Jambmaths question:
The maximum value of the function $f(x)=2+x-{{x}^{2}}$is
Option A:
$\frac{3}{2}$
Option B:
$\frac{1}{2}$
Option C:
$\frac{9}{4}$
Option D:
$\tfrac{7}{4}$
Jamb Maths Solution:
$\begin{align} & f(x)=2+x-{{x}^{2}} \\ & {{f}^{1}}(x)=1-2x \\ & \text{At stationary point }{{f}^{1}}(x)=0 \\ & 1-2x=0 \\ & x=\tfrac{1}{2} \\ & {{f}^{11}}(x)=\frac{d}{dx}(1-2x)=-2 \\ & \text{Since }{{f}^{11}}(x)<0\text{ At }x=\tfrac{1}{2}\text{ is a max}\text{. point} \\ & \text{Maximum value of the function} \\ & f(\tfrac{1}{2})=2+\tfrac{1}{2}-{{(\tfrac{1}{2})}^{2}}=2+\tfrac{1}{2}-\tfrac{1}{4}=\tfrac{8+2-1}{4}=\tfrac{9}{4} \\\end{align}$
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