JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.
Problem
If the domain of the function $\log_5(18x – x^2 – 77)$ is $(\alpha, \beta)$ and the domain of the function $\log_{(x-1)}\left(\dfrac{2x^2+3x-2}{x^2-3x-4}\right)$ is $(\gamma, \delta)$, then $\alpha^2+\beta^2+\gamma^2$ is equal to:
Key insight. A logarithm with a variable base, like $\log_{(x-1)}(\ldots)$, quietly carries two extra conditions beyond “argument positive” — the base itself must be positive, and it can’t equal 1. Missing either one is the most common way to get this kind of domain question wrong.
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Approach
Handle the two functions separately. The first is a straightforward “argument must be positive” quadratic inequality. The second needs three simultaneous conditions — base positive, base $\neq 1$, and argument positive — and the answer is whatever $x$-values satisfy all three at once.
Solution
Step 1 — Domain of $\log_5(18x-x^2-77)$
We need $18x – x^2 – 77 > 0$. Multiplying by $-1$ flips the inequality:
$$x^2 – 18x + 77 < 0 \implies (x-7)(x-11) < 0 \implies 7 < x < 11$$
So $(\alpha, \beta) = (7, 11)$.
Step 2 — Base conditions for $\log_{(x-1)}(\ldots)$
The base $x – 1$ must be positive and not equal to 1:
$$x – 1 > 0 \implies x > 1, \qquad x – 1 \neq 1 \implies x \neq 2$$
Step 3 — Argument condition
We need $\dfrac{2x^2+3x-2}{x^2-3x-4} > 0$. Factorising both:
$$2x^2+3x-2 = (2x-1)(x+2), \qquad x^2-3x-4 = (x-4)(x+1)$$
$$\frac{(2x-1)(x+2)}{(x-4)(x+1)} > 0$$
The critical points, in order, are $-2, -1, \tfrac{1}{2}, 4$. Testing the sign in each interval (wavy-curve method) shows the expression is positive on:
$$x < -2, \qquad -1 < x < \tfrac{1}{2}, \qquad x > 4$$
Step 4 — Combine with the base conditions
We also need $x > 1$ and $x \neq 2$. Intersecting this with the positive-argument regions from Step 3, only $x > 4$ survives (the other two positive regions lie entirely below $x=1$):
$$(\gamma, \delta) = (4, \infty)$$
So $\gamma = 4$.
Step 5 — Compute the final sum
$$\alpha^2 + \beta^2 + \gamma^2 = 7^2 + 11^2 + 4^2 = 49 + 121 + 16 = 186$$
Answer
$$186$$
Common mistakes
- Forgetting the base $\neq 1$ condition. It’s easy to check “base positive” and move straight to the argument, silently dropping $x \neq 2$ — which happens to fall inside the surviving domain here and would otherwise slip through unnoticed.
- Sign errors in the wavy-curve method from mismatching which side of a repeated or nearby root is positive. Testing one convenient point in each interval (rather than trusting memory of the pattern) avoids this.
Practise next
- Find the domain of $\log_{(x-2)}\left(\dfrac{x^2-5x+6}{x^2-1}\right)$, applying the same three-condition check (base positive, base ≠ 1, argument positive).
Show answer
$(3,\infty)$. Three conditions, all of which must hold.
Base positive: $x-2>0$, so $x>2$. Base not $1$: $x\neq3$. Argument positive: $\dfrac{(x-2)(x-3)}{(x-1)(x+1)}>0$.
Once $x>2$, the factors $x-2$, $x-1$ and $x+1$ are all positive, so the sign of the argument is the sign of $x-3$ — forcing $x>3$. That also disposes of $x\neq3$.
So the domain is $x>3$. The base condition is the one most often forgotten, and here it is what rules out everything to the left of $2$ before the argument is even considered.
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