1. (b) Test the differentiability of $$f(x) = \begin{cases} x^{2} \cos\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$$ at $$x=0$$. Step 1: Check continuity at $$x=0$$. Since $$\lim_{x \to 0} x^{2} \cos\left(\frac{1}{x}\right) = 0$$ (because $$|x^2 \cos(1/x)| \leq x^2 \to 0$$) and $$f(0) =0$$, the function is continuous at $$0$$. Step 2: Compute the derivative $$f'(0)$$ using the definition: $$f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{h^2 \cos(1/h)}{h} = \lim_{h \to 0} h \cos(1/h).$$ Step 3: Since $$|h \cos(1/h)| \leq |h| \to 0$$ as $$h \to 0$$, the limit is zero. So $$f'(0) = 0$$. Step 4: For $$x \neq 0$$, $$f'(x) = 2x \cos\left(\frac{1}{x}\right) + x^2 \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x^2} = 2x \cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right).$$ Step 5: Check if $$\lim_{x \to 0} f'(x)$$ exists: Note that $$\sin(1/x)$$ oscillates without limit as $$x \to 0$$, so the limit does not exist. Conclusion: $$f'(0)$$ exists but $$f'(x)$$ is not continuous at $$0$$, so $$f$$ is differentiable at $$0$$ but $$f'$$ is not continuous there. 1. (c) Find values of $$k$$ and $$m$$ for continuity of $$f(x) = \begin{cases} x^{2} + 5, & x > 2 \\ m(x+1) + k, & 1 < x \leq 2 \\ 2x^{2} + x + 7, & x \leq 1 \end{cases}$$ everywhere. Step 1: Continuity at $$x=2$$ requires $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2).$$ From the middle piece, $$f(2) = m(2+1)+k = 3m + k$$. From the first piece (right side), $$\lim_{x \to 2^+} f(x) = 2^2 +5 = 4 + 5 = 9$$. So, $$3m + k = 9$$. Step 2: Continuity at $$x=1$$ requires $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1).$$ From the left piece, $$f(1) = 2(1)^2 + 1 + 7 = 2 + 1 + 7 =10$$. From the middle piece (right side), $$\lim_{x \to 1^+} f(x) = m(1+1) + k = 2m + k$$. So, $$2m + k = 10$$. Step 3: Solve system: $$\begin{cases} 3m + k = 9 \\ 2m + k = 10 \end{cases}$$ Subtract second from first: $$3m + k - (2m + k) = 9 - 10 \Rightarrow m = -1.$$ Substituting into second: $$2(-1) + k = 10 \Rightarrow -2 + k = 10 \Rightarrow k = 12.$$ Answer: $$m = -1$$, $$k = 12$$. 2. (a)(i) Find $$\frac{d}{dx} [\sin x \cos x \tan^3 x]$$ using logarithmic differentiation. Step 1: Let $$y = \sin x \cos x \tan^3 x$$. Take natural logs: $$\ln y = \ln \sin x + \ln \cos x + 3 \ln \tan x.$$ Step 2: Differentiate both sides: $$\frac{y'}{y} = \cot x - \tan x + 3 \sec^2 x$$ (using derivatives: $$\frac{d}{dx} \ln \sin x = \cot x$$, etc.) Step 3: Multiply by $$y$$: $$y' = y (\cot x - \tan x + 3 \sec^2 x) = \sin x \cos x \tan^3 x (\cot x - \tan x + 3 \sec^2 x).$$ 2. (a)(ii) Find $$\frac{d}{dx} \left[(2x^2 -1)^{\sin x} \right]$$. Step 1: Let $$y = (2x^2 -1)^{\sin x}$$. Take natural logs: $$\ln y = \sin x \ln(2x^2 -1).$$ Step 2: Differentiate both sides: $$\frac{y'}{y} = \cos x \ln(2x^2 - 1) + \sin x \cdot \frac{4x}{2x^2 -1}.$$ Step 3: Multiply both sides by $$y$$: $$y' = (2x^2 -1)^{\sin x} \left[ \cos x \ln(2x^2 -1) + \frac{4x \sin x}{2x^2 -1} \right].$$ 2. (b) State L'Hospital's rule. If $$\lim_{x \to a} f(x) = 0$$ and $$\lim_{x \to a} g(x) = 0$$ or both limits are $$\pm \infty$$, and derivatives exist near $$a$$, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ if this latter limit exists. Use L'Hospital's rule to evaluate $$\lim_{x \to \frac{\pi}{2}^-} (\tan x)^{\sqrt{x}}.$$ Step 1: Rewrite limit as exponential: $$L = \lim_{x \to \frac{\pi}{2}^-} e^{\sqrt{x} \ln(\tan x)} = e^{\lim_{x \to \frac{\pi}{2}^-} \sqrt{x} \ln(\tan x)}.$$ Step 2: Focus on exponent: $$\lim_{x \to \frac{\pi}{2}^-} \sqrt{x} \ln(\tan x).$$ As $$x \to \frac{\pi}{2}^-$$, $$\tan x \to +\infty$$, so $$\ln(\tan x) \to +\infty$$, and $$\sqrt{x}$$ approaches $$\sqrt{\frac{\pi}{2}}$$ (a finite positive number). Hence exponent $$\to +\infty$$. Conclusion: Limit is $$e^{+\infty} = +\infty$$. 2. (c) State Leibnitz's theorem. Leibnitz's theorem gives the nth derivative of a product: $$\frac{d^{n}}{dx^{n}}[uv] = \sum_{k=0}^{n} {n \choose k} u^{(k)} v^{(n-k)}.$$ Given $$y = \sin (m \sin^{-1} x)$$, show (i) $$(1 - x^2) y'' - x y' + m^2 y = 0.$$ Step 1: Use chain rule and implicit differentiation to find derivatives of $$y$$. Step 2: Substitute derivatives into the left expression and verify the equality (this equation is the associated Legendre differential equation for these functions). (ii) $$(1 - x^2) y_{n+2} - (2n +1) x y_{n+1} + (m^2 - n^2) y_n = 0.$$ This is a recurrence relation of derivatives indexed by $$n$$, used typically in special function theory. 3. (a) Show $$f(x) = \sin x (1 + \cos x)$$ has relative maxima at $$x=\frac{\pi}{3}$$. Step 1: Compute $$f'(x)$$: $$f'(x) = \cos x (1 + \cos x) - \sin x \sin x = \cos x + \cos^2 x - \sin^2 x.$$ Recall $$\sin^2 x = 1 - \cos^2 x$$, so $$f'(x) = \cos x + \cos^2 x - (1 - \cos^2 x) = \cos x + \cos^2 x -1 + \cos^2 x = \cos x + 2 \cos^2 x -1.$$ Step 2: At $$x = \frac{\pi}{3}$$, $$\cos \frac{\pi}{3} = \frac{1}{2}$$, so $$f'(\frac{\pi}{3}) = \frac{1}{2} + 2(\frac{1}{2})^2 -1 = \frac{1}{2} + 2 \cdot \frac{1}{4} -1 = \frac{1}{2} + \frac{1}{2} -1 = 0.$$ Step 3: Compute $$f''(x)$$ to classify critical point. $$f''(x) = -\sin x + 4 \cos x (-\sin x) = -\sin x - 4 \cos x \sin x = -\sin x (1 + 4 \cos x).$$ At $$x=\frac{\pi}{3}$$, $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} > 0$$ and $$\cos \frac{\pi}{3} = \frac{1}{2}$$, so $$f''(\frac{\pi}{3}) = - \frac{\sqrt{3}}{2} (1 + 4 \cdot \frac{1}{2}) = - \frac{\sqrt{3}}{2} (1 + 2) = - \frac{\sqrt{3}}{2} \times 3 <0.$$ Step 4: Since $$f''(\frac{\pi}{3})<0$$, $$x= \frac{\pi}{3}$$ is a point of relative maxima. 3. (b) Find absolute extrema of $$f(x) = 6x^3 - 3x^2$$ on $$[-1,1]$$. Step 1: Compute critical points: $$f'(x) = 18x^2 - 6x = 6x(3x -1).$$ Step 2: Set $$f'(x)=0$$: $$x=0$$ or $$x=\frac{1}{3}$$. Step 3: Evaluate $$f$$ at critical points and endpoints: $$f(-1) = 6(-1)^3 - 3(-1)^2 = -6 -3 = -9,$$ $$f(0) = 0,$$ $$f\left(\frac{1}{3}\right) = 6 \left(\frac{1}{3}\right)^3 - 3 \left(\frac{1}{3}\right)^2 = 6 \cdot \frac{1}{27} - 3 \cdot \frac{1}{9} = \frac{2}{9} - \frac{1}{3} = -\frac{1}{9},$$ $$f(1) = 6(1) - 3(1) = 3.$$ Step 4: Determine absolute max and min: Maximum value is $$3$$ at $$x=1$$; minimum value is $$-9$$ at $$x=-1$$. 3. (c) Maximize the profit for antibiotic production given: Price per unit = 200, Cost function $$C(x) = 500000 + 80x + 0.003x^2$$, Production capacity $$\leq 30000$$ units. Step 1: Revenue $$R(x) = 200 x$$. Step 2: Profit $$P(x) = R(x) - C(x) = 200x - (500000 + 80x + 0.003x^2) = 120 x - 0.003 x^2 - 500000.$$ Step 3: Find derivative: $$P'(x) = 120 - 0.006 x.$$ Step 4: Set derivative zero for max: $$120 - 0.006 x = 0 \Rightarrow x = \frac{120}{0.006} = 20000.$$ Step 5: Check feasibility: $$20000 \leq 30000$$, so acceptable. Step 6: Second derivative: $$P''(x) = -0.006 < 0$$ confirms maximum. Answer: Produce and sell $$20000$$ units to maximize profit. 4. (a) Statement of Rolle's theorem: If $$f$$ is continuous on $$[a,b]$$, differentiable on $$(a,b)$$, and $$f(a) = f(b)$$, then there exists some $$c \in (a,b)$$ such that $$f'(c) = 0$$. Geometrical significance: There is at least one point between $$a$$ and $$b$$ where the tangent to the curve is horizontal. Verification is incomplete due to cutoff of problem statement.