Question 5

loading...

Prev/Next links

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: 

$\begin{align}  & f(x)=2+x-{{x}^{2}} \\ & {{f}^{1}}(x)=1-2x \\ & \text{At stationary point }{{f}^{1}}(x)=0 \\ & 1-2x=0 \\ & x=\tfrac{1}{2} \\ & {{f}^{11}}(x)=\frac{d}{dx}(1-2x)=-2 \\ & \text{Since }{{f}^{11}}(x)<0\text{ At }x=\tfrac{1}{2}\text{ is a max}\text{. point} \\ & \text{Maximum value of the function} \\ & f(\tfrac{1}{2})=2+\tfrac{1}{2}-{{(\tfrac{1}{2})}^{2}}=2+\tfrac{1}{2}-\tfrac{1}{4}=\tfrac{8+2-1}{4}=\tfrac{9}{4} \\\end{align}$

Jamb Maths Topic: 
Year of Exam: