Question

The curve y(x) = ax³ + bx² + cx + 5 touches the x-axis at the point P(–2, 0) and cuts the y-axis at the point Q, where y′ is equal to 3. Then the local maximum value of y(x) is

• A. \(\frac{27}{4}\)
• B. \(\frac{29}{4}\)
• C. \(\frac{37}{4}\)
• D. \(\frac{9}{2}\)

Answer 1

The Correct Option is A

 To solve this problem, we analyze the given function and the points where it interacts with the axes. The function given is:

\(y(x) = ax^3 + bx^2 + cx + 5\)

1. Since the curve touches the x-axis at point P(-2, 0), it implies that y(-2) = 0 and the curve has a triple root at this point. So, we have:
2. Furthermore, since the curve is tangent at this point, the first derivative y'(x) should be zero at x = -2:
3. We also know the curve cuts the y-axis at a point where y' is equal to 3. This condition provides:
4. Now substituting \(c = 3\) into the previous equations, we derive:
   • From Equation 1: \(4b = 8a + 1 \Rightarrow b = 2a + \frac{1}{4}\)
   • Substitute this in Equation 2:
   • \(12a - 4(2a + \frac{1}{4}) + 3 = 0\)
   • \(12a - 8a - 1 + 3 = 0 \Rightarrow 4a + 2 = 0 \Rightarrow a = -\frac{1}{2}\)
   • Substitute back to find \(b\):
   • \(b = 2(-\frac{1}{2}) + \frac{1}{4} = -1 + \frac{1}{4} = -\frac{3}{4}\)
5. Therefore, the function can be rewritten as:
6. The local maximum occurs when the derivative changes sign from positive to negative, and we find it using y'(x) = 0:
7. We discard \(x = -2\) since it was used in initial conditions, and compute y(1):