IBDP Maths SL 5.7 The second derivative AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
A horizontal tangent occurs where the gradient is zero, so set \(f'(x)=0\).
\(3x^2+12x-15=0\)
Divide by \(3\).
\(x^2+4x-5=0\)
Factorize the quadratic.
\((x+5)(x-1)=0\)
Therefore:
\(x=-5\) or \(x=1\)
Since \(a<b\):
✅ Answer: \(a=-5\) and \(b=1\)
(b)
At \(x=a=-5\), the sign of \(f'(x)\) changes from positive to negative.
This means that \(f\) is increasing before \(x=-5\) and decreasing after \(x=-5\).
Therefore, the graph turns from rising to falling.
✅ Answer: There is a local maximum point at \(x=a\).
(c)
Differentiate \(f'(x)=3x^2+12x-15\) to find the second derivative.
\(f”(x)=6x+12\)
The second derivative is zero at \(x=c\), so:
\(6c+12=0\)
\(6c=-12\)
\(c=-2\)
✅ Answer: \(c=-2\)
(d)
The sign diagram shows that \(f”(x)\) changes from negative to positive at \(x=c\).
Therefore, the graph changes from concave down to concave up. A change in concavity confirms the presence of a point of inflexion.
Simply having \(f”(c)=0\) is not enough by itself; the change of sign is the important condition.
✅ Answer: Yes, there is a point of inflexion at \(x=c\).
(e)
To find \(f(x)\), integrate \(f'(x)\).
\(f(x)=\displaystyle\int(3x^2+12x-15)\,dx\)
\(f(x)=x^3+6x^2-15x+C\)
Use the condition \(f(-2)=36\).
\(36=(-2)^3+6(-2)^2-15(-2)+C\)
\(36=-8+24+30+C\)
\(36=46+C\)
\(C=-10\)
Therefore:
✅ Answer: \(f(x)=x^3+6x^2-15x-10\)