1. (a) Given gradient function $g(x) = 3x^2 - 7$ and curve passes through $(-1, 12)$, find the equation of the curve. Integrate the gradient to find $y$: $$y = \int (3x^2 - 7) dx = x^3 - 7x + C$$ Use point $(-1,12)$ to find $C$: $$12 = (-1)^3 - 7(-1) + C = -1 + 7 + C = 6 + C \implies C = 6$$ Equation is: $$y = x^3 - 7x + 6$$ 2. (b) Given $z = 1 + i$, find $z^8$. Write $z$ in polar form: $r = \sqrt{1^2 +1^2} = \sqrt{2}$, $\theta = \arctan(1) = \frac{\pi}{4}$ Use De Moivre's theorem: $$z^8 = r^8 \left( \cos(8\theta) + i \sin(8\theta) \right) = (\sqrt{2})^8 (\cos(2\pi) + i \sin(2\pi)) = 16 (1 + 0i) = 16$$ 3. (c) Solve inequality: $$\frac{x}{2x + 3} \geq 2$$ Bring all terms to one side: $$\frac{x}{2x + 3} - 2 \geq 0 \implies \frac{x - 2(2x + 3)}{2x + 3} \geq 0 \implies \frac{x - 4x - 6}{2x + 3} \geq 0 \implies \frac{-3x -6}{2x + 3} \geq 0$$ Rewrite numerator: $$\frac{-3(x + 2)}{2x + 3} \geq 0$$ Determine critical points: $x = -2$ and $x = -\frac{3}{2}$ Test intervals around these points to solve the inequality: - For $x < -2$, plug $x=-3$: $$\frac{-3(-3 + 2)}{2(-3) + 3} = \frac{-3(-1)}{-6 + 3} = \frac{3}{-3} = -1 < 0$$ - For $-2 < x < -\frac{3}{2}$, plug $x=-1.6$: $$\frac{-3(-1.6 + 2)}{2(-1.6) + 3} = \frac{-3(0.4)}{-3.2 + 3} = \frac{-1.2}{-0.2} = 6 > 0$$ - For $x > -\frac{3}{2}$, plug $x=0$: $$\frac{-3(0 + 2)}{2(0) + 3} = \frac{-6}{3} = -2 < 0$$ Inequality holds where expression $\geq 0$, so: $$-2 < x < -\frac{3}{2}$$ Since denominator cannot be zero, exclude $x = -\frac{3}{2}$. Thus solution in interval notation: $$(-2, -\frac{3}{2})$$ 4. (d) Derive MacLaurin series of $$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... + a_n x^n$$ MacLaurin series expands function around $x=0$: $$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + ...$$ Calculate derivatives at 0: $$f(0) = a_0$$ $$f'(x) = a_1 + 2a_2 x + 3a_3 x^2 + ... \implies f'(0) = a_1$$ $$f''(x) = 2a_2 + 6a_3 x + 12 a_4 x^2 + ... \implies f''(0) = 2a_2$$ $$f^{(3)}(x) = 6a_3 + 24 a_4 x + ... \implies f^{(3)}(0) = 6 a_3$$ $$f^{(4)}(x) = 24 a_4 + ... \ o f^{(4)}(0) = 24 a_4$$ Therefore MacLaurin series up to $x^4$ term is: $$f(x) = a_0 + a_1 x + \frac{2a_2}{2!} x^2 + \frac{6 a_3}{3!} x^3 + \frac{24 a_4}{4!} x^4 = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$$ Essentially original polynomial truncated to degree 4. 5. (e) Evaluate $$\int \frac{x^5 + x -1}{x^4 - x^3} dx$$ Simplify denominator: $$x^4 - x^3 = x^3 (x - 1)$$ Rewrite integrand: $$\frac{x^5 + x -1}{x^3(x-1)} = \frac{x^5}{x^3(x-1)} + \frac{x}{x^3(x-1)} - \frac{1}{x^3(x-1)} = \frac{x^2}{x-1} + \frac{1}{x^2 (x-1)} - \frac{1}{x^3 (x-1)}$$ Focus on partial fractions or separate terms: Better to perform polynomial division: Divide numerator by denominator: $$\frac{x^5 + x -1}{x^4 - x^3} = x + \frac{x -1}{x^3 (x-1)} = x + \frac{x -1}{x^3 (x-1)}$$ Cancel $(x-1)$ in numerator and denominator in fraction: $$= x + \frac{1}{x^3}$$ So the integral becomes: $$\int \left(x + \frac{1}{x^3}\right) dx = \int x dx + \int x^{-3} dx = \frac{x^2}{2} - \frac{1}{2 x^2} + C$$ 6. B7 (a) Prove $\sqrt{2}$ is irrational. Assume $\sqrt{2} = \frac{p}{q}$ with $p,q$ integers with no common factors. Then $$2 = \frac{p^2}{q^2} \implies p^2 = 2 q^2$$ So $p^2$ even, implies $p$ even, so $p=2k$. Substitute: $$4k^2 = 2 q^2 \implies 2 k^2 = q^2$$ Then $q^2$ even, implies $q$ even. This contradicts assumption $p,q$ coprime. Hence $\sqrt{2}$ is irrational. 7. (b) Find Taylor series of $e^x$ at $x=1$ up to $x^3$ term. Taylor series at $x=a$: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n$$ Here $f(x) = e^x$, all derivatives $f^{(n)}(x) = e^x$. At $x=1$: $$f^{(n)}(1) = e$$ So series up to $n=3$: $$e + e (x-1) + \frac{e}{2} (x-1)^2 + \frac{e}{6} (x-1)^3$$ 8. (c) Find all solutions of $$z^4 - 9 z^3 + 26 z^2 - 9 z - 65 = 0$$ Given root $z = 3 + 2 i$. Conjugate root $3 - 2 i$ also root (coefficients real). Factor quadratic from these roots: $$ (z - (3+ 2 i))(z - (3 - 2 i)) = z^2 - 6 z + 13$$ Divide polynomial by $z^2 -6 z + 13$: Long division gives quotient: $$z^2 - 3$$ Solve $z^2 - 3 = 0$: $$z = \pm \sqrt{3}$$ Roots: $$3 + 2 i, 3 - 2 i, \sqrt{3}, -\sqrt{3}$$ 9. (d) Find critical points of $$f(x)= x^3 - 12 x + 16$$ Find derivative: $$f'(x) = 3 x^2 - 12$$ Set $f'(x) = 0$: $$3 x^2 - 12 = 0 \implies x^2 = 4 \implies x = \pm 2$$ Classify with second derivative: $$f''(x) = 6 x$$ At $x=2$: $$f''(2) = 12 > 0$$ minimum At $x=-2$: $$f''(-2) = -12 < 0$$ maximum 10. (e) Evaluate $$\int \frac{1}{x^2 + k^2} dx$$ Use substitution: $$x = k \tan \theta, \quad dx = k \sec^2 \theta d\theta$$ Then $$x^2 + k^2 = k^2 \tan^2 \theta + k^2 = k^2 (\tan^2 \theta + 1) = k^2 \sec^2 \theta$$ Integral becomes: $$\int \frac{k \sec^2 \theta d\theta}{k^2 \sec^2 \theta} = \int \frac{1}{k} d\theta = \frac{\theta}{k} + C$$ Back-substitute: $$\theta = \arctan \frac{x}{k}$$ Answer: $$\frac{1}{k} \arctan \frac{x}{k} + C$$