Question

If \(A^4 + B^4 = 100\), then the greatest possible value of "A" lies between

• A. 0 and 3
• B. 3 and 6
• C. 6 and 9
• D. 9 and 12
• E. 12 and 15

Hint:

When trying to maximize one variable in an equation like \(x^n + y^n = k\) (where n is even), set the other variable's term to its minimum possible value, which is usually 0.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are asked to find the greatest possible value of \(A\) given the equation \(A^4 + B^4 = 100\). To maximize the value of \(A\), we must minimize the value of \(B\).
Step 2: Detailed Explanation:
The term \(B^4\) represents a number raised to an even power. The result of raising any real number (positive, negative, or zero) to an even power is always non-negative (\(\geq 0\)).
To maximize \(A^4\), we must choose the smallest possible value for \(B^4\). The minimum value of \(B^4\) is 0, which occurs when \(B=0\).
Substitute \(B=0\) into the given equation:
\[ A^4 + 0^4 = 100 \]
\[ A^4 = 100 \]
To find the greatest possible value of \(A\), we take the fourth root of 100:
\[ A = \sqrt[4]{100} \]
The fourth root can be calculated as a nested square root:
\[ A = \sqrt{\sqrt{100}} = \sqrt{10} \]
Now we need to estimate the value of \(\sqrt{10}\) to determine which range it lies in.
We know that \(3^2 = 9\) and \(4^2 = 16\).
Since \(9<10<16\), it follows that \(\sqrt{9}<\sqrt{10}<\sqrt{16}\).
Therefore, \(3<\sqrt{10}<4\).
Step 3: Final Answer:
The greatest possible value of \(A\) is \(\sqrt{10}\), which is approximately 3.16. This value lies between 3 and 6. This corresponds to option (B).