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CSS Pure Mathematics QUESTION #4094
Question 1
In applying the Mean Value Theorem to prove \(e^x\geq1+x\) for all \(x\in\mathbb{R}\), define \(g(t)=e^t-1-t\). Then \(g'(t)=e^t-1\). For \(t>0\), \(g'(t)\):
  • Is negative, so \(g\) is decreasing
  • Is zero, so \(g\) is constant
  • Is positive, so \(g\) is increasing, implying \(g(x)>g(0)=0\)✔️
  • Is undefined
Correct Answer Explanation
For \(t>0\): \(g'(t)=e^t-1>0\). So \(g\) is strictly increasing for \(t>0\), meaning \(g(x)>g(0)=0\) for \(x>0\), i.e., \(e^x-1-x>0\), i.e., \(e^x>1+x\). A similar argument with \(g\) decreasing for \(t<0\) completes the proof for all \(x\in\mathbb{R}\).