Question

If $f(x,y) = 2x^2 + 3xy- 3y^2+3$ , then what is the value of f(f(3, 2), f(–2, –1))?

• A. 1257
• B. 1527
• C. 1787
• D. 1997
• E. 2007

Answer 1

Answer: (E)

To find the value of \(f(f(3, 2), f(-2, -1))\), we need to calculate the function values \(f(3,2)\) and \(f(-2,-1)\) first, and then use these results to find the final answer.

Step 1: Calculate \(f(3, 2)\)

The function is given by \(f(x,y) = 2x^2 + 3xy - 3y^2 + 3\). 

Substitute \(x = 3\) and \(y = 2\) into the function:

\(f(3, 2) = 2(3)^2 + 3(3)(2) - 3(2)^2 + 3\)
\(= 2 \times 9 + 3 \times 6 - 3 \times 4 + 3\)
\(= 18 + 18 - 12 + 3\)
\(= 27\)

Step 2: Calculate \(f(-2, -1)\)

Substitute \(x = -2\) and \(y = -1\) into the function:

\(f(-2, -1) = 2(-2)^2 + 3(-2)(-1) - 3(-1)^2 + 3\)
\(= 2 \times 4 + 3 \times 2 - 3 \times 1 + 3\)
\(= 8 + 6 - 3 + 3\)
\(= 14\)

Step 3: Calculate \(f(f(3, 2), f(-2, -1)) = f(27, 14)\)

Finally, substitute \(x = 27\) and \(y = 14\) into the function:

\(f(27, 14) = 2(27)^2 + 3(27)(14) - 3(14)^2 + 3\)
\(= 2 \times 729 + 3 \times 378 - 3 \times 196 + 3\)
\(= 1458 + 1134 - 588 + 3\)
\(= 2007\)

Thus, the value of \(f(f(3, 2), f(-2, -1))\) is 2007.

Therefore, the correct option is 2007.