Mathematics Past Paper PDF

### Name: MEREZ CHAN WAW CHING ### ID: SUOL2400132 ### NO: ### DATE: ## Question 1 $3^x - 4(3^x + 1) + 27 = 0$: $3^x - (12)(3^x) + 27 = 0$ Let $y=3^x$ $y^2 - 12y + 27 = 0$ $(y - 3)(y - 9) = 0$ $y = 3$ or $y = 9$ Sub $y = 3^x$, $y = 3$ or $y = 9$ $3^x = 3$ $x = 1$ $3^x = 9$ $x = 2$ $x = 1, x = 2$...

### Name: MEREZ CHAN WAW CHING ### ID: SUOL2400132 ### NO: ### DATE: ## Question 1 $3^x - 4(3^x + 1) + 27 = 0$: $3^x - (12)(3^x) + 27 = 0$ Let $y=3^x$ $y^2 - 12y + 27 = 0$ $(y - 3)(y - 9) = 0$ $y = 3$ or $y = 9$ Sub $y = 3^x$, $y = 3$ or $y = 9$ $3^x = 3$ $x = 1$ $3^x = 9$ $x = 2$ $x = 1, x = 2$ ## Question 2 $\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$ $\frac{1 * (2 + \sqrt{3})}{2 - \sqrt{3}} * \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$ $\frac{2(2 + \sqrt{3}) + 1(2 + \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}$ $\frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}$ $\frac{(2 + \sqrt{3})^2}{(2 + \sqrt{3})(2 - \sqrt{3})}$ $\frac{4 + 2\sqrt{3} + 2\sqrt{3} + 3}{4 - 2\sqrt{3} + 2\sqrt{3} - 3}$ $\frac{4 + 4\sqrt{3} + 3}{4 - 3}$ $\frac{4 + 4\sqrt{3} + 3}{1}$ $= 4 + 4\sqrt{3} + 3$ $= 7 + 4\sqrt{3}$ ## Question 3 $log_3x - 2log_3x = 1$ $log_3x - 2log_3x = 1$ Let $y = log_3x$ $y - 2y = 1$ $-y = 1$ $(y - 2)(y + 1) = 0$ $y = 2$ or $y = 1$ Sub $y = log_3x$, $y = 2$, or $y = 1$ $log_3x = 2$ $x = 3^2$ $ x = 9$ $log_3x = 1$ $x = 3$ ## Question 4 $2x^4 - 5x^3 - 0x^2 + 0x - 6$ $2x^2(x^2 - x + 2)$ $-3x(x^2 - x + 2)$ $x^2 - x + 2|2x^4 - 5x^3 - 0x^2 + 0x - 6$ $2x^4 - 2x^3 + 4x^2$ $- - 3x^2 - 4x^2 + 0x - 6$ $- - 3x^2 + 3x^2 - 6x$ $- -7x^2 + 6x - 6$ $- -7x^2 + 7x - 14$ $- -x + 8$ $- -x + 8$ Standard: $2x^2 - 3x - 7 + x^2 - x + 2$ quotient: $2x^2 - 3x - 7$ remainder: $8 - x$ ## Question 5 $P(x) = x^3 + x - 2, P(1) = 1$ $P(1) = (1)^3 + 1 - 2 = 0$, therefore $x - 1$ is a factor of $x^3 + x - 2$ $P(x) = (x - 1) (x^2 +x + 2)$ $P(3) = (3 - 1)(3^2 + 3 + 2)$