Question:
Let \(f(x)=x^2+ax+b\) and \(g(x)=f(x+1)-f(x-1)\). If \(f(x)≥0\) for all real \(x\), and \(g(20)=72\), then the smallest possible value of \(b\) is
Let \(f(x)=x^2+ax+b\) and \(g(x)=f(x+1)-f(x-1)\). If \(f(x)≥0\) for all real \(x\), and \(g(20)=72\), then the smallest possible value of \(b\) is
Updated On: Sep 17, 2024
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The Correct Option is D
Solution and Explanation
The correct answer is (D): \(4\)
\(f(x)=x^2+ax+b\)
\(g(x)=f(x+1)-f(x-1)\)
=\({(x+1)^2+a(x+1)+b}-{(x-1)^2+a(x-1)+b}\)
\(g(x)=4x+2a\)
\(g(20)=72\)
\(80+2a=72 ⇒ a=-4\)
\(∴ f(x)=x^2-4x+b\)
\(f(x)=(x-2)^2+b-4\)
When \(b≥4 f(x)≥0\) for all \(x\)
\(∴\) The minimum value of b is \(4\)
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