IB Mathematics SL 5.3 The derivative of functions AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)(i)
First, write the fractional term using a negative power:
\(B(x)=20+32x^{-1}+2x\).
Differentiating each term gives
\(B'(x)=0-32x^{-2}+2\).
Therefore,
\(B'(x)=-\dfrac{32}{x^2}+2\).
✅ Answer: \(B'(x)=-\dfrac{32}{x^2}+2\)
(a)(ii)
At a minimum point, the gradient is zero. Therefore,
\(B'(x)=0\).
\(-\dfrac{32}{x^2}+2=0\)
\(\dfrac{32}{x^2}=2\)
\(2x^2=32\)
\(x^2=16\).
This gives \(x=\pm4\). Since the domain is \(x\geq1\), only \(x=4\) is valid.
Substitute \(x=4\) into \(B(x)\):
\(B(4)=20+\dfrac{32}{4}+2(4)\)
\(B(4)=20+8+8\)
\(B(4)=36\).
Since the question states that there is no local maximum point over the domain, this stationary value is the minimum.
✅ Answer: The minimum value of \(B(x)\) is \(36\), occurring when \(x=4\).
(b)
The preparation time is less than \(40\) minutes, so
\(T(k)<40\).
\(20+\dfrac{32}{k}+2k<40\).
Since \(k\in\mathbb{Z}^{+}\), we know that \(k>0\), so multiplying both sides by \(k\) does not reverse the inequality:
\(20k+32+2k^2<40k\).
\(2k^2-20k+32<0\).
Dividing by \(2\):
\(k^2-10k+16<0\).
Factorizing gives
\((k-2)(k-8)<0\).
The product is negative between its two roots, so
\(2<k<8\).
Since \(k\) must be a positive integer,
\(k\in\{3,4,5,6,7\}\).
✅ Answer: \(k=3,4,5,6\) or \(7\)