Question:medium

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

Updated On: Nov 25, 2025
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The Correct Option is

Solution and Explanation

The correct answer is option (E):
2007

Let's break this problem down step by step to understand why the answer is 2007.

First, we need to find the value of f(3, 2). To do this, we substitute x = 3 and y = 2 into the function f(x, y) = 2x^2 + 3xy - 3y^2 + 3:

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

Next, we need to find the value of f(-2, -1). We substitute x = -2 and y = -1 into the function f(x, y):

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

Now, we know that f(3, 2) = 27 and f(-2, -1) = 14. We need to find the value of f(f(3, 2), f(-2, -1)), which means we need to find f(27, 14). We substitute x = 27 and y = 14 into the function f(x, y):

f(27, 14) = 2(27)^2 + 3(27)(14) - 3(14)^2 + 3
f(27, 14) = 2(729) + 3(378) - 3(196) + 3
f(27, 14) = 1458 + 1134 - 588 + 3
f(27, 14) = 2592 - 588 + 3
f(27, 14) = 2004 + 3
f(27, 14) = 2007

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