A function F is defined on the set R, of real numbers by \(f : x \to px^{2} + qx + 2\), where p and q are constants. If \(f(-2) = 0\) and \(f(1) = 3\), find \(f(-4)\).
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Correct Answer: Option n
Explanation:
\(f(x) = px^{2} + qx + 2\)
\(f(-2) = p(-2)^{2} + q(-2) + 2 = 0\)
\(4p - 2q + 2 = 0\)
\(\implies 4p - 2q = -2 ... (1)\)
\(f(1) = p(1)^{2} + q(1) + 2 = 3\)
\(p + q + 2 = 3\)
\(\implies p + q = 1 ... (2)\)
From (2), \(q = 1 - p\). Substitute for q in (1).
\(4p - 2(1 - p) = -2 \implies 4p - 2 + 2p = -2\)
\(6p = 0 \implies p = 0\)
\(q = 1 - p = 1 - 0 = 1\)
\(\therefore f(x) = x + 2\)
\(f(-4) = -4 + 2 = -2\)
\(f(x) = px^{2} + qx + 2\)
\(f(-2) = p(-2)^{2} + q(-2) + 2 = 0\)
\(4p - 2q + 2 = 0\)
\(\implies 4p - 2q = -2 ... (1)\)
\(f(1) = p(1)^{2} + q(1) + 2 = 3\)
\(p + q + 2 = 3\)
\(\implies p + q = 1 ... (2)\)
From (2), \(q = 1 - p\). Substitute for q in (1).
\(4p - 2(1 - p) = -2 \implies 4p - 2 + 2p = -2\)
\(6p = 0 \implies p = 0\)
\(q = 1 - p = 1 - 0 = 1\)
\(\therefore f(x) = x + 2\)
\(f(-4) = -4 + 2 = -2\)