Question 18

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Jambmaths question: 

Find the maximum value of y in the equation $y=1-2x-3{{x}^{2}}$

Option A: 

$\tfrac{5}{4}$

Option B: 

$\tfrac{5}{3}$

Option C: 

$\tfrac{3}{4}$

Option D: 

$\tfrac{4}{3}$

Jamb Maths Solution: 

$\begin{align}  & y=1-2x-3{{x}^{2}} \\ & \frac{dy}{dx}=-2-6x \\ & \text{At }\frac{dy}{dx}=0,\text{ } \\ & -2-6x=0 \\ & x=-\frac{1}{3} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=-6<0\text{   (Test for Maximum point fulfilled)} \\ & so\text{ }at\text{ }x=-\frac{1}{3}\text{ will give maximum value for }y \\ & \text{The value of the maximum value is }y \\ & y=1-2(-\tfrac{1}{3})-3{{(-\tfrac{1}{3})}^{2}}=1+\tfrac{2}{3}-\tfrac{1}{3} \\ & y=\tfrac{3+2-1}{3}=\tfrac{4}{3} \\\end{align}$

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