2021 H2 Math Paper 1 Answers
a = 4 , b = − 6 , c = 0 , d = 7. {a=4, b=-6, c=0, d=7.} a = 4 , b = − 6 , c = 0 , d = 7. x = 1 + 5 or x = − 1 + 7 . {x=1 + \sqrt{5} \textrm{ or } x=- 1 + \sqrt{7}.} x = 1 + 5 or x = − 1 + 7 . y = 1 2 x + 7 2 . {y = \frac{1}{2} x + \frac{7}{2}.} y = 2 1 x + 2 7 . ∣ z ∣ = 1 , arg z = 1 4 π . {|z|=1, \; \arg z = \frac{1}{4} \pi.} ∣ z ∣ = 1 , arg z = 4 1 π . z 2 = i . {z^2 = \mathrm{i}.} z 2 = i . − 1 x + 2 + 3 x + 3 + − 2 x + 4 {\displaystyle \frac{- 1}{x+2} + \frac{3}{x+3} + \frac{- 2}{x+4}} x + 2 − 1 + x + 3 3 + x + 4 − 2 1 6 + 1 n + 3 − 2 n + 4 . {\displaystyle \frac{1}{6} + \frac{1}{n+3} - \frac{2}{n+4}.} 6 1 + n + 3 1 − n + 4 2 . 1 6 . {\displaystyle \frac{1}{6}.} 6 1 .
x = 0 , x = 4 a . {x=0, \; x=4a.} x = 0 , x = 4 a . y = 1 2 a . {y=\frac{1}{2a}.} y = 2 a 1 . Translation of 2 a {2a} 2 a x {x} x
e sin − 1 x = 1 + x + 1 2 x 2 + 1 3 x 3 + … {\mathrm{e}^{\sin^{-1}x} = 1 + x + \frac{1}{2} x^2 + \frac{1}{3} x^3 + \ldots} e s i n − 1 x = 1 + x + 2 1 x 2 + 3 1 x 3 + … − x + 2 z = − 3 {- x + 2 z = - 3} − x + 2 z = − 3 λ = − 1 , μ = 1. {\lambda = -1, \; \mu = 1.} λ = − 1 , μ = 1. ( 1 5 − 1 ) , ( − 3 3 1 ) . {\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}, \; \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}.} ⎝ ⎛ 1 5 − 1 ⎠ ⎞ , ⎝ ⎛ − 3 3 1 ⎠ ⎞ . Maximum point ( 1 4 π , 1 2 2 e 1 4 π ) {\displaystyle \left( \frac{1}{4} \pi, \; \frac{1}{2} \sqrt{2} \mathrm{e}^{\frac{1}{4} \pi} \right)} ( 4 1 π , 2 1 2 e 4 1 π )
π 8 ( e π − 3 ) . {\displaystyle \frac{\pi}{8} \left(\mathrm{e}^\pi - 3 \right).} 8 π ( e π − 3 ) . d N d t = k N . {\displaystyle \frac{\mathrm{d}N}{\mathrm{d}t} = kN.} d t d N = k N . N = A e k t . {N = A \mathrm{e}^{kt}.} N = A e k t . k = 1 2 ln 3. {k = \frac{1}{2} \ln 3.} k = 2 1 ln 3. 2 ln 10 ln 3 min. {\displaystyle \frac{2 \ln 10}{\ln 3} \textrm{ min.}} ln 3 2 ln 10 min. d N d t = k N − d . {\displaystyle \frac{\mathrm{d}N}{\mathrm{d}t} = kN - d.} d t d N = k N − d . t = 2 ln 3 ln ( N ln 3 − 2 d 100 ln 3 − 2 d ) . {\displaystyle t = \frac{2}{\ln 3} \ln \left( \frac{N \ln 3 - 2d}{100 \ln 3 - 2d} \right).} t = ln 3 2 ln ( 100 ln 3 − 2 d N ln 3 − 2 d ) . N = 2 ln 3 ( d + ( 50 ln 3 − d ) e t ln 3 2 ) . {\displaystyle N = \frac{2}{\ln 3} \Big( d + (50 \ln 3 - d) \mathrm{e}^{\frac{t\ln 3}{2}} \Big).} N = ln 3 2 ( d + ( 50 ln 3 − d ) e 2 t l n 3 ) . d > 50 ln 3 {d > 50 \ln 3} d > 50 ln 3 5.35 min. {5.35 \textrm{ min.}} 5.35 min. 2 α ( π − β + sin β ) . {2\alpha(\pi - \beta + \sin \beta).} 2 α ( π − β + sin β ) . β = 1.90. {\beta = 1.90.} β = 1.90. Greatest height = 22.9 m. {\textrm{Greatest height } = 22.9 \textrm{ m.}} Greatest height = 22.9 m. Least height = 15.1 m. {\textrm{Least height } = 15.1 \textrm{ m.}} Least height = 15.1 m. Solutions Full solutions (with working)
The actual questions are the copyright of UCLES and MOE. These answers are my own and any errors therein are mine alone.
Desmos was used to generate the sketch of the curves in Q2 and Q6.