If \( f(x) = ax^2 + bx + c \) is an even function and \( g(x) = px^3 + qx^2 + rx \) is an odd function, and if \( h(x) = f(x) + g(x) \) and \( h(-2) = 0 \), then \( 8p + 4q + 2r = \)?
Show Hint
- \( 4a + 3b + 2c \)
- \( a + b + c \)
- \( 4a + 2b + c \)
\( 8a + 4b + 2c \)
The Correct Option is C
Solution and Explanation
Since \(f(x)\) is an even function, it satisfies \(f(-x) = f(x)\). For \(g(x)\), being an odd function, it satisfies \(g(-x) = -g(x)\). Given \(h(-2) = 0\), substituting \(-2\) into the function \(h(x)\), we have: \[ h(-2) = f(-2) + g(-2) = f(2) - g(2) = 0 \] This implies \(f(2) = g(2)\). To express \(f(2)\) and \(g(2)\) in terms of their coefficients, we use: \[ f(2) = 4a + 2b + c, \quad g(2) = 8p + 4q + 2r \] Given \(f(2) = g(2)\), it follows that: \[ 4a + 2b + c = 8p + 4q + 2r \] Therefore, \(8p + 4q + 2r\) equals \(4a + 2b + c\), matching the correct answer option.
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