UAS 23-24 FAD1014 Mathematics II
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UNIVERSITI MALAYA UNIVERSITI MALAYA
PEPERIKSAAN ASASI SAINS HAYAT / FIZIKAL EXAMINATION FOR FOUNDATION IN LIFE / PHYSICAL SCIENCES
SESI AKADEMIK 2023/2024 : SEMESTER 2 ACADEMIC SESSION 2023/2024 : SEMESTER 2
FAD 1014 : Matematik II Mathematics II
Mei/Jun 2024 Masa : 2 jam May/June 2024 Time : 2 hours
ARAHAN KEPADA CALON : INSTRUCTIONS TO CANDIDATES :
Kertas soalan ini dibahagikan kepada dua bahagian: Bahagian A dan Bahagian B. Calon dikehendaki menjawab semua soalan.
This question paper is divided into two parts: Part A and Part B. Candidate is required to answer all questions.
[University of Malaya crest/logo: shield with tiger motifs, hibiscus flower, and banner reading "ILMU PUNCA KEMAJUAN"]
(Kertas soalan ini mengandungi 8 soalan dalam 8 halaman yang dicetak) (This question paper consists of 8 questions on 8 printed pages)
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FAD 1014
FORMULA
Properties of Integrals: $$\int k f(u),du = k \int f(u),du$$ $$\int_a^a f(x),dx = 0$$ $$\int_a^c f(x),dx = \int_a^b f(x),dx + \int_b^c f(x),dx$$ $$\int [f(u) \pm g(u)],du = \int f(u),du \pm \int g(u),du$$ $$\int_a^b f(x),dx = -\int_b^a f(x),dx$$ $$\int u,dv = uv - \int v,du$$
Integration rules: $$\int du = u + C$$ $$\int u^n,du = \frac{u^{n+1}}{n+1} + C$$ $$\int \frac{du}{u} = \ln|u| + C$$ $$\int e^u,du = e^u + C$$ $$\int a^u,du = \frac{1}{\ln a} a^u + C$$ $$\int \sin u,du = -\cos u + C$$ $$\int \cos u,du = \sin u + C$$ $$\int \sec^2 u,du = \tan u + C$$ $$\int \csc^2 u,du = -\cot u + C$$ $$\int \csc u \cot u,du = -\csc u + C$$ $$\int \sec u \tan u,du = \sec u + C$$ $$\int \frac{du}{a^2+u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$$ $$\int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C$$ $$\int \frac{du}{u\sqrt{u^2-a^2}} = \frac{1}{a}\sec^{-1}\left(\frac{|u|}{a}\right) + C$$
Powers of Natural Numbers: $$\sum_{k=1}^{n} k = \frac{1}{2}n(n+1)$$ $$\sum_{k=1}^{n} k^2 = \frac{1}{6}n(n+1)(2n+1)$$ $$\sum_{k=1}^{n} k^3 = \frac{1}{4}n^2(n+1)^2$$
Binomial series: $$(a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2 + \frac{n(n-1)(n-2)}{3!}a^{n-3}x^3 + \cdots$$ $$= a^n + \binom{n}{1}a^{n-1}x + \binom{n}{2}a^{n-2}x^2 + \binom{n}{3}a^{n-3}x^3 + \cdots$$
Taylor and Maclaurin Series: $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \cdots + \frac{f^{(n-1)}(a)(x-a)^{n-1}}{(n-1)!}$$
Conic sections formulas: $$(x-h)^2 = 4a(y-k)$$ $$(y-k)^2 = 4a(x-h)$$ $$(x-h)^2 + (y-k)^2 = r^2$$ $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$ $$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$$ $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$ $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
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[Top right header:] FAD 1014
BAHAGIAN A / PART A
1. Nyatakan BENAR atau PALSU bagi pernyataan-pernyataan di bawah.
(a) Jika $\displaystyle\int \sec^{2} u , du = \tan u + c$, maka $\displaystyle\int \sec^{2} 2x , dx = \tan 2x + c$.
(b) Satu-satunya cara untuk menyelesaikan $\displaystyle\int \sin x \cos x , dx$ ialah dengan menggunakan penggantian $u = \sin x$.
(c) Cara terbaik untuk menyelesaikan $\displaystyle\int x \cos x , dx$ dengan menggunakan teknik pengamiran bahagian demi bahagian adalah dengan memilih $u = \cos x$ dan $dv = x$.
(d) Cara terbaik untuk menggunakan formula pengamiran bahagian demi bahagian adalah dengan memilih $u$ dan $dv$ yang mana $\displaystyle\int v , du$ adalah lebih mudah daripada $\displaystyle\int u , dv$.
(e) $\displaystyle\int_{1}^{2} \frac{2x^{3}}{\sqrt{x^{2}+1}},dx = \frac{2(2)^{3}}{\sqrt{2^{2}+1}} - \frac{2(1)^{3}}{\sqrt{1^{2}+1}}$
(f) Untuk menyelesaikan $\displaystyle\int \frac{2x-1}{(x-3)^{2}+4},dx$, pada mulanya, gunakan penggantian $x - 3 = 2 \tan \theta$.
State TRUE or FALSE for the statements below.
(a) If $\displaystyle\int \sec^{2} u , du = \tan u + c$, then $\displaystyle\int \sec^{2} 2x , dx = \tan 2x + c$.
(b) The only way to solve $\displaystyle\int \sin x \cos x , dx$ is by using substitution $u = \sin x$.
(c) The best way to solve $\displaystyle\int x \cos x , dx$ is by using integration by parts method via choosing $u = \cos x$ and $dv = x.dx$
(d) The best way to use integration by parts formula is by choosing $u$ and $dv$ such that $\displaystyle\int v , du$ is simpler than $\displaystyle\int u , dv$.
(e) $\displaystyle\int_{1}^{2} \frac{2x^{3}}{\sqrt{x^{2}+1}},dx = \frac{2(2)^{3}}{\sqrt{2^{2}+1}} - \frac{2(1)^{3}}{\sqrt{1^{2}+1}}$
(f) To solve $\displaystyle\int \frac{2x-1}{(x-3)^{2}+4},dx$ , first use substitution $x - 3 = 2 \tan \theta$
(6 markah/marks)
2. (a) Nyatakan peringkat dan darjah bagi persamaan pembeza berikut:
(i) $\displaystyle\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{4} + y = x$.
(ii) $\displaystyle\left(\frac{d^{3}y}{dx^{3}}\right)^{2} - 3\frac{d^{2}y}{dx^{2}} + 2\left(\frac{dy}{dx}\right)^{4} = y^{4}$.
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[Top right header:] FAD 1014
(b) Tulis sebutan ke-$k$, $a_{k}$, bagi jujukan / siri berikut:
(i) $\displaystyle 1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots$
(ii) $\displaystyle 1 + \frac{1}{2},; 2 + \frac{1}{3},; 3 + \frac{1}{4},; \dots$
(iii) $3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \cdots$
(a) State the order and degree for the following differential equations:
(i) $\displaystyle\frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{4} + y = x$.
(ii) $\displaystyle\left(\frac{d^{3}y}{dx^{3}}\right)^{2} - 3\frac{d^{2}y}{dx^{2}} + 2\left(\frac{dy}{dx}\right)^{4} = y^{4}$.
(b) Write the $k^{\text{th}}$ term, $a_{k}$, of the following sequences / series:
(i) $\displaystyle 1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots$
(ii) $\displaystyle 1 + \frac{1}{2},; 2 + \frac{1}{3},; 3 + \frac{1}{4},; \dots$
(iii) $3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \cdots$
(7 markah/marks)
3. Nyatakan sama ada pernyataan-pernyataan di bawah adalah BENAR atau PALSU. Jika PALSU, berikan justifikasi jawapan anda.
(a) Jumlah bagi $(1+x)^{n}$ adalah sah untuk semua nilai $x$, bagi $n$ ialah integer positif atau bukan positif.
(b) Untuk pengembangan binomial $(1+x)^{n}$, jika $n$ ialah integer positif, siri ini ialah siri terhingga di mana ia berakhir pada istilah $x^{n}$ dan jika $n$ ialah integer bukan positif, siri ini ialah siri tak terhingga dan menumpu.
(c) $(a+b)^{6}$ boleh dikembangkan sebagai
$(a+b)^{6} = \binom{6}{1}a^{6}b^{1} + \binom{6}{2}a^{5}b^{2} + \binom{6}{3}a^{4}b^{3} + \binom{6}{4}a^{3}b^{4} + \binom{6}{5}a^{2}b^{5} + \binom{6}{6}a^{1}b^{6}$ .
(d) Pekali sebutan adalah simetri dan mengikut corak dalam segitiga Pascal dalam pengembangan $(a+b)^{n}$ di mana $n$ ialah integer positif.
(e) Binomial $(1+x)^{n}$ boleh dikembangkan sebagai $(1+x)^{n} = 1 + nx + \dfrac{n(n-1)}{2!}x^{2} + \dfrac{n(n-1)(n-2)}{3!}x^{3} + \cdots$ bagi $n$ ialah integer bukan positif.
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