Question:
If \(g(x) = x^2+x-2\) and \(\frac{1}{2}(g\circ f)(x) = 2x^2-5x+2\) then \(f(x)\) is equal to
If \(g(x) = x^2+x-2\) and \(\frac{1}{2}(g\circ f)(x) = 2x^2-5x+2\) then \(f(x)\) is equal to
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Write g(f(x)) = f^2 + f - 2 and compare with the given polynomial.
Updated On: Oct 1, 2026
- \(2x+3\)
- \(2x-3\)
- \(2x^2+3x+2\)
- \(2x^2-3x-2\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the composite:
\(g(f(x))=[f(x)]^2+f(x)-2\). The condition says \(\tfrac12 g(f(x))=2x^2-5x+2\), so \(g(f(x))=4x^2-10x+4\).
Step 2: Form an equation in f:
\(f^2+f-2=4x^2-10x+4\), so \(f^2+f=4x^2-10x+6\).
Step 3: Test the options:
Try \(f=2x-3\): \(f^2=4x^2-12x+9\), so \(f^2+f=4x^2-12x+9+2x-3=4x^2-10x+6\). This matches.
Step 4: Why the other options are wrong.
\(f=2x+3\) gives \(4x^2+14x+12\). The quadratic options for \(f\) give a degree 4 expression, which cannot equal a degree 2 polynomial.
Final Answer:
f(x) = 2x - 3. \[ \boxed{\text{(B) }2x-3} \]
\(g(f(x))=[f(x)]^2+f(x)-2\). The condition says \(\tfrac12 g(f(x))=2x^2-5x+2\), so \(g(f(x))=4x^2-10x+4\).
Step 2: Form an equation in f:
\(f^2+f-2=4x^2-10x+4\), so \(f^2+f=4x^2-10x+6\).
Step 3: Test the options:
Try \(f=2x-3\): \(f^2=4x^2-12x+9\), so \(f^2+f=4x^2-12x+9+2x-3=4x^2-10x+6\). This matches.
Step 4: Why the other options are wrong.
\(f=2x+3\) gives \(4x^2+14x+12\). The quadratic options for \(f\) give a degree 4 expression, which cannot equal a degree 2 polynomial.
Final Answer:
f(x) = 2x - 3. \[ \boxed{\text{(B) }2x-3} \]
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