Soal Latihan PAT Kelas XI - Matematika Wajib

💦Soal No.1

\(\mathop {\lim }\limits_{h \to 0} \frac{{{\textstyle{t \over {x + h}}} - {\textstyle{1 \over x}}}}{h}\) merupakan definisi turunan dari fungsi f(x) = ...

(A) \(\frac{1}{x}\)
(B) x2
(C) - x
(D) -\(\frac{1}{x}\)
(E) x

Pembahasan :

Rumus:
\({f^1}\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x
\right)}}{h}\)
maka
\(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {\frac{1}{{x + h}}} \right) - f\left( {\frac{1}{x}}
\right)}}{h} = {f^1}\left( x \right)\)
dimana
\({f^1}\left( x \right) = - {x^{ - 2}} = \frac{1}{{{x^2}}}\)

💥Kunci Jawaban : A

💦Soal No.2

Jika \(f\left( x \right) = \sqrt x + \sqrt[3]{{{x^2}}}\) maka f'(x) = ...

(A) \(\frac{2}{{\sqrt x }} + \frac{3}{{\sqrt[3]{x}}}\)

(B) \(\frac{1}{{2\sqrt x }} + \frac{2}{{3\sqrt[{}]{x}}}\)

(D) \(\frac{1}{{2\sqrt x }} + \frac{2}{{3\sqrt[3]{x}}}\)

(E) \(\frac{1}{{2\sqrt x }}\_ + \frac{2}{{3\sqrt[{}]{{{x^2}}}}}\)

Pembahasan :

Rumus:
 \)y = c \to {y^1} = 0\)
 \(y = a{x^n} \to {y^1} = an{x^{n - 1}}\)
 \(y = {\left( {f\left( x \right)} \right)^n} \to {y^1} = n \cdot {\left( {f\left( x \right)} \right)^{n -
1}} \cdot {f^1}\left( x \right)\)
 \(y = u \times v \to {y^1} = {u^1} \times v + u \times {v^1}\)
 \(y = \frac{u}{v} \to {y^1} = \frac{{{u^1} \times v - u \times {v^1}}}{{{v^2}}}\)
\(f\left( x \right) = \sqrt x + \sqrt[3]{{{x^2}}}\)
\(\begin{array}{l}f\left( x \right) = {x^{1/2}} + {x^{1/2}}\\{f^1}\left( x \right) = \frac{1}{2}{x^{1/2}} + \frac{2}{3}{x^{ - \frac{1}{3}}}\end{array}\)
\( = \frac{1}{{2\sqrt x }} + \frac{2}{{3\sqrt[3]{x}}}\)

💥Kunci Jawaban : B

💦Soal No.3

Jika \(f\left( x \right) = \frac{{{x^2} - 2x + 1}}{{\sqrt x }}\) maka f'(x) = ...

(A) \(\frac{3}{2}\sqrt x + \frac{1}{{\sqrt x }} - \frac{2}{{x\sqrt x }}\)

(B) \(\frac{3}{2}x\sqrt x - \frac{1}{{x\sqrt x }} - \frac{1}{{2{x^2}\sqrt x }}\)

(D) \(\frac{3}{2}\sqrt x - \frac{1}{{x\sqrt x }} + \frac{1}{{2{x^2}\sqrt x }}\)

(E) \(\frac{3}{2}\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{2x\sqrt x }}\)

Pembahasan :

\(f\left( x \right) = \frac{{{x^2} - 2x + 1}}{{\sqrt x }}\)
\(\begin{array}{l} = \frac{{{x^2}}}{{{x^{1/2}}}} - \frac{{2x}}{{{x^{1/2}}}} + \frac{1}{{{x^{1/2}}}}\\ = {x^{2\frac{1}{2}}} - 2{x^{1 - 1/2}} + {x^{ - 1/2}}\end{array}\)
\( = {x^{\frac{3}{2}}} - 2{x^{\frac{1}{2}}} + {x^{ - \frac{1}{2}}}\)
\({f^1}\left( x \right) = \frac{3}{2}{x^{\frac{1}{2}}} - {x^{\frac{1}{2}}} + {x^{ - \frac{1}{2}}}\)
\( = \frac{3}{2}\sqrt x - \frac{1}{{\sqrt x }} - \frac{1}{{2x\sqrt x }}\)

💥Kunci Jawaban : E

💦Soal No.4

Jika \(f\left( x \right) = ax + b{x^2}\) dan \(f'(1) = 5\) dan \(f'(2) = 7\) maka nilai a+b = ....

(A) 25

(B) 16

(D) 8

(E) 4

Pembahasan :
\(f\left( x \right) = ax + b{x^2}\)
\({f^1}\left( x \right) = a + 2bx\)
\({f^1}\left( x \right) = 5 \equiv a + 2b...\left( 1 \right)\)
\({f^1}\left( 2 \right) = 7 \equiv a + 4b...\left( 2 \right)\)
eliminasi
\(\frac{\begin{array}{l}2a + a = 5\\4b + a = 7\end{array}}{{2b = 2 \to b = 1}} - \)
\( \Rightarrow 2b + a = 5\)
a = 5 – 2 = 3
maka a + b = 1 + 3 = 4
💥Kunci Jawaban : C

💦Soal No.5

Jika \(y = \frac{{x - 1}}{{{x^2} + 1}}\) maka \(\frac{{dy}}{{dx}}\) = ....

(A) \(\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)

(B) \(\frac{{2{x^2} - {x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)

(D) \(\frac{{ - 3{x^3}}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)

(E) \(\frac{{ - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)

Pembahasan :
\(y = \frac{{x - 1}}{{{x^2} + 1}} \equiv \frac{u}{v}\)
 \(u = x - 1 \to {u^1} = 1\)
 \(v = {x^2} + 1 \to {v^1} = 2x\)
maka
\({y^1} = \frac{{1\left( {{x^2} + 1} \right) - \left( {x - 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 1}
\right)}^2}}}\)
\({y^1} = \frac{{{x^2} + 1 - 2{x^2} + 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)
\({y^1} = \frac{{ - {x^2} + 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)
💥Kunci Jawaban : C

💦Soal No.6

Fungsi \(f(x) = \frac{{{x^2} - 3x}}{{{x^2} + 2x + 1}}\) maka f'(2) = ...

(A) \( - \frac{2}{9}\)

(B) \(\frac{1}{9}\)

(D) \(\frac{7}{{27}}\)

(E) \(\frac{{ - 1}}{{27}}\)

Pembahasan :
\(f\left( x \right) = \frac{{{x^2} - 3x}}{{{x^2} + 2x + 1}} = \frac{u}{v}\)
 \({u^1} = 2x - 3\)
 \({v^1} = 2x + 2\)
\({f^1}\left( x \right) = \frac{{\left( {2x - 3} \right)\left( {{x^2} + 2x + 1} \right) - \left( {{x^2} - 3x}
\right)\left( {2x + 2} \right)}}{{{{\left( {{x^2} + 2x + 1} \right)}^2}}}\)
\({f^1}\left( 2 \right) = \frac{{\left( {4 - 3} \right)\left( {4 + 4 + 1} \right) - \left( {4 - 6} \right)\left(
{4 + 2} \right)}}{{{{\left( {4 + 4 + 1} \right)}^2}}}\)
\( = \frac{{9 + 12}}{{{9^2}}} = \frac{{21}}{{81}} = \frac{7}{{27}}\)
💥Kunci Jawaban : E

💦Soal No.7

Diketahui f(x) = \(\frac{1}{{{{\left( {1 - {x^2}} \right)}^2}}}\) maka= ...

(A) \(\frac{{4x}}{{{{\left( {1 - {x^2}} \right)}^3}}}\)

(B) \(\frac{{ - 2x}}{{{{\left( {1 - {x^2}} \right)}^3}}}\)

(D) \(\frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^3}}}\)

(E) \(\frac{{ - 4x}}{{{{\left( {1 - {x^2}} \right)}^3}}}\)

Pembahasan :
\(f\left( x \right) = \frac{1}{{{{\left( {1 - {x^2}} \right)}^2}}} = {\left( {1 - {x^2}} \right)^{ - 2}}\)
\({f^1}\left( x \right) = - 2{\left( {1 - {x^2}} \right)^{ - 3}} \cdot \left( { - 2x} \right)\)
\( = \frac{{4x}}{{{{\left( {1 - {x^2}} \right)}^3}}}\)
💥Kunci Jawaban : A

💦Soal No.8

Jika \(f(2) = 3,{\rm{ }}{f^{\,/}}(2) = 6,{\rm{ }}g(2) = 1,{\rm{ }}g{\,^/}(2) = 4\) dan

\({\rm{h(x) = }}\frac{{f(x)g(x)}}{{f(x) - g(x)}}\) maka h'(2) = ...

(A) \(\frac{{15}}{2}\)

(B) 6

(D) 9

(E) 12

Pembahasan :
\(h\left( x \right) = \frac{{f\left( x \right) \cdot g\left( x \right)}}{{f\left( x \right) - g\left( x
\right)}} = \frac{u}{v}\)
\({h^1}\left( x \right) = \frac{{\left( {{f^1}\left( x \right) \cdot {g^1}\left( x \right)} \right)\left(
{f\left( x \right) - g\left( x \right)} \right) - \left( {f\left( x \right) \cdot g\left( x \right)} \right)\left(
{{f^1}\left( x \right) - {g^1}\left( x \right)} \right)}}{{{{\left( {f\left( x \right) - g\left( x \right)}
\right)}^2}}}\)
\({h^1}\left( 2 \right) = \frac{{\left( {{f^1}\left( 2 \right) \cdot {g^1}\left( 2 \right)} \right)\left(
{f\left( 2 \right) - g\left( 2 \right)} \right) - \left( {f\left( 2 \right) \cdot g\left( 2 \right)}
\right)\left( {{f^1}\left( 2 \right) - {g^1}\left( 2 \right)} \right)}}{{{{\left( {f\left( 2 \right) - g\left( 2
\right)} \right)}^2}}}\)
\({h^1}\left( 2 \right) = \frac{{\left( {6 \cdot 4} \right)\left( {3 - 1} \right) - \left( {3 \cdot 1}
\right)\left( {6 - 4} \right)}}{{{{\left( {3 - 1} \right)}^2}}}\)
\( = \frac{{48 - 6}}{4} = \frac{{42}}{2} = \frac{{21}}{2}\)
💥Kunci Jawaban : A

💦Soal No.9

Jika \(y = \frac{{a{x^2} + bx + c}}{{a{x^2} + bx}}\) dengan a, b dan c konstan, maka \(\frac{{dy}}{{dx}} = ...\)

(A) \(\frac{{2acx + bc}}{{{{\left( {a{x^2} + bx} \right)}^2}}}\)

(B) \(\frac{{ - 2acx - bc}}{{{{\left( {a{x^2} + bx} \right)}^2}}}\)

(D) \(\frac{{2acx + bc}}{{a{x^2} + bx}}\)

(E) \(\frac{{2abc}}{{{{\left( {a{x^2} + bx} \right)}^2}}}\)

Pembahasan :
\(y = \frac{{a{x^2} + bx + c}}{{a{x^2} + bx}} \equiv \frac{u}{v}\)
\(\frac{{dy}}{{dx}} = \frac{{\left( {2ax + b} \right)\left( {a{x^2} + bx} \right) - \left( {a{x^2} + bx + c}
\right)\left( {2ax + b} \right)}}{{{{\left( {a{x^2} + bx} \right)}^2}}}\)
\( = \frac{{\left( {2ax + b} \right)\left( {a{x^2} + bx - a{x^2} - bx - c} \right)}}{{{{\left( {a{x^2} + bx}
\right)}^2}}}\)
\( = \frac{{ - 2acx - bc}}{{{{\left( {a{x^2} + bx} \right)}^2}}}\)
💥Kunci Jawaban : B

💦Soal No.10

Jika f(x) = \(\sqrt[3]{{{x^2} + 2}}\) maka f'(x) = ...

(A) \(\frac{{ - 2x}}{{3\sqrt {{x^2} + 2} }}\)

(B) \(\frac{{ - 2x}}{{3\sqrt[3]{{{{\left( {{x^2} + 2} \right)}^2}}}}}\)

(D) \(\frac{{3x}}{{3\sqrt[3]{{{{\left( {{x^2} + 2} \right)}^2}}}}}\)

(E) \(\frac{{ - 3x}}{{3\sqrt[3]{{{{\left( {{x^2} + 2} \right)}^2}}}}}\)

Pembahasan :

\(f\left( x \right) = {\left( {{x^2} + 2} \right)^{1/3}}\)
\({f^1}\left( x \right) = \frac{1}{3}{\left( {{x^2} + 2} \right)^{ - \frac{2}{3}}} \cdot \left( {2x}
\right)\)
\( = \frac{{2x}}{{3\sqrt[3]{{{{\left( {{x^2} + 2} \right)}^2}}}}}\)

💥Kunci Jawaban : C

💦Soal No.11

Jika \(y = {U^2} + 2\) dan \(U = {x^3} + 2{x^2} + x - 1\) maka \(\frac{{dy}}{{dx}} = ....\)

(A) \(2\left( {{x^3} + 2{x^2} + x - 1} \right)\)

(B) \( - 2\left( {3{x^2} + 4x + 1} \right)\left( {{x^3} + 2{x^2} + x + 1} \right)\)

(D) \(2\left( {3{x^2} + 4x + 1} \right)\left( {{x^3} + 2{x^2} + x - 1} \right)\)

(E) 2U

Pembahasan :
 \(y = {u^2} + 2\)
\(\frac{{dy}}{{du}} = 2u\)
 \(U = {x^3} + 2{x^2} + x - 1\)
\(\frac{{du}}{{dx}} = 3{x^2} + 4x + 1\)
\( \Rightarrow \frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\)
\(\frac{{dy}}{{dx}} = \left( {2u} \right)\left( {3{x^2} + 4x + 1} \right)\)
\( = 2\left( {3{x^2} + 4x + 1} \right)\left( {{x^3} + 2{x^3} + x - 1} \right)\)
💥Kunci Jawaban : D

💦Soal No.12

Persamaan garis singgung kurva \(y = {x^2} + 2x - 1\) pada titik berabsis 2 adalah ...

(A) y = 11x - 14

(B) y = 2x + 11

(D) y = 6x + 5

(E) y = 6x - 5

Pembahasan :
\(y = f\left( x \right) = {x^2} + 2x - 1\)
 \(y = f\left( 2 \right) = {2^2} + 2\left( 2 \right) - 1\)
\( = 4 + 4 - 1 = 7\)
titik singgung (2, 7)
 \(m = {y^1} = 2x + 2 = 4 + 2 = 6\)
= persamaan garis singgungnya:
\(y - {y_1} = m\left( {x - {x_1}} \right)\)
\(y - 7 = 6\left( {x - 2} \right)\)
\(y = 6x - 12 + 7\)
\(y = 6x - 5\)
💥Kunci Jawaban : E

💦Soal No.13

Persamaan garis singgung kurva \(y = {x^2} + x - 1\) pada titik berordinat 5 adalah ....

(A) y = -5x - 11

(B) y = -5x + 19

(D) y = 5x - 6 dan y = -5x + 19

(E) y = 5x - 5 dan y = -5x - 10

Pembahasan :
\(y = f\left( x \right) = {x^2} + x - 1\)
 ordinat = 5
\(5 = {x^2} + x - 1\)
\({x^2} + x - 6 = 0\)
\(\left( {x - 2} \right)\left( {x + 3} \right) = 0\)
titik singgungnya ada dua (-3, 5) dan (2, 5)
 gradien =
= persamaan garis I
m = 2 (-3) + 1 = -5
y – 5 = -5 (x + 3)
y = -5x – 10
= persamaan garis II
m = 2 (2) + 1 = 5
y – 5 = 5 (x – (-2))
y – 5 = 5x – 10
y = 5x – 5
💥Kunci Jawaban : E

💦Soal No.14

Garis y = 4x + 1 menyinggung kurva \(y = a{x^2} + bx\) pada titik berabsis 2. Nilai b yang memenuhi adalah ...

(A) 2

(B) 3

(D) 1

(E) 4

Pembahasan :
y = 4x + 1 menyinggung \(y = a{x^2} + bx\) di titik yang berabsis ( x = 2) maka:
 y = 4 (2) + 1 = 9
jadi titik singgungnya (2, 9)
 gradient y = 4x + 1 adalah 4, gradient garis singgung = y1
2ax + b = 4 … (1)
4a + b = 4 …. (2)
 (2, 9) : 9 = a (2)2 + b (2)
9 = 4a + 2b …. (2)
eliminasi
\(\frac{\begin{array}{l}4a + 2b = 9\\4a + b = 4\end{array}}{{b = 5}} - \)
💥Kunci Jawaban : C

💦Soal No.15

Jika garis y = x+3 menyinggung kurva \(y = {x^2} - 3x + 7\) maka koordinat titik singgung nya adalah ...

(A) (2,7)

(B) (1,5)

(D) (2,5)

(E) (1,4)

Pembahasan :
.
 gradient garis y = x + 3 adalah 1
 gradient garis singgung (m)
\(m = {y^1} = 2x - 3\)
= 1 = 2x – 3 \( \to \) x = 2
y = f (x) = f (2) = 2 + 3 = 5
jadi titik singgungnya (2, 5)
💥Kunci Jawaban : D

💦Soal No.16

Grafik fungsi \(y = \frac{2}{3}{x^3} - 3{x^2} + 4x + 2\) akan naik pada interval ...

(A) 2 < x < 4

(B) x < 2 atau x > 4

(D) x < 1 atau x > 2

(E) -2 < x < 4

Pembahasan :
\(y = \frac{2}{3}{x^3} - 3{x^2} + 4x + 2\)
akan naik pada \({y^1} > 0\)
\(2{x^2} - 6x + 4 > 0\)
\({x^2} - 3x + 2 > 0\)
\(\left( {x - 2} \right)\left( {x - 1} \right) > 0\)

HP = {x < 1 atau x > 2}
💥Kunci Jawaban : D

💦Soal No.17

Grafik fungsi \(y = {x^3} - 3{x^2} - 72x + 1\) akan turun pada interval ...

(A) a < -6 atau x > 4

(B) -6 < x < 4

(D) 4 < x < 6

(E) -4 < x < 6

Pembahasan :
\(y = {x^3} - 2{x^2} - 7x + 1\)
akan turun pada \({y^1} < 0\)
\(3{x^2} - 4x - 7 < 0\)
\(\left( {3x - 7} \right)\left( {x + 1} \right) < 0\)

HP = \(\left\{ { - 1 < x < \frac{7}{3}} \right\}\)
💥Kunci Jawaban : E

💦Soal No.18

Grafik fungsi \(y = {x^3} - 3{x^2} - 9x\) dan fungsi \(y = {x^3} - 12{x^2} + 1\) akan sama-sama turun pada interval ...

(A) 1 < x < 2

(B) 2 < x < 3

(D) -2 < x < 3

(E) -1 < x < 2

Pembahasan :
grafik fungsi turun:
 \(y = {x^3} - 3{x^3} - 9x\)
\({y^1} \equiv 3{x^2} - 6x - 9 < 0\)
\({x^2} - 2x - 3 < 0\)
\(\left( {x - 3} \right)\left( {x + 1} \right) < 0\)
\(1 - 1 < x < 3...\left( 1 \right)\)
 \(y = {x^3} - 12x + 1\)
\({y^1} \equiv 3{x^2} - 12x < 0\)
\(3x\left( {x - 4} \right) < 0\)
\(0 < x < 4...\left( 2 \right)\)
Irisannya

HP = {0 < x < 3}
💥Kunci Jawaban : C

💦Soal No.19

Agar grafik fungsi \(y = {x^3} - 3{x^2} - ax\) naik pada interval x < -2 atau x > 4 maka nilai a harus sama dengan ...

(A) 4

(B) 24

(D) 48

(E) 8

Pembahasan :
\(y = {x^3} - 3{x^2} - ax\)
\({y^1} > 0\)
\(3{x^2} - 6x - a > 0\)
penyelesaiannya adalah x < -2 atau x > 4 artinya akar dari \(3{x^2} - 6x - a\) adalah x = -2 dan x =
4
(x + 2) (x – 4) > 0
\({x^2} - 2x - 8 > 0\) (kali 3)
\(3{x^2} - 6x - 24 > 0\) maka a = + 24
💥Kunci Jawaban : B

💦Soal No.20

Nilai minimum dari \(f(x) = \frac{1}{3}{x^3} + {x^2} + x + 5\) dalam interval \(2 \le \times \le 4\) adalah ...

(A) \(46\frac{1}{3}\)

(B) \(13\frac{2}{3}\)

(D) \(4\frac{2}{3}\)

(E) \(4\frac{1}{3}\)

Pembahasan :
\(f\left( x \right) = \frac{1}{3}{x^3} + {x^2} + x + 5\)
 \({f^1}\left( x \right) = 0\)
\({x^2} + 2x + 1 = 0\)
\(\left( {x + 1} \right)\left( {x + 1} \right) = 0\)
\(x = - 1\)
 nilai minimum terjadi saat nilai x = -1 yaitu f (-1)
\(f\left( { - 1} \right) = \frac{1}{3}{\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1}
\right) + 5\)
\( = - \frac{1}{3} + 5 = 4\frac{2}{3}\)
💥Kunci Jawaban : B

💦Soal No.21

Perhatikan gambar berikut!

Luas daerah yang digambar akan mencapai maksimum Jika koordinat titik R adalah:

(A) (1,\(\frac{2}{5}\))

(B) (1,\(\frac{2}{5}\))

(D) (\(\frac{1}{2}\),15)

(E) (\(\frac{1}{2}\),\(\frac{15}{4}\))

Pembahasan :
persamaan garis tersebut adalah :
5x + 2y = 10
2y = 10 – 5x
y = \(5 - \frac{5}{2}x\)
persegi panjang tersebut memiliki panjangnya x dan lebarnya y = \(5 - \frac{5}{2}x\) maka
luasnya = p x l
\(\alpha = x\left( {5 - \frac{5}{2}x} \right)\)
\(f\left( x \right) = 5x - \frac{5}{2}{x^2}\)
\({f^1}\left( x \right) \equiv 5 - 5x = 0\)
\(5x = 5\)
\(x = 1\)
\(f\left( x \right) = f\left( 1 \right) = 5 - \frac{5}{2} = \frac{5}{2}\)
jadi luas maksimumnya adalah \(\frac{5}{2}\), dan titik R (1, \(\frac{5}{2}\)
💥Kunci Jawaban : B

💦Soal No.22

Dua kandang berdampingan masing-masing dengan ukuran x m dan y m dan luasnya 12 m2. Agar panjang pagar yang diperlukan sesedikit mungkin, maka panjang x dan y berturut-turut adalah ...

(A) 2 m dan 6 m

(B) 6 m dan 2 m

(D) 3 m dan 4 m

(E) \(2\sqrt 3 \)m dan \(3\sqrt 2 \)m

Pembahasan :
Dari gambar tersebut panjang pagar yang dibutuhkan adalah 3x + 4y = p
 luas persegi panjangnya = 12
xy = 12
y = \(\frac{{12}}{x}\)
\( \Rightarrow p = 3x + 4\left( {\frac{{12}}{x}} \right)\)
\(f\left( x \right) = 3x + 48{x^{ - 1}}\)
\({f^1}\left( x \right) = 0\)
\(3 - 48{x^{ - 2}} = 0\)
\(3 = \frac{{48}}{{{x^2}}}\)
\({x^2} = 16 \to x = 4\)
\(y = \frac{{12}}{x} = \frac{{12}}{4} = 3\)
💥Kunci Jawaban : D

💦Soal No.23

Jika suatu proyek dikerjakan dalam x hari maka biaya proyek perhari menjadi \(3x + \frac{{1.200}}{x} - 60\) ribu rupiah. Biaya proyek minim adalah ...

(A) 1.200 ribu rupiah

(B) 900 ribu rupiah

(D) 750 ribu rupiah

(E) 720 ribu rupiah

Pembahasan :
 biaya total proyek adalah
\(B\left( x \right) = x\left( {3x + \frac{{1200}}{x} - 60} \right)\)
\( = 3{x^2} + 1200 - 60x\)
 \({B^1}\left( x \right) = 0\)
\(6x - 60 = 0 \to x = 10\)
maka biaya total minimum adalah
\(B\left( {10} \right) = 3\left( {{{10}^2}} \right) + 1200 - 60\left( {10} \right)\)
\( = 300 + 1200 - 600\)
= 900
💥Kunci Jawaban : B

💦Soal No.24

\(\int {\frac{2}{{x\sqrt[3]{{{x^2}}}}}} dx = \)...

(A) \(\frac{3}{{\sqrt[3]{{{x^2}}}}} + c\)

(B) \(\frac{{ - 3}}{{\sqrt[3]{{{x^2}}}}} + c\)

(D) \(3\sqrt[3]{{{x^2}}} + c\)

(E) \( - 3\,\sqrt[3]{{{x^2}}} + c\)

Pembahasan :

\(\int {\frac{2}{{x\sqrt[3]{{{x^2}}}}}dx} \)
\( = \int {\frac{2}{{x \cdot {x^{2/3}}}}} dx = \int {\frac{2}{{{x^{5/3}}}}} dx\)
\( = \int {2{x^{ - 5/2}}} dx = \frac{2}{{ - \frac{5}{3} + 1}}{x^{ - \frac{5}{3} + 1}} + c\)
\( = \frac{2}{{ - \frac{2}{3}}}{x^{ - \frac{2}{3}}} + c = - 3{x^{ - \frac{2}{3}}} + c\)
\( = \frac{{ - 3}}{{{x^{2/3}}}} + c = \frac{{ - 3}}{{\sqrt[3]{{{x^2}}}}} + c\)

💥Kunci Jawaban : B

💦Soal No.25

\(\int {{{(x - \frac{1}{x})}^2}dx} \) = ....

(A) \(\frac{1}{3}{x^3} - 2x + \frac{1}{x} + c\)

(B) \(\frac{1}{3}{x^3} - 2x - \frac{1}{x} + c\)

(D) \(\frac{1}{3}{x^3} + 2x - \frac{1}{x} + c\)

(E) \(\frac{1}{3}{\left( {x + \frac{1}{x}} \right)^3} + c\)

Pembahasan :
\({\int {{{\left( {x - \frac{1}{x}} \right)}^2}dx = \int {{x^2} - 2\left( x \right)\left( {\frac{1}{x}}
\right)} + \left( {\frac{1}{x}} \right)} ^2}dx\)
\( = \int {{x^2} - 2 + {x^{ - 2}}dx} \)
\( = \frac{1}{3}{x^2} - 2x + \frac{1}{{ - 1}}{x^{ - 1}} + c\)
\( = \frac{1}{3}{x^3} - 2x - \frac{1}{x} + c\)
💥Kunci Jawaban : B

💦Soal No.26

Sebuah kurva mempunyai persamaan y = F(x). Jika \({F^/}(x) = 3{x^2} + 2\) dan titik (2, 5) terletak pada kurva y = F(x), maka rumus fungsi y = F(x) adalah ...

(A) \({x^3} + 2x - 7\)

(B) \({x^3} + 2x - 6\)

(D) \(3{x^3} + 2x - 32\)

(E) \({x^3} + 2x - 17\)

Pembahasan :
 \(f\left( x \right) = \int {{f^1}\left( x \right)dx} \)
\(f\left( x \right) = \int {3{x^2} + 2dx} \)
\( = {x^3} + 2x + c\)
 kurva melalui titik (2, 5)
\(5 = {2^3} + 2\left( 2 \right) + c\)
\(5 = 12 + c \to c = - 7\)
maka \(f\left( x \right) = {x^3} + 2x + \left( { - 7} \right)\)
\(f\left( x \right) = {x^3} + 2x - 7\)
💥Kunci Jawaban : A

💦Soal No.27

Sebuah kurva melalui titik (3, 30) dan mempunyai gradien garis singgung di setiap titik pada kurva tersebut dinyatakan dengan 6x + 4. Rumus fungsi kurva tersebut adalah ...

(A) F(x) = \(3{x^2}\) + 4x + 2

(B) F(x) = \(3{x^2}\) + 4x – 2

(D) F(x) = \(3{x^2}\) + 4x + 9

(E) F(x) = \(3{x^2}\) – 4x – 2

Pembahasan :
 gradient garis singgung
\(\left( m \right) = {f^1}\left( x \right) = 6x + 4\)
\(f\left( x \right) = \int {6x + 4dx} \)
\(f\left( x \right) = 3{x^2} + 4x + c\)
 Kurva f (x) melalui titik (3, 30)
\(30 = 3{\left( 3 \right)^2} + 4\left( 3 \right) + c\)
\(30 = 27 + 12 + c\)
\(c = - 9\)
\( \Rightarrow f\left( x \right) = 3{x^2} + 4x - 9\)
💥Kunci Jawaban : C

💦Soal No.28

\(\int {\frac{{4x}}{{\sqrt[3]{{{x^2} - 1}}}}dx} = ...\)

(A) - 6 \(\sqrt[3]{{{{({x^2} - 1)}^2}}} + c\)

(B) - 3 \(\sqrt[3]{{{{({x^2} - 1)}^2}}} + c\)

(D) \(\frac{4}{3}\) \(\sqrt[3]{{{{({x^2} - 1)}^2}}} + c\)

(E) 3 \(\sqrt[3]{{{{({x^2} - 1)}^2}}} + c\)

Pembahasan :

\(\int {\frac{{4x}}{{\sqrt[3]{{{x^2} - 1}}}}dx = \int {4x{{\left( {{x^2} - 1} \right)}^{ - 1/3}}dx} } \)
misal \(U = {x^2} - 1 \to du = 2xdx\)
\( \Rightarrow \int {{{\left( {{x^2} - 1} \right)}^{ - \frac{1}{3}}} \cdot 2 \cdot 2xdx} \)
\( = \int {{U^{ - \frac{1}{3}}}} \cdot 2 \cdot du = \int {2{U^{ - \frac{1}{3}}}} du\)
\( = \frac{2}{{ - \frac{1}{3} + 1}}{U^{ - \frac{1}{3} + 1}} + c\)
\( = \frac{2}{{\frac{2}{3}}}{U^{\frac{2}{3}}} + c = 3\sqrt[3]{{{U^2}}} + c\)
\( = 3\sqrt[3]{{{{\left( {{x^2} - 1} \right)}^2}}} + c\)

💥Kunci Jawaban : E

💦Soal No.29

Hasil dari \(\int\limits_0^4 {x\sqrt {{x^2} + 9} } \,dx\)

(A) \(12\frac{2}{3}\)

(B) \(18\frac{2}{3}\)

(D) \(32\frac{2}{3}\)

(E) \(33\frac{2}{3}\)

Pembahasan :

\(\int\limits_0^4 {x\sqrt {{x^2} + 9} dx} = \int\limits_0^4 {{{\left( {{x^2} + 9} \right)}^{1/2}}} \times dx\)
misal \({x^2} + 9 = U \to du = 2 \times dx\)
\( \Rightarrow \int\limits_0^4 {\frac{1}{2}} \cdot {\left( {{x^2} + 9} \right)^{1/2}} \cdot 2 \times dx = \int\limits_0^4 {\frac{1}{2}} \,{U^{1/2}}du\)
\( = \left. {\frac{{\frac{1}{2}}}{{1 + \frac{1}{2}}}{U^{\frac{1}{2} + 1}}} \right]_0^4 = \left.
{\frac{1}{3}{U^{\frac{3}{2}}}} \right]_0^4\)
\( = \left. {\frac{1}{3}{{\left( {{x^2} + 9} \right)}^{3/2}}} \right]_0^4\)
\( = \left( {\frac{4}{3}{{\left( {16 + 9} \right)}^{\frac{3}{2}}}} \right) - \left( {\frac{4}{3}{{\left( {0 +
9} \right)}^{\frac{3}{2}}}} \right)\)
\( = \frac{1}{3}{\left( {{5^2}} \right)^{\frac{3}{2}}} - \frac{1}{3}{\left( {{3^2}} \right)^{\frac{3}{2}}}\)
\( = \frac{1}{3}\left( {125} \right) - \frac{1}{3}\left( {27} \right)\)
\( = \frac{1}{3}\left( {125 - 27} \right) = \frac{1}{3}\left( {98} \right)\)
\( = 32\frac{2}{3}\)

💥Kunci Jawaban : D

💦Soal No.30

Hasil dari \(\int {x\sqrt {2x + 1} \,dx} = ...\)

(A) \(\frac{x}{3}{(2x + 1)^{\frac{3}{2}}} - \frac{1}{{30}}{(2x + 1)^{\frac{5}{2}}} + c\)

(B) \(\frac{x}{3}{(2x + 1)^{ - \frac{1}{2}}} - \frac{1}{5}{(2x + 1)^{\frac{1}{2}}} + c\)

(D) \(\frac{2}{3}{(2x + 1)^{\frac{3}{2}}} - \frac{1}{{30}}{(2x + 1)^{\frac{1}{2}}} + c\)

(E) \(\frac{x}{3}{(2x + 1)^{\frac{3}{2}}} - \frac{1}{{15}}{(2x + 1)^{\frac{5}{2}}} + c\)

Pembahasan :

\(\int {x\sqrt {2x + 1} dx} \)
misal 2x + 1 = U
2x = u – 1
\(x = \frac{1}{2}u - \frac{1}{2}\)
\(dx = \frac{1}{2}du\)
\( \Rightarrow \int {\left( {\frac{1}{2}u - \frac{1}{2}} \right){U^{\frac{1}{2}}}} \frac{1}{2}du\)
\( = \int {\frac{1}{4}{U^{\frac{3}{2}}} - \frac{1}{4}{U^{\frac{1}{2}}}} du\)
\( = \frac{{\frac{1}{4}}}{{\frac{5}{2}}}{U^{\frac{5}{2}}} -
\frac{{\frac{1}{4}}}{{\frac{3}{2}}}{U^{\frac{3}{2}}} + c\)
\( = \frac{1}{{10}}{U^{\frac{5}{2}}} - \frac{1}{6}{U^{\frac{3}{2}}} + c\)
\( = \frac{1}{{10}}{\left( {2x + 1} \right)^{\frac{5}{2}}} - \frac{1}{6}{\left( {2x + 1}
\right)^{\frac{3}{2}}} + c\)

💥Kunci Jawaban : C