A curve has the equation y = ax^3 + bx^2 + 12x + 1, where a and b are constants. If the cu… — JAMB UTME

A curve has the equation $y = ax^3 + bx^2 + 12x + 1$ , where $a$ and $b$ are constants. If the curve has stationary points at $x = 1$ and $x = 2$ , what is the value of $a$ ?

$2$ Answer
B
$6$
C
$-2$
D
$-6$

Answer

The value of

a

2

To find

a

, take the first derivative of the curve, yielding

\frac{dy}{dx} = 3ax^2 + 2bx + 12

. Setting this to zero gives a quadratic equation with roots

x = 1

and

x = 2

. The product of roots for a quadratic equation

Ax^2 + Bx + C = 0

\frac{C}{A}

. Therefore,

1 \times 2 = \frac{12}{3a}

, which simplifies to

2 = \frac{4}{a}

, giving

a = 2

Step-by-Step Solution

Find the derivative of the given curve with respect to

x

\frac{dy}{dx} = 3ax^2 + 2bx + 12

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

Set the derivative to zero and substitute the stationary point locations

x = 1

and

x = 2

The roots of the quadratic equation

3ax^2 + 2bx + 12 = 0

are

x_1 = 1

and

x_2 = 2

Since stationary points are given at

x=1

and

x=2

, these values satisfy

\frac{dy}{dx} = 0

Apply the product of roots formula for a quadratic equation

Ax^2 + Bx + C = 0

, where

x_1 \cdot x_2 = \frac{C}{A}

1 \times 2 = \frac{12}{3a} \implies 2 = \frac{4}{a}

Equating the product of roots

1 \times 2 = 2

\frac{12}{3a}

isolates parameter

a

Solve for

a

a = 2

Multiplying both sides by

a

gives

2a = 4

, so

a = 2

Key Concept

Stationary points occur where

\frac{dy}{dx} = 0

. For a cubic curve, the derivative is a quadratic equation whose roots correspond to the

x

-coordinates of the stationary points.

Estimated Time:2m 0s

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Step-by-Step Solution

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