Kertas Model SPM Matematik Tambahan (Soalan dan Jawapan)

Kertas Model SPM Matematik Tambahan

Bahagian B

[40 markah]

Jawab mana-mana empat soalan daripada bahagian ini.

Soalan 8

Rajah 4 menunjukkan sebahagian daripada suatu lengkung x = y² + 2. Kecerunan garis lurus QR
ialah –1.

Cari

(a) persamaan garis lurus PQ, [2 markah]

(b) luas rantau berlorek, [4 markah]

Jawapan serta penyelesaian:

(a)

y– 2 = –1 (x – 6)

y= –x + 6 + 2

y = –x + 8

(b)

y= –x + 8, pada paksi-y, x= 0

y= 8

Titik R = (0, 8)

Luas rantau berlorek

= Luas bawah lengkung + luas segi tiga

\begin{array}{l} = \int_{0}^{2} x d y + \frac{1}{2} (8 - 2) (6) \\ = \int_{0}^{2} (y^{2} + 2) d y + \frac{1}{2} (6) (6) \\ = {[\frac{y^{3}}{3} + 2 y]}_{0}^{2} + 18 \\ = [\frac{2^{3}}{3} + 2 (2) - 0] + 18 \\ = \frac{20}{3} + 18 \\ = 24 \frac{2}{3} u n i t^{2} \end{array}

(c)

Isipadu kisaran

= Isipadu bawah lengkung + Isipadu kon

\begin{array}{l} = π \int_{0}^{2} x^{2} d y + \frac{1}{3} π j^{2} t \\ = π \int_{0}^{2} {(y^{2} + 2)}^{2} d y + \frac{1}{3} π {(6)}^{2} (6) \\ = π \int_{0}^{2} (y^{4} + 4 y^{2} + 4) d y + 72 π \\ = π {[\frac{y^{5}}{5} + \frac{4 y^{3}}{3} + 4 y]}_{0}^{2} + 72 π \\ = π [\frac{2^{5}}{5} + \frac{4 {(2)}^{3}}{3} + 4 (2) - 0] + 72 π \\ = π [\frac{32}{5} + \frac{32}{3} + 8] + 72 π \\ = \frac{376}{15} π + 72 π \\ = 97 \frac{1}{15} π {unit}^{3} \end{array}