Question

If x is a positive integer satisfying \(64 ≤ x ≤ 121\) and \(y = \frac{x^2 + 4\sqrt{x}(x + 16) + 256}{x + 8\sqrt{x} + 16}\) then which of the following is satisfied by y?

• A. \(39 ≤ y ≤ 68\)
• B. \(38 < y < 94\)
• C. \(40 < y ≤ 68\)
• D. \( 42 ≤ y ≤ 78\)
• E. \( 52 ≤ y ≤ 88\)

Answer 1

The Correct Option is B

Step 1: Understand the problem.
We are given that \( x \) is a positive integer satisfying \( 64 \leq x \leq 121 \), and we are asked to find the range for \( y \), where \( y \) is defined by the expression:
\( y = \frac{x^2 + 4\sqrt{x}(x + 16) + 256}{x + 8\sqrt{x} + 16} \)

Step 2: Simplify the expression for \( y \).
The given expression for \( y \) is:
\( y = \frac{x^2 + 4\sqrt{x}(x + 16) + 256}{x + 8\sqrt{x} + 16} \)
Let's try to simplify it. Begin by observing that both the numerator and denominator contain terms involving \( x \) and \( \sqrt{x} \). The goal is to see if we can recognize a pattern or simplify further.

Let's try substituting some values of \( x \) within the given range \( 64 \leq x \leq 121 \) to find a general pattern for \( y \).

Step 3: Calculate \( y \) for specific values of \( x \).
- For \( x = 64 \):
\( y = \frac{64^2 + 4\sqrt{64}(64 + 16) + 256}{64 + 8\sqrt{64} + 16} = \frac{4096 + 4(8)(80) + 256}{64 + 64 + 16} = \frac{4096 + 2560 + 256}{144} = \frac{6912}{144} = 48 \)
- For \( x = 81 \):
\( y = \frac{81^2 + 4\sqrt{81}(81 + 16) + 256}{81 + 8\sqrt{81} + 16} = \frac{6561 + 4(9)(97) + 256}{81 + 72 + 16} = \frac{6561 + 3492 + 256}{169} = \frac{10309}{169} ≈ 61 \)
- For \( x = 100 \):
\( y = \frac{100^2 + 4\sqrt{100}(100 + 16) + 256}{100 + 8\sqrt{100} + 16} = \frac{10000 + 4(10)(116) + 256}{100 + 80 + 16} = \frac{10000 + 4640 + 256}{196} = \frac{14896}{196} ≈ 76 \)
- For \( x = 121 \):
\( y = \frac{121^2 + 4\sqrt{121}(121 + 16) + 256}{121 + 8\sqrt{121} + 16} = \frac{14641 + 4(11)(137) + 256}{121 + 88 + 16} = \frac{14641 + 6032 + 256}{225} = \frac{20929}{225} ≈ 93 \)

Step 4: Conclusion.
From the calculations, we observe that for values of \( x \) within the range \( 64 \leq x \leq 121 \), the value of \( y \) lies between 48 and 93.
Therefore, the range for \( y \) is approximately:
\( 38 < y < 94 \)

Final Answer:
The correct option is (B): \( 38 < y < 94 \).