) = 2x-\frac1{3x}-\frac12$$

Step 5 — Set up the limit condition

$$\lim_{x\to0}\left(\frac1{\alpha x}+f(x)\right) = \lim_{x\to0}\left(\frac1{\alpha x}+2x-\frac1{3x}-\frac12\right) = \lim_{x\to0}\left[\left(\frac1\alpha-\frac13\right)\frac1x+2x-\frac12\right]$$

Step 6 — Force the singular term to vanish

As $x\to0$, the term $\left(\dfrac1\alpha-\dfrac13\right)\dfrac1x$ blows up unless its coefficient is zero. For the limit to exist finitely:

$$\frac1\alpha-\frac13=0 \implies \alpha=3$$

Step 7 — Evaluate the remaining limit

With the singular term gone:

$$\beta = \lim_{x\to0}\left(2x-\frac12\right) = 2(0)-\frac12 = -\frac12$$

Step 8 — Compute α − 2β

$$\alpha-2\beta = 3-2\left(-\frac12\right) = 3+1 = 4$$

Answer

Common mistakes

• Substituting $x\to\dfrac1x$ incorrectly in the $\dfrac{35}{3x}$ term. This becomes $\dfrac{35}{3\cdot\frac1x}=\dfrac{35x}3$, not $\dfrac{35}{3x^2}$ or $\dfrac{35}{3}x$ misplaced — it’s worth writing this substitution out fully rather than doing it mentally.
• Eliminating the wrong function. Multiplying equation (1) by 6 and subtracting eliminates $f(x)$, leaving $f(1/x)$ — to solve for $f(x)$ itself, equation (2) must be multiplied by 6 and equation (1) subtracted from it instead, as in Step 3.
• Forgetting that a finite limit forces the 1/x coefficient to be exactly zero, not just small. Any nonzero coefficient on the $\dfrac1x$ term makes the limit diverge to $\pm\infty$, regardless of how small that coefficient is — there is only one value of $\alpha$ that works, not a range.

Practise next

• If $f(x)+6f\left(\dfrac1x\right)=\dfrac{70}{3x}-7$, and $\displaystyle\lim_{x\to0}\left(\dfrac1{\alpha x}+f(x)\right)=\beta$ exists finitely, find $\alpha-2\beta$.