Kunci Matematika Kelas Xi Sma Persiapan Ujian Akhir Semester 2

KUNCI PERSIAPAN UJIAN AKHIR SEMESTER 2 SMA KELAS XI MATEMATIKA

1.

Jawaban: c

− 4 = .2( 5) 2 − 4 = 50 − 4 = 46

lim lim 2x 2 x →5

Pembahasan: 2.

Jawaban: c

lim lim

x →3

−9 ( x - 3) ( x + 3) x +3 3+3 6 = lim = = = lim lim lim − x − 6 x →3 ( x − 3) ( x + 2) x →3 x + 2 3 + 2 5

x2 x

2

Pembahasan:

3.

Jawaban: a

1

−

x Pembahasan:

−

1

( 2) 2

x2

f’(x) = (x -1)’ = -1x -2- =

f(x) = 1

=−

1

→

4

f’(2) = 4.

Jawaban: b Pembahasan:

f(x) = tan x sin x, maka f(x)' = u'v + uv' 1 cos 2 x

u = tan x maka u’ = sec 2 x = v = sin x maka v’ = cosx f’(x) = u'v + uv' 1 sin x + tan x cos x cos 2 = .

=

sin x cos

2

+

sin x cos x

cos x

=

sin x + sin x cos

2

=

2 sin x cos 2 x

Jawaban: c Pembahasan:

y y’

= x4 – 8x2 –  = 4x3 – 1!x " # = 4x (x – 2) ( x + 2) " # – –

+ +

-2 !.

– –

0

x < - 2 Jawaban: c

++

2 0 < x < 2

Pembahasan:

f(x) = x3 + 3x2 + $ f’(x) = 3x2 + !x " # = 3x (x + 2) " # ++

– –

-2

+ +

0

-2 < x < 0

1

%. c. 8.

–1 " x " 1 atau 1 " x " 3

Jawaban: e Pembahasan:

f(x) = 2x 3 – x2 + 12x f’(x) = !x2 – 18x + 12 & # x2 – 3x + 2 & # (x – 1) (x – 2) & # + +

– –

1

++

2

x < 1

.

x >2


x(t) = t y(t) = t - 4t + $ e*aak minimum *ika y'(t) = # 2t - 4 = # t =2 t = 2 → x(2) = 2 = 4 ai, semut e*aak minimum ke sumu x *ika *aak ai sumu y = 4 1#.


f'(x) = 3x + 12x - 1$ funsi tuun *ika f'(x) " # 3x + 12x - 1$ " # x + 4x - $ " # (x + $) (x - 1) " # x " -$ atau x " 1

11.

Jawaban: e Pembahasan:

(x) = (3x + 4) $ (2x - 1)4 u = (3x + 4)$ → u' = $(3x + 4) 4 !x = 3#x(3x + 4) 4 v = (2x - 1)4 → v' = 4(2x - 1)3 . 2 = 8 (2x - 1) 3 (x) = uv '(x) = u'v + v'u '(x) = 3#x(3x + 4) 4 . (2x - 1)4 + 8 (2x - 1)3 . (3x + 4) $ '(x) = (3x + 4) 4(3#x . (2x - 1) 4 + 8 (2x - 1)3 . (3x + 4)) '(x) = (3x + 4) 4(2x - 1)3 ( 3#x . (2x - 1) + 8 . (3x + 4)) '(x) = (3x + 4) 4 (2x - 1)3 ( !#x - 3#x + 24x + 32) '(x) = (3x + 4) 4 (2x - 1)3 (84x - 3#x + 32) 12.

Jawaban: a

$im x →∞

x3 5x 2

+1

 ka"ena #angkat te"tinggi !(x)

< g(x) maka hasilnya = 0

Pembahasan:

2

13.

Jawaban: a

$im x →∞

4 x +1

< g(x) maka hasilnya = 0

 ka"ena #angkat te"tinggi !(x)

Pembahasan: "4. Jawaban: e

+x  (sem&anya %ibagikan %engan x 2 ) maka x →∞ 1 − 2 x − x 2 x2

$im

hasilnya

1+ 0 0 - 0 -1

= −1

Pembahasan:

1$.

Jawaban: # Pembahasan:

lim

θ→ 0

sin 5θ tan 3θ

= lim θ→ 0

sin 5θ 5θ

5θ

θ→0 /ntuk lim

θ→ 0

3θ

1

tan 3θ 3θ

= lim θ →0

3θ → 0

eakiat

sin 5θ

sin 5θ 5θ

θ→ 0

3θ tan 3θ

. lim

θ →0

5θ 3θ

5θ → 0

an

sin 5θ = 5lim . lim θ→0 5θ 3θ→0 tan 3θ

. lim

3θ

, se0ina

. lim

5θ

tan 3θ θ→0 3θ

= 1.1. 5 = 5 3

3

.

1!.

Jawaban: a

sin x

lim x

→0

=1

sin x ≈ x

x

untuk x keci maka

Pembahasan:

"$. Jawaban: a

lim

x →0

tg x x

=1

Pembahasan:

lim

x →0

tg x x

= lim x →0

sin x cos x

= x →0

maka lim

sin x x

. lim

x →0

1 cos x

= 1.1 = 1

"%. Jawaban: c Pembahasan:

lim

x →0

sin 3x x

= lim x →0

3 sin 3x 3x

= 3 3lim x →0

sin 3x 3x

= 3 .1 = 3

"&. Jawaban: b Pembahasan:

3

1 − cos 2 x

1 − cos 2x   2  sin 2 x  = lim lim     = 2 lim x →0 x →0 x2  2   x 2  x →0 x 2 2 2  sin x  = 2 lim sin x  = 2.12 = 2.1 = 2 = 2 lim    x →0  x →0 x   x  

2'. Jawaban: e

tg3x

lim

x →0

6x

 1 1 tg3x 1  tg3x    6x  = lim = . = lim .    x →0  3x   tg6x  2 2 3x →0 3x lim tg6 x 6 x →0

1 2

.1.1 =

1 2

6x

Pembahasan: 2". Jawaban: e

lim

tg6x

x →0

12x

 1 1 tg6x  tg6x    12x  . = lim . = lim    x →0  6x   tg12x  2 2 6 x →0 6x

lim

12 x →0

1 tg12x

=

1 2

.1.1 =

1 2

12x

Pembahasan:

22.

Jawaban: e. Pembahasan:

(f o g )( x )

23.

= f((x)) = f (2 x + 3) = (2 x + 3)2 –3(2 x + 3) – 4 = 4 x 2 + 12x +  – !x –  – 4 = 4x2 + !x – 4

Jawaban: b. ( Pembahasan:

1

5

2

(f o g )(x)

= f((x)) 1

(f o g )(2)

x

=f(

) 1

1

x

x 2

= 2(

) + $( 1

x

=2 24.

2 2 =2. + 1 5 + 2 2 =

)

=3

5

2

x

+

Jawaban: a. Pembahasan: x

+1

f(x) = 2 ⇔ y =x+1 ⇔ x = y2 – 1 -1 2 ⇔ x = f (y)= y – 1 -1 ⇔ f (x) = x2 – 1

4

− x + 2 3

25.

Jawaban: c. Pembahasan:

f(x) = 2 – 3x ⇔ y = 2 – 3x ⇔ 3x =–y+2 − y + 2 3 -1 ⇔ x = f (y)= − x + 2 3 -1 ⇔ f (x) = 2!. Jawaban: #. 4x – 12 Pembahasan:

y =x+3 ⇔ –  x =–y+3 ⇔  x =y–3 ⇔ x = 4(y – 3) ⇔ x = 4y – 12 -1 ⇔ x = f (y)= 4y – 12 -1 ⇔ f (x) = 4x – 12 ⇔

5 x + 1 x

2%.


1 x

−5

f(x) = 1 x

−5

y = ⇔ y(x – $) = 1 1 ⇔

y

x – $

⇔

= 1 y

x

= +$ 1 + 5 y 5 y + 1

⇔

=

y

y

-1

x = f (y)=

⇔

5 x + 1 x ⇔

28.

-1

f (x)

=

x2 + x Pembahasan: ( o f) (x) Jawaban: #.

= 4x2 - 2x 5

= 4x2 - 2x

(2x- 1)

y

+1 2

isakan y = 2x – 1, maka x = 2

 y + 1     2  (y)

2.

Jawaban: b.

 y + 1     2 

=4 - 2 2 = y + 2y + 1 – y – 1

x+

Pembahasan:

(f –1 o f –1)(x)

= f-1(f -1(x))  x + 3 

 

 

2

= f -1

 x + 3 + 3     2  =  2     

x

+3+6 2 2

= x

+9 2 2

= x + 9

=

− x + 2 5

3#.


f(x) = 2 – $x ⇔ y = 2 – $x ⇔ $x =–y+2 − y + 2 5 -1 ⇔ x = f (y)= − x + 2 5 -1 ⇔ f (x) =

31.

Jawaban: #.

4x2 –12x + 1#

Pembahasan:

( f o  )(x)

32.

= f((x)) = f(2 x – 3) = (2 x – 3)2 + 1 = 4x2 –12x +  + 1 = 4x2 –12x + 1#

Jawaban: a.

x2 – % 6

Pembahasan: x

+'

f(x) = 2 ⇔ y =x+% ⇔ x = y2 – % -1 2 ⇔ x = f (y)= y – %

33.

Jawaban: e.

4x – 8

Pembahasan:

y =x+2 ⇔ –  x =–y+2 ⇔  x =y–2 ⇔ x = 4(y – 2) ⇔ x = 4y – 8 -1 ⇔ x = f (y)= 4y – 8 -1 ⇔ f (x) = 4x – 8 ⇔

(4. Pembahasan:

f(x) = $ + 1$ x + x2 + x3 f’(x) = # + 1$ + 18x + 3x2 & # x2 + !x + $ & # (x + $) (x + 1) & #

++

– –

-5

x < -5

+ +

-1

x > -1

(. Jawaban: # Pembahasan

f(x) = (2x - 2)5 f '(x) = 3(2x - 2) . 4x = 12x(2x - 2) = 12x (2(x - 1)) = 12x . 4 . (x - 1) = 48 x (x + 1) (x - 1) x = #, atau x = -1 atau x = 1 f '(x) = # (niai stasione) /ntuk x = #  f(x) = (2x - 2)5 f '(#) = (# - 2)5 = -8 /ntuk x = -1  f(x) = (2x - 2)5 f '(-1) = (2 . 1 - 2)5 = #5 = # /ntuk x = 1  f(x) = (2x - 2)5 f '(-1) = (2 . 1 - 2)5 = #5 = # ai, niai maksimumnya = # (!. Jawaban: c Pembahasan:

6ece7atan aaa0 tuunan 7etama ai aak (). 9ece7atan aaa0 tuunan keua ai aak ()  = t5 - 3t. ' = 3t - 3 '' = !t 9aa saat kece7atannya #  '

v = 3t - 3 # = 3t - 3 3t = 3 t = 1 t=1 ai 9ece7atannya ('') = !t = ! x 1 = ! m:et ($. Jawaban: e Pembahasan:

(x) = (3x - 4x + !) x (x) = x5 - 2x + !x + ; (-1) = -1 - 2 - ! + ; = # ;= ai, (x) = x5 - 2x + !x +  (%. Jawaban: a Pembahasan:

f(x) = 3x cos x maka u = 3x an u’ = 3 v = cos x an v’ = - sin x f’(x) = u'v + uv' = 3 cos x + 3x (-sin x) = 3 cos x – 3x sin x (&. Jawaban: a Pembahasan:

f(x) = (3x + 2) sin x, maka f’(x) = 3 sin x + (3x + 2) cos x 4'. Jawaban: e Pembahasan: f(x)

= (!x 2 – 1) cos x, maka f’(x) = 12x cos x – (!x2 – 1) sin x

(

Kunci Matematika Kelas Xi Sma Persiapan Ujian Akhir Semester 2

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