1. **State the problem:** We have a function $$f(x) = ax^2 + \frac{b}{x^2}$$ with $$x>0$$ and constants $$a, b$$ unknown. Point $$P(1,5)$$ lies on the graph of $$f(x)$$. Tasks: (a) Find expression for $$f'(x)$$. (b) Given gradient of tangent at $$P$$ is 3, find $$a$$ and $$b$$. (c) Find equation of normal to $$f(x)$$ at $$P$$. Then a second function $$g(x) = \frac{k}{x^3}$$ with constant $$k$$ and $$x>0$$. (d) Tangents of $$f$$ and $$g$$ are parallel at $$x=1$$. (e) Normal to $$g(x)$$ at $$x=1$$ intersects x-axis at $$S$$. Find area of triangle $$ORS$$, where $$O$$ is origin. 2. **(a) Find $$f'(x)$$:** Given $$f(x) = ax^2 + b x^{-2}$$. Differentiate term by term: $$f'(x) = 2ax - 2b x^{-3} = 2ax - \frac{2b}{x^3}$$. 3. **(b) Use $$P(1,5)$$ and gradient at $$P$$ to find $$a$$ and $$b$$:** Since $$P=(1,5)$$ is on $$f$$, $$5 = a(1)^2 + \frac{b}{(1)^2} = a + b$$. Gradient at $$x=1$$ is 3: $$f'(1) = 2a(1) - \frac{2b}{1^3} = 2a - 2b = 3$$. From these two equations: (1) $$a + b = 5$$ (2) $$2a - 2b = 3 \Rightarrow a - b = \frac{3}{2}$$. Add (1) and (2): $$a + b + a - b = 5 + \frac{3}{2} \Rightarrow 2a = \frac{13}{2} \Rightarrow a = \frac{13}{4} = 3.25$$. Use (1): $$3.25 + b = 5 \Rightarrow b = 5 - 3.25 = 1.75$$. 4. **(c) Find equation of normal to $$f$$ at $$P$$:** Gradient of tangent at $$P$$ is $$3$$, so gradient of normal is negative reciprocal: $$m_{normal} = -\frac{1}{3}$$. Equation of normal line passing through $$P(1,5)$$: $$y - 5 = -\frac{1}{3}(x - 1)$$ $$\Rightarrow y = -\frac{1}{3}x + \frac{1}{3} + 5 = -\frac{1}{3}x + \frac{16}{3}$$. 5. **(d) Tangents to $$f$$ and $$g$$ are parallel at $$x=1$$:** Function $$g(x) = \frac{k}{x^3} = kx^{-3}$$. Derivative: $$g'(x) = -3k x^{-4} = -\frac{3k}{x^4}$$. At $$x=1$$, $$g'(1) = -3k$$. Tangent to $$f$$ at $$x=1$$ has slope 3, tangent to $$g$$ at $$x=1$$ has slope $$-3k$$. They are parallel, so slopes equal: $$3 = -3k \Rightarrow k = -1$$. 6. **(e) Normal to $$g$$ at $$x=1$$ and area of triangle $$ORS$$:** Gradient of $$g$$ at $$x=1$$: $$m_t = g'(1) = -3k = -3(-1) = 3$$. Gradient of normal to $$g$$ is negative reciprocal: $$m_n = -\frac{1}{3}$$. Point on curve at $$x=1$$: $$g(1) = k/(1)^3 = -1$$. Equation of normal to $$g$$ at $$x=1, y=-1$$: $$y - (-1) = -\frac{1}{3}(x - 1) \Rightarrow y + 1 = -\frac{1}{3}x + \frac{1}{3}$$ $$\Rightarrow y = -\frac{1}{3}x + \frac{1}{3} - 1 = -\frac{1}{3}x - \frac{2}{3}$$. Find x-intercept of this line (point $$S$$): Set $$y=0$$: $$0 = -\frac{1}{3}x - \frac{2}{3}$$ Multiply both sides by 3: $$0 = -x - 2 \Rightarrow x = -2$$. So, $$S = (-2,0)$$. 7. **Area of triangle $$ORS$$:** Points: $$O=(0,0)$$, $$R=(1,0)$$ (from $$f$$'s point?), $$S=(-2,0)$$ is on x-axis. We need to clarify point $$R$$: The problem says triangle $$ORS$$, where $$O$$ is origin and $$S$$ is intersection with x-axis of normal to $$g$$. Point $$R$$ might be the point on curve $$f$$ at $$x=1$$, which is $$P(1,5)$$. Triangle formed by points $$O(0,0)$$, $$R(1,5)$$, and $$S(-2,0)$$. Area of triangle with vertices $$(x_1,y_1), (x_2,y_2), (x_3,y_3)$$ is: $$\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$. Plug values: $$= \frac{1}{2} |0(5 - 0) + 1(0 - 0) + (-2)(0 - 5)| = \frac{1}{2} |0 + 0 + (-2)(-5)| = \frac{1}{2} \times 10 = 5$$. **Final answers:** \(a) f'(x) = 2ax - \frac{2b}{x^3} \) \(b) a = \frac{13}{4} = 3.25, b = \frac{7}{4} = 1.75 \) \(c) \text{Normal equation at } P: y = -\frac{1}{3} x + \frac{16}{3} \) \(d) k = -1 \) \(e) \text{Area of triangle } ORS = 5 \)