A curve is defined by the equation y = x^3 + px^2 + qx + 5, where p and q are constants. I… — JAMB UTME

A curve is defined by the equation $y = x^3 + px^2 + qx + 5$ , where $p$ and $q$ are constants. If the curve has a stationary point with a local maximum at $x = -1$ and a local minimum at $x = 3$ , what is the value of $p + q$ ?

$-12$ Answer
B
$12$
C
$-15$
D
$-6$

Answer

The value of

p + q

-12

The derivative of

y = x^3 + px^2 + qx + 5

\frac{dy}{dx} = 3x^2 + 2px + q

. Setting

\frac{dy}{dx} = 0

at the stationary points

x = -1

and

x = 3

means

\frac{dy}{dx} = 3(x + 1)(x - 3) = 3x^2 - 6x - 9

. Comparing coefficients gives

2p = -6 \implies p = -3

and

q = -9

. Summing these constants gives

p + q = -12

. Evaluating the second derivative

\frac{d^2y}{dx^2} = 6x + 2p = 6x - 6

confirms a maximum at

x = -1

(

\frac{d^2y}{dx^2} = -12 < 0

) and a minimum at

x = 3

(

\frac{d^2y}{dx^2} = 12 > 0

Step-by-Step Solution

Find the derivative of the given function.

\frac{dy}{dx} = 3x^2 + 2px + q

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

Use the stationary points

x = -1

and

x = 3

to form a quadratic equation for the derivative.

\frac{dy}{dx} = 3(x + 1)(x - 3) = 3(x^2 - 2x - 3) = 3x^2 - 6x - 9

Since

x = -1

and

x = 3

are roots of

\frac{dy}{dx} = 0

, the derivative must factor as

3(x - (-1))(x - 3)

Equate coefficients of the two derivative expressions to solve for

p

and

q

2p = -6 \implies p = -3

and

q = -9

Matching corresponding terms gives

2p = -6

and

q = -9

Calculate the required sum

p + q

p + q = -3 + (-9) = -12

Summing the calculated constants yields

-12

Key Concept

Determining parameters of a polynomial function from given stationary points using differentiation and coefficient matching.

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