1. **Express the following in factors:** (i) Simplify $\frac{(n+2)!}{(n+2)!} = 1$ because any nonzero number divided by itself equals 1. (ii) Factorize $(n-1)! - (n-2)!$: Note: $(n-1)! = (n-1)(n-2)!$ So, $(n-1)! - (n-2)! = (n-1)(n-2)! - (n-2)! = ((n-1) - 1)(n-2)! = (n-2)(n-2)!$ (iii) Express $(n+2)! + (n+1)! + n!$ in factors: Recall $(n+2)! = (n+2)(n+1)n!$ and $(n+1)! = (n+1)n!$, so: $$(n+2)! + (n+1)! + n! = (n+2)(n+1)n! + (n+1)n! + n!$$ Factor $n!$ out: $$n![(n+2)(n+1) + (n+1) + 1] = n![(n+2)(n+1) + (n+1) + 1]$$ Simplify inside brackets: $$(n+2)(n+1) = n^2 + 3n + 2$$ So sum inside brackets: $$n^2 + 3n + 2 + n + 1 + 1 = n^2 + 4n + 4 = (n+2)^2$$ Hence: $$(n+2)! + (n+1)! + n! = n!(n+2)^2$$ 2. **Determine the value of n:** (i) Solve $\binom{n}{6} = \binom{n}{8}$. Recall symmetry property: $$\binom{n}{k} = \binom{n}{n-k}$$ So $\binom{n}{6} = \binom{n}{8} \implies 6 = n - 8 \implies n = 14$. (ii) Solve $2^n P_3 = n+1 P_3$ Permutations formula: $nP_r = \frac{n!}{(n-r)!}$ So: $$2^n \times \frac{3!}{0!} = \frac{(n+1)!}{(n-2)!}$$ Since $P_3 = n(n-1)(n-2)$ for integer $n \geq 3$, rewrite: $$2^n \times 3! = (n+1)n(n-1)$$ Substitute $3! = 6$: $$6 \times 2^n = (n+1)n(n-1)$$ Try integer values: For $n=3$: LHS $= 6 \times 8 =48$; RHS $= 4 \times 3 \times 2=24$ no For $n=4$: LHS $= 6 \times 16 =96$; RHS $= 5 \times 4 \times 3=60$ no For $n=5$: LHS $= 6 \times 32 =192$; RHS $= 6 \times 5 \times 4=120$ no For $n=6$: LHS $=6 \times 64 =384$; RHS $= 7 \times 6 \times 5=210$ no For $n=7$: LHS $= 6 \times 128 =768$; RHS $= 8 \times 7 \times 6=336$ no For $n=8$: LHS $= 6 \times 256=1536$; RHS $= 9 \times 8 \times 7=504$ no No integer solutions satisfy equality exactly, so no integer $n$ satisfies the equation. 3. **Expand $ (2x + 3)^5 $** using binomial theorem: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ Here, $a=2x, b=3, n=5$ Expand terms: $$\binom{5}{0}(2x)^5 3^0 + \binom{5}{1}(2x)^4 3^1 + \binom{5}{2}(2x)^3 3^2 + \binom{5}{3}(2x)^2 3^3 + \binom{5}{4}(2x)^1 3^4 + \binom{5}{5}(2x)^0 3^5$$ Calculate stepwise: 1. $1 \times 32x^5 \times 1 = 32x^5$ 2. $5 \times 16x^4 \times 3 = 240x^4$ 3. $10 \times 8x^3 \times 9 = 720x^3$ 4. $10 \times 4x^2 \times 27 = 1080x^2$ 5. $5 \times 2x \times 81 = 810x$ 6. $1 \times 1 \times 243 = 243$ Final expression: $$32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243$$ 4. Simplify $\frac{\log_{10}8 - \log_{10}4}{\log_{10}4 - \log_{10}2}$ Use log subtraction rule: $\log a - \log b = \log \frac{a}{b}$ Numerator: $\log_{10}\frac{8}{4} = \log_{10}2$ Denominator: $\log_{10}\frac{4}{2} = \log_{10}2$ So expression = $\frac{\log_{10}2}{\log_{10}2} = 1$ 5. Find value of $y$ given pentagon angles: Angles are $y^\circ, (y+10)^\circ, (y+20)^\circ, (y+40)^\circ, (y+50)^\circ$ Sum of interior angles of pentagon = $(5 - 2) \times 180 = 540^\circ$ Sum angles: $$y + (y+10) + (y+20) + (y+40) + (y+50) = 540$$ Simplify: $$5y + 120 = 540 \implies 5y = 420 \implies y = 84$$ 6. Find sum of first six terms of sequence $T_n = 5n + 1$ Sum: $$S_6 = \sum_{n=1}^6 (5n + 1) = 5 \sum_{n=1}^6 n + \sum_{n=1}^6 1 = 5 \times \frac{6 \times 7}{2} + 6 = 5 \times 21 + 6 = 105 + 6 = 111$$ 7. Integrate: (i) $\int \sqrt{16z^2} dz = \int 4|z| dz$ Assuming $z \geq 0$ for simplicity: $$= \int 4z dz = 2z^2 + C$$ (ii) $\int (4x + 3x^2) dx = \int 4x dx + \int 3x^2 dx = 2x^2 + x^3 + C$ (iii) $\int_0^3 (x+1)(2x+1) dx$ First expand: $$(x+1)(2x+1) = 2x^2 + x + 2x + 1 = 2x^2 + 3x + 1$$ Integrate termwise: $$\int_0^3 (2x^2 + 3x + 1) dx = \left[\frac{2}{3}x^3 + \frac{3}{2}x^2 + x \right]_0^3$$ Calculate at 3: $$\frac{2}{3} \times 27 + \frac{3}{2} \times 9 + 3 = 18 + 13.5 + 3 = 34.5$$ 8. Differentiate with respect to $x$: (i) $y = (3x - 2)(x + 2)$ Use product rule: $$y' = (3x - 2)'(x+2) + (3x - 2)(x+2)' = 3(x+2) + (3x - 2)(1) = 3x + 6 + 3x - 2 = 6x + 4$$ (ii) $y = (3x^2 + 4)^2$ Use chain rule: $$y' = 2(3x^2 +4) \times (6x) = 12x(3x^2 + 4)$$ (iii) $y = 3x (x + 2)^{-1}$ Rewrite $y = 3x (x+2)^{-1}$ Use product rule: $$y' = 3 (x+2)^{-1} + 3x \times (-1)(x+2)^{-2} = \frac{3}{x+2} - \frac{3x}{(x+2)^2} = \frac{3(x+2) - 3x}{(x+2)^2} = \frac{3x + 6 - 3x}{(x+2)^2} = \frac{6}{(x+2)^2}$$ 9. Check handwritten derivative: Given $y = (3x)^3 = 27x^3$ Derivative: $$dy/dx = 3 \times (3x)^{3-1} \times 3 = 27 \times 3x^2 = 81x^2$$ But hand note says: $dy = 3(3x)^{3-1} = 3(3x)^2 = 27x^2$, missing derivative of inside function (chain rule effect). Correct derivative is $81x^2$.