Question:medium

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?

Updated On: Jan 13, 2026
  • \(39 ≤ y ≤ 68\)
  • \(38 < y < 94\)
  • \(40 < y ≤ 68\)
  • \( 42 ≤ y ≤ 78\)
  • \( 52 ≤ y ≤ 88\)
Show Solution

The Correct Option is B

Solution and Explanation

The correct answer is option (B): (38 < y < 94)

Let's simplify the expression for y. We are given that
y = (x^2 + 4 * sqrt(x) * (x + 16) + 256) / (x + 8 * sqrt(x) + 16)

The numerator can be rewritten by letting t = √x. Then
y = (t^4 + 4t^3 + 64t + 256) / (t^2 + 8t + 16).

Factor the numerator:
t^4 + 4t^3 + 64t + 256 = (t + 4)^2 (t^2 - 4t + 16)
and the denominator is (t + 4)^2. Canceling the common factor gives
y = t^2 - 4t + 16 = x - 4√x + 16.

Since √x = t ranges from 8 to 11 when x is between 64 and 121,
y(8) = 48 and y(11) = 93, so
48 ≤ y ≤ 93 (or 48 < y < 93 for open endpoints).

Therefore option (B) is correct.

Final Answer: 38 < y < 94

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