Question

The number of distinct integers $n$ for which $\log_{\left(\frac14\right)}(n^2 - 7n + 11)>0$ is:

• A. \(1\)
• B. \(2\)
• C. \(20\)
• D. infinite

Hint:

Complete the square on the quadratic first -- since n is an integer, the distance from the vertex is always a half-integer, so you only need to check which of those few possible squared-distances fits the range the log condition demands.

Answer 1

The Correct Option is B

We must solve: \[ \log_{\left(\frac14\right)}(n^2 - 7n + 11)>0. \] Two conditions must hold: --- Step 1: Domain condition The logarithm’s argument must be positive: \[ n^2 - 7n + 11>0. \] Solve the quadratic inequality. Roots: \[ n=\frac{7\pm\sqrt{49-44}}{2} = \frac{7\pm\sqrt{5}}{2} \approx 2.38,\; 4.62. \] Since the parabola opens upward: \[ n<\frac{7-\sqrt5}{2} \quad \text{or} \quad n>\frac{7+\sqrt5}{2}. \] Thus the integer values allowed by the domain are: \[ \{\ldots,0,1,2\} \cup \{5,6,7,\ldots\}. \] ---
Step 2: Inequality from the logarithm Since the base \(\frac14\) is less than 1, the inequality reverses when removing the log: \[ \log_{\left(\frac14\right)}(A)>0 \quad\Longleftrightarrow\quad A<1. \] Thus: \[ n^2 - 7n + 11<1, \] \[ n^2 - 7n + 10<0, \] \[ (n - 2)(n - 5)<0. \] Thus: \[ 2<n<5, \] with integer candidates: \[ n = 3,4. \] --- 
Step 3: Combine with domain condition We test whether \(n=3,4\) satisfy the \emph{positivity} of the argument: \[ f(n) = n^2 - 7n + 11. \] \[ f(3) = 9 - 21 + 11 = -1\quad (\text{invalid}). \] \[ f(4) = 16 - 28 + 11 = -1\quad (\text{invalid}). \] Thus no integer satisfies both conditions if the inequality is strict \(>0\). --- Interpretation used in the answer key The official answer key indicates that the intended inequality was effectively: \[ \log_{\left(\frac14\right)}(n^2 - 7n + 11) \ge 0. \] This gives: \[ n^2 - 7n + 11 \le 1, \] \[ n^2 - 7n + 10 \le 0, \] \[ (n-2)(n-5) \le 0, \] \[ 2 \le n \le 5. \] Testing domain condition: - \(n=2: f(2) = 1>0 \Rightarrow \log = 0\) (valid) - \(n=3: f(3) = -1\) invalid - \(n=4: f(4) = -1\) invalid - \(n=5: f(5) = 1>0 \Rightarrow \log = 0\) (valid) Valid integers: \[ n = 2, 5. \] Thus there are exactly: \[ \boxed{2} \] solutions.

Answer 2

Approach: Rather than solving the inequality symbolically, just evaluate \(f(n) = n^2 - 7n + 11\) at integers near its minimum and read off which ones qualify. This sidesteps solving two separate quadratic inequalities.

Step 1: Tabulate \(f(n)\).
\(f(0)=11,\ f(1)=5,\ f(2)=1,\ f(3)=-1,\ f(4)=-1,\ f(5)=1,\ f(6)=5,\ f(7)=11.\)
For \(n \le -1\) or \(n \ge 8\), \(f(n)\) only grows larger since the parabola opens upward, so no new candidates appear there.

Step 2: Apply the domain condition. The logarithm needs \(f(n) > 0\), which rules out \(n = 3, 4\) (both give \(f(n) = -1\)).

Step 3: Apply the value condition. We also need \(f(n) \le 1\) (the reading consistent with the answer key), which is satisfied only at \(n = 2\) and \(n = 5\) (both give \(f(n) = 1\)); every other domain-valid \(n\) gives \(f(n) \ge 5\), too large.

Note: Taken as a strict inequality \(f(n) < 1\), no integer works at all, since \(n=3,4\) fail the domain and no other integer gets \(f(n)\) below 1. The answer key's count of 2 matches the non-strict reading \(f(n) \le 1\).

Two integers, \(n = 2\) and \(n = 5\), option (2).

Answer 3

Concept:

• Writing the quadratic in completed-square (vertex) form shows that for any integer $n$, the expression $n-3.5$ is always a half-integer ($\pm0.5,\pm1.5,\pm2.5,\dots$) -- so instead of solving inequalities separately, just check which half-integer distances from the vertex land inside the required range.

Step 1: Complete the square for $f(n)=n^2-7n+11$.
$f(n) = \left(n-\dfrac{7}{2}\right)^2 - \dfrac{49}{4}+11 = (n-3.5)^2 - 1.25$.

Step 2: Translate the log condition into a range on $f(n)$.
Since the base $\frac14$ is less than $1$, $\log_{\frac14}(f(n)) \ge 0$ (the reading matching the answer key) means $0 < f(n) \le 1$.

Step 3: Convert that range into a range on $(n-3.5)^2$.
$0 < (n-3.5)^2 - 1.25 \le 1 \;\Rightarrow\; 1.25 < (n-3.5)^2 \le 2.25$.

Step 4: Use the half-integer property.
For integer $n$, $n-3.5$ is always $\pm0.5,\pm1.5,\pm2.5,\dots$, so $(n-3.5)^2$ can only be $0.25, 2.25, 6.25,\dots$. The only one of these values that falls inside $(1.25,\,2.25]$ is exactly $2.25$.

Step 5: Solve for $n$.
$(n-3.5)^2 = 2.25 \;\Rightarrow\; n-3.5 = \pm1.5 \;\Rightarrow\; n=5 \text{ or } n=2$.

Final Answer: There are exactly $2$ integers ($n=2$ and $n=5$).