IBDP Maths SL 5.6 Derivatives of xn , sinx , cosx , tanx and lnx AA HL Paper 1- Exam Style Questions- New Syllabus

(a)
The distance \( l \) from the origin \( (0,0) \) to a point \( (x, y) \) is \( \sqrt{x^2 + y^2} \).
Substituting \( y = f(x) = \sqrt{x^2 \ln x + 4 – x^2} \):
\( l = \sqrt{x^2 + \left(\sqrt{x^2 \ln x + 4 – x^2}\right)^2} \)
\( l = \sqrt{x^2 + x^2 \ln x + 4 – x^2} \)
\( l = \sqrt{x^2 \ln x + 4} \).
\( \boxed{l = \sqrt{x^2 \ln x + 4}} \)

(b)
To minimize the distance \( l \), it is sufficient to minimize the expression under the radical, \( L(x) = x^2 \ln x + 4 \).
Differentiating \( L(x) \) with respect to \( x \) using the product rule:
\( L'(x) = \dfrac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \dfrac{d}{dx}(\ln x) + 0 \)
\( L'(x) = 2x \ln x + x^2 \cdot \left(\dfrac{1}{x}\right) = 2x \ln x + x \).
Setting the derivative to zero for stationary points:
\( x(2\ln x + 1) = 0 \).
Since \( x > 0 \), we solve \( 2\ln x + 1 = 0 \):
\( \ln x = -\dfrac{1}{2} \implies x = e^{-1/2} = \dfrac{1}{\sqrt{e}} \).
To verify this is a minimum, consider the second derivative:
\( L”(x) = 2 \ln x + 2 + 1 = 2 \ln x + 3 \).
At \( x = e^{-1/2} \), \( L”(e^{-1/2}) = 2(-1/2) + 3 = 2 > 0 \), confirming a local minimum.
\( \boxed{x = \frac{1}{\sqrt{e}}} \)