CAT 2025 Slot 1 QA Question 21

The question is: "The number of distinct integers $n$ for which $\log_{1/4}(n^2 - 7n + 11) > 0$ is..." and the options are A) 0, B) 1, C) infinite, D) 2.

Hmm, logarithms can sometimes be tricky, especially with different bases. the base here is $1/4 , which is less than 1. I think that affects the inequality direction when we convert the logarithmic inequality to its exponential form. Let me recall the properties of logarithms.

First, the logarithm $\log_b(a) > 0$ depends on the base $b$. If $b > 1 , then$a > 1makes the log positive. But if &#36;0 < b < 1 , then &#36;0 < a < 1 makes the log positive. So in this case, since the base is $1/4 , which is less than 1, the inequality$\log_{1/4}(n^2 - 7n + 11) > 0$implies that the argument$n^2 - 7n + 11$ must be between 0 and 1.

So, I can write that as:

Wait, but $n^2 - 7n + 11$ is a quadratic expression. if it can ever be between 0 and 1. Maybe I should solve the inequalities step by step.

First, let's solve the left part of the inequality: And then the right part:

Starting with the first inequality: $$n^2 - 7n + 11 > 0$ I can find the discriminant to see if this quadratic ever crosses zero. The discriminant $D = b^2 - 4ac = (-7)^2 - 4*1*11 = 49 - 44 = 5$. Since the discriminant is positive, the quadratic has two real roots. Calculate them.

The roots are: and

So the quadratic $n^2 - 7n + 11$ is positive outside the interval (2.382, 4.618). Since we're dealing with integers $n , the values where the quadratic is positive are$n ≤ 2$and$n ≥ 5$.

Now, moving to the second inequality: Subtract 1 from both sides: Which gives: and

So the quadratic $n^2 - 7n + 10$ factors as (n - 2)(n - 5) , and it's less than zero between its roots, i.e., for &#36;2 < n < 5.

Putting both inequalities together, we have: From the first inequality: $n \leq 2$ or $n \geq 5$ From the second inequality: $2 < n < 5$

So the intersection of these two is... Hmm, . The first inequality says $n \leq 2$ or n \geq 5 , and the second says &#36;2 < n < 5. So the overlap is only at $n = 2$ and $n = 5$. But wait, at $n = 2 , the second inequality is$n < 5 , but the first inequality includes $n \leq 2$. So actually, the overlapping region is only at $n = 2$ and $n = 5$. But

Wait, no, because the second inequality is $2 < n < 5 , so it doesn't include 2 or 5. The first inequality includes $n \leq 2$ and $n \geq 5$. So the intersection is only the points where both conditions are satisfied. But since the second inequality is strict (doesn't include 2 or 5), there's no integer $n$ that satisfies both $n \leq 2$ or $n \geq 5$ and $2 < n < 5$. So does that mean there are no integers that satisfy both inequalities?

Wait, that can't be right. Let me think again. The first inequality is $n^2 - 7n + 11 > 0 , which is true for$n ≤ 2$and$n ≥ 5$. The second inequality is$n^2 - 7n + 11 < 1 , which is true for $2 < n < 5$. So the values of $n$ that satisfy both are the integers that are in both regions. But since $n$ must be an integer, let's list the possible integers in each interval.

From the first inequality, the integers are $n \leq 2$ (so n=2,1,0,...) and $n \geq 5$ (n=5,6,7,...). From the second inequality, the integers are n=3,4.

So the overlap is only if there are integers that are both in $n \leq 2$ or $n \geq 5$ and in \( 2