ethiopian university entrance examination (euee

Post on 17-Nov-2021

3 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Ethiopian University Entrance Examination (EUEE) Mathematics

for Natural Science Ginbot 2011/June 2019

BOOKLET CODE: 17 SUBJECT CODE:02

Number Of Items:65 Time Allowed:3 HOURS

DIRECTIONS: For each of the following problems. Choose the best answer from

the given alternative and carefully blacken the letter of your best choice on the

separate answer sheet provided.

1. A man wants to fence two identical area of each enclosure that he can make with 192

mater fencing material if the side along the river does not need a fence?

What is the maximum area of each enclosure that he can make with 192 meter fencing material if

the side along the river does not need a fence?

(A) 1530 m2

(B) 1564 m2

(C) 1536 m2

(D) 1664 m2

Let x and y be the width and depth of rectangle. A inters of x and y becomes

A= xy

The total length of the fencing is 192 m

3y= 192

X= 96 โˆ’3

2y

A (y) = (96- 3

2 ๐‘ฆ) ๐‘ฆ = 96โˆ’

3

2๐‘ฆ2 = ; 0โ‰ค ๐‘ฆ โ‰ค 64

A (y)= 96y -3

2๐‘ฆ2 is contain [0,64] and diff. on (0,64)

Aโ€™(y)= 0 โŸน 96-3y=0

โŸนy=32m

Hence, y=32 is a critical number to get the max area at y=32

A(0)=0 A=64 and A (32)= 1536 ๐‘š2

Ans. C

2. If ๐‘“(๐‘ฅ)= ๐‘ฅ3 -2 ๐‘ฅ+1, then which one of the following is NOT true?

(A) ๐‘“(๐‘)= 1

4 for some โˆ โˆˆ [โˆ’1, 0].

(B) ๐‘“(๐‘)= 1

2 for some โˆ โˆˆ [0, 1].

(C) ๐‘“(๐‘)= 0 for some โˆ โˆˆ [โˆ’2, 0].

(D) ๐‘“(๐‘)= 3 for some โˆ โˆˆ [1, 2].

3. If ) ๐‘“(๐‘ฅ) = In (2tan ๐‘ฅ), then what is the value of ๐‘“โ€ฒ(0)?

(A) 1

(B) -2ln 2

(C) ln 2

2

(D) ln2

4. If ๐‘“(๐‘ฅ) = ๐›ผ๐‘ฅ โˆ’ ๐‘ and ๐‘“โˆ’1 (๐‘ฅ + 1)=1

2๐‘ฅ+2, for each ๐‘ฅ ๐œ–, then what must be the value of

๐›ผ and ๐‘?

(A) ๐›ผ = ยฝ, ๐‘= -2

(B) ๐›ผ =2, ๐‘ =3

Let CโˆˆR such that

๐‘“(๐‘ฅ) = ๐‘ฅ2 โˆ’ 2๐‘ฅ + 1

Ans. A is using intermediate value theorem

If C โˆˆ [โˆ’1,0]

๐‘“(โˆ’1) = 2 ๐‘Ž๐‘›๐‘‘ ๐‘“(0) = 1

Hence, 1

4 is not included b/n that values

๐‘“(๐‘ฅ) = ln(2tan ๐‘ฅ)

๐‘“(๐‘ฅ) = ๐‘”(โ„Ž(๐‘ฅ))

Where, g(x) - ln ๐‘ฅ, ๐‘”โ€ฒ(๐‘ฅ) =1

๐‘ฅ

h(x) = 2tan ๐‘ฅ , hโ€ฒ(x) = 2tan ๐‘ฅ , ๐‘ก๐‘Ž ๐‘› 2 ๐‘ ๐‘’๐‘2๐‘ฅ

๐‘“โ€ฒ(๐‘ฅ) = 1

2tan ๐‘ฅ . ln 2 ๐‘ ๐‘’๐‘2 ๐‘ฅ

๐‘“โ€ฒ(0) = 1

2tan 0. ln 2 ๐‘ ๐‘’๐‘2 ๐‘ฅ

= ๐ฅ๐ง 2

Ans. D

(C) ๐›ผ =1, ๐‘=1

(D) ๐›ผ =2, ๐‘=2

(E)

5. If 2(2๐‘ฅ ๐‘ฅโˆ’5 โˆ’3

) -1 = (3 2

โˆ’5 โˆ’4), then what is the value of ๐‘ฅ?

(A) 2

(B) -2

(C) 1

2

(D) - 1

2

6. What is the area of the region bounded by the lines ๐‘ฅ = 0, ๐‘ฅ = 2 and ๐‘ฆ = 1 and the

curve ๐‘ฆ = ๐‘’2๐‘ฅ?

(A) ๐‘’2

2 -

1

2

(B) ๐‘’4

2 -

5

2

(C) ๐‘’3

3 -

1

3

(D) - ๐‘’4

2 +

5

2

๐‘“(๐‘ฅ) = ๐‘Ž๐‘ฅ โˆ’ ๐‘

๐‘“โˆ’1(๐‘ฅ) =๐‘ฅ + ๐‘

๐‘Ž

๐‘“โˆ’1(๐‘ฅ + 1) =๐‘ฅ + 1 + ๐‘

๐‘ , ๐‘๐‘ข๐‘ก ๐‘“โˆ’1(๐‘ฅ + 1) =

1

2๐‘ฅ + 2 =

๐‘ฅ + 4

2

๐‘ฅ + 1 + ๐‘

๐‘Ž=

๐‘ฅ + 4

2

โˆด ๐‘Ž = 2 ๐‘Ž๐‘›๐‘‘ 1 + ๐‘ = 4 ๐‘ = 3

Ans. B

2(2๐‘ฅ ๐‘ฅ โˆ’5 โˆ’3

)โˆ’1

= (3 2โˆ’5 โˆ’4

) , ๐ดโˆ’1 =1

2 ( 3 2

โˆ’5 โˆ’4)

Let ๐ดโˆ’1 =1

2( 1 2

โˆ’5 โˆ’4 ) ๐ดโˆ’1 =

1

2 (3 2

โˆ’5 โˆ’4)

โˆ’1

A=( 4 2โˆ’5 โˆ’3

) , 2x = 4

X= 2

Ans, A

7. If ๐‘“ if the greatest integer function and g is the absolute value function, the what is the

value of (๐‘“ ๐‘œ g) (3

2) + (g ๐‘œ ๐‘“) (โˆ’

4

3) ?

(A) 1

(B) 2

(C) -1

(D) 3

๐’š = ๐’†๐’™๐Ÿ

The area b/n the line x=0, x=2 and y=1 is

๐‘’4

y = 1

y = 0 x = 2

A โˆซ ๐‘’2๐‘ฅ2

0 ๐‘‘๐‘ฅ , let u = 2๐‘ฅ , ๐‘ฅ = 0, u =

๐‘ฅ = 2, ๐‘ข = 4

๐‘‘๐‘› = 2 ๐‘‘๐‘ฅ

1

2 ๐‘‘๐‘› = ๐‘‘๐‘ฅ

A1

2 โˆซ ๐‘’44

0 ๐‘‘๐‘ฆ

= 1

2[๐‘’4]4

=1

2 [๐‘’4 โˆ’ ๐‘’0] =

1

2 [๐‘’4 โˆ’ 1]

๐‘’4

2โˆ’

1

2 Area of rectangle

๐‘’4

2โˆ’

1

2โˆ’ 2 =

๐‘’4

2โˆ’

5

2

Ans, B

8. Which one of following is true about the graph of ๐‘“(๐‘ฅ)๐‘ฅ2+5๐‘ฅ+6

๐‘ฅ2โˆ’4 +3?

(A) A horizontal asymptote of the graph of the graph is ๐‘ฆ =4.

(B) The vertical asymptotes of the graph are ๐‘ฅ = -2 and ๐‘ฅ =2.

(C) The graph has a hole at ๐‘ฅ =2.

(D) The graph has ๐‘ฆ- intercept at (0, โˆ’3

2)

9. What is the solution of โˆซ 4๐‘ฅ (โ„“๐‘› ๐‘ฅ +1

๐‘ฅ2)d๐‘ฅ ?

(A) 4๐‘ฅ2(โ„“๐‘› ๐‘ฅ + 1)-2๐‘ฅ2+c

(B) ๐‘ฅ2(2โ„“๐‘› ๐‘ฅ โˆ’ 1)+4 ๐‘™๐‘› ๐‘ฅ +c

(C) ๐‘ฅ2(2โ„“๐‘› ๐‘ฅ + 1)+4 ๐‘™๐‘› ๐‘ฅ +c

(D) ๐‘ฅ2(4โ„“๐‘› ๐‘ฅ โˆ’ 1)+2 ๐‘™๐‘› ๐‘ฅ +c

Give โ€“f is the greatest integer function and g is the absolute value function , then (fog) (3/2)+ (gof)(-

4/3)= 1+2=3

(๐‘“๐‘œ๐‘”) (3/2) = ๐‘“(๐‘” (3/2))

= ๐‘“(3/2) = 1

(๐‘”๐‘œ๐‘“) (โˆ’4/3) = ๐‘”(๐‘“(โˆ’4/3))

= ๐‘”(โˆ’2)

= 2

Ans. D

๐’‡(๐’™) ๐’™๐Ÿ+๐Ÿ“๐’™+๐Ÿ”

๐’™๐Ÿโˆ’๐Ÿ’ +3

=๐’™๐Ÿ+๐Ÿ“๐’™+๐Ÿ”+๐Ÿ‘(๐’™๐Ÿโˆ’๐Ÿ’)

๐’™๐Ÿโˆ’๐Ÿ’ =

๐Ÿ”

โˆ’๐Ÿ’+ ๐Ÿ‘

=๐’™๐Ÿ+๐Ÿ“๐’™+๐Ÿ”+๐Ÿ‘๐’™ ๐Ÿโˆ’๐Ÿ๐Ÿ

๐’™๐Ÿโˆ’๐Ÿ’ =

โˆ’๐Ÿ‘

๐Ÿ+ ๐Ÿ‘ โˆ’

๐Ÿ‘

๐Ÿ

= ๐Ÿ’๐’™๐Ÿ+๐Ÿ“๐’™โˆ’๐Ÿ”

๐’™๐Ÿโˆ’๐Ÿ’

Hence ๐’Ž = ๐’, โ‡’ ๐’š = ๐Ÿ’ is a H.A

๐‘ฟ = ๐Ÿ is a V.A

Its y intercept is ( ๐ŸŽ,๐Ÿ‘

๐Ÿ)

Makes a hole at ๐’™ = โˆ’๐Ÿ

Ans. A

10. The earthโ€™s orbit has a semi-major axis ๐›ผ โ‰ˆ 149.6 Gm (gigameters) and an eccentricity

of ๐‘’ โ‰ˆ 0.017. What is the approximate value of the semi-minor axis?

(A) 149.58 Gm

(B) 145.32 Gm

(C) 149.06 Gm

(D) 152.14 Gm

โˆซ 4๐‘ฅ (โ„“๐‘› ๐‘ฅ +1

๐‘ฅ2 ) ๐‘‘๐‘ฅ

= โˆซ 4๐‘ฅ โ„“๐‘› ๐‘ฅ +4

๐‘ฅ ๐‘‘๐‘ฅ

= โˆซ 4๐‘ฅ โ„“๐‘› ๐‘ฅ ๐‘‘๐‘ฅ + โˆซ4

๐‘ฅ ๐‘‘๐‘ฅ

= 4 [โˆซ ๐‘ฅ โ„“๐‘› ๐‘ฅ ๐‘‘๐‘ฅ + โˆซ1

๐‘ฅ ๐‘‘๐‘ฅ]

= 4 [๐‘ฅ2โ„“๐‘› ๐‘ฅ

2โˆ’

๐‘ฅ2

4+ โ„“๐‘› ๐‘ฅ] + ๐‘

=2๐‘ฅ2โ„“๐‘› ๐‘ฅ โˆ’ ๐‘ฅ2 + 4 โ„“๐‘› ๐‘ฅ + ๐‘

= ๐‘ฅ2(2โ„“๐‘› ๐‘ฅ โˆ’ 1) + 4 โ„“๐‘› ๐‘ฅ + ๐‘

Ans. B

๐‘Ž โ‰ˆ 149 ๐บ๐‘€

๐‘’ โ‰ˆ 0.017

๐‘’๐‘

๐‘Ž

โ‡’ ๐‘ = ๐‘’๐‘Ž

= 0.017 x149.6 G.M

= 2.5432 G.M

From the relation

๐‘Ž2 = ๐‘2+๐‘2

๐‘2 = ๐‘Ž2-๐‘2

=(149.6)2 โˆ’ (2.5432)2

๐‘2 = 22,373.69 G.M2

๐‘ = 149.58๐‘š

Ans. A

11. A water tank is a rectangular parallelepiped with has length 3 m, width 2 m and height 2.5

m. If water is flowing into the tank at the rate of 0.12 m3/sec, then how fast does the level of

water rise up in the tank?

(A) 0.04 m/sec

(B) 0.03m/sec

(C) 0.02 m/sec

(D) 0.06 m/sec

12. What is the sum of the infinite series โˆ‘ 5โˆž๐‘˜=0

2๐‘˜ + 5๐‘˜

10 ๐‘˜

(A) 15

(B) 25

4

(C) 65

4

(D) 5

L= 3m.w = 2m h=25m rate 0.12m3 /sec

V= Ab h , Ab= Lw=6m2

๐‘‘๐‘ฃ

๐‘‘๐‘ก= (๐ด๐‘โ„Ž)

๐‘‘โ„Ž

๐‘‘๐‘ก ,

๐‘‘๐‘ฃ

๐‘‘๐‘ก= 0.12๐‘š2/๐‘ ๐‘’๐‘

๐‘‘๐‘ฃ

๐‘‘๐‘ก= ๐ด๐‘

๐‘‘โ„Ž

๐‘‘๐‘ก

โˆด๐‘‘โ„Ž

๐‘‘๐‘ก=

๐‘‘๐‘ฃ

๐‘‘๐‘ก ,

1

๐ด๐‘

=0.12๐‘š3/๐‘ ๐‘’๐‘

6๐‘š2= 0.02๐‘š/๐‘ ๐‘’๐‘

Ans. C

Sum of โˆ‘ 5 (2๐‘˜+5๐‘˜

10๐‘˜ )โˆž๐ฟ=0

โ‡’ 5 [โˆ‘2๐‘˜

10๐‘˜

โˆž

๐‘˜=0

+ โˆ‘5๐‘˜

10๐‘˜

โˆž

๐‘˜=0

]

โ‡’ 5 [5

4+ 2]

โ‡’ 5 [13

4]

= 65

4

Ans. C

13. The marks of 50 students is given below:

Marks 0-10 10-20 20-30 30-40 40-50

No. of students 5 8 ๐‘“1 10 ๐‘“2

If the median of the data is 26, what are the values of ๐‘“1๐‘Ž๐‘›๐‘‘ ๐‘“2 ?

(A) ๐‘“1 = 20, ๐‘“2 = 7

(B) ๐‘“1 = 12, ๐‘“2 = 15

(C) ๐‘“1 = 15, ๐‘“2 = 12

(D) ๐‘“1 = 7, ๐‘“2 = 20

14. There are three children in a room with, ages four, five, and six. If a five-year-old child enters

the room, then which of the following statement is correct?

(A) Mean age will stay the same but the standard deviation will increase.

(B) Mean age will stay the same but the standard deviation will decrease.

(C) Mean age and standard deviation will increase.

(D) Mean age and standard deviation will stay the same.

Data, 4, 5, 6, ๏ฟฝฬ…๏ฟฝ = 5

New data, 4, 5, 5, 6, ๏ฟฝฬ…๏ฟฝ = 5 mean stays

S.D (data) =2

3= 0.67

S.D (new data)= 2

4= 0.5 It decreases

Ans. B

Given ๐‘› = 50, ๐ฟ๐ถ๐ต = 19.5 ๐ถ + ๐‘ = 11

๐‘“๐‘ = ๐‘“1 , ๐‘š๐‘’๐‘‘๐‘–๐‘Ž๐‘› = 26

Median = LCB + (๐‘›

2 โˆ’๐ถ๐‘“๐‘

๐‘“๐‘) ๐‘–

26 = 19.5 + (25 โˆ’ 13

๐‘“1) 11

6.5 =12.11

๐‘“1

๐‘“1 =132

6.5= 20 ๐‘“2 = 50 โˆ’ 5 + 8 + 20 + 10 = 7

Ans.A

15. For real numbers x and y, which one of the following is NOT true?

(A) (โˆ€๐‘ฅ)(โˆ€๐‘ฆ) (๐‘ฆ2 + ๐‘ฅ2 โ‰ฅ โˆ’1)

(B) (โˆƒ๐‘ฅ)(โˆ€๐‘ฆ) (๐‘ฆ โ‰ฅ ๐‘ฅ2 + 1)

(C) (โˆ€๐‘ฅ)(โˆƒ๐‘ฆ) (๐‘ฆ โ‰ฅ ๐‘ฅ2 + 1)

(D) (โˆƒ๐‘ฅ)(โˆƒ๐‘ฆ) (๐‘ฆ โ‰ฅ ๐‘ฅ2 + 1)

16. If the second and fifth terms of geometric progression are - 1

2 and

1

16 ,

(A) 128

85

(B) 255

256

(C) 85

128

(D) 256

255

17. Which one of the following is a valid argument?

(A) If I donโ€™t change my oil regularly, my engine will die. My engine died, Thus, I didnโ€™t

change my oil regularly.

(B) If you do very problem in the book, then you will learn the subject. You learned the

subject. Thus , you did every problem in the book,

(C) If I am literate, then I can read and write. I can read but I canโ€™t write. Thus I am not

literate.

(D) If it rains or snows, then my roof leaks. My roof is leaking. Thus, it is raining and

snowing.

Ans. B

๐‘ฎ๐Ÿ = โˆ’๐Ÿ๐Ÿโ„ , ๐‘ฎ๐Ÿ“ = ๐Ÿ

๐Ÿ๐Ÿ”โ„

๐‘ฎ๐Ÿ

๐‘ฎ๐Ÿ“

=๐’“๐‘ฎ๐Ÿ

๐’“๐Ÿ’ ๐‘ฎ๐Ÿ

, ๐’“๐Ÿ‘ =โˆ’๐Ÿ

๐Ÿโ„๐Ÿ

๐Ÿ๐Ÿ”โ„โ„ =

โˆ’๐Ÿ

๐Ÿ–

โˆด ๐‘ฎ๐Ÿ = ๐Ÿ ๐’“ =โˆ’๐Ÿ

๐Ÿ

๐‘บ๐Ÿ– =๐‘ฎ๐Ÿ(๐Ÿโˆ’๐’“๐Ÿ–)

๐Ÿ โˆ’ ๐’“= ๐Ÿ–๐Ÿ“

๐Ÿ๐Ÿ๐Ÿ–โ„

Ans. C

18. What is the value of โˆซ๐‘ฅ๐‘’โˆ’1 + ๐‘’๐‘ฅโˆ’1

๐‘ฅ๐‘’ + ๐‘’๐‘ฅ d๐‘ฅ?

(A) 1

๐‘’ ln(๐‘ฅ๐‘’ + ๐‘’๐‘ฅ+1)+ c

(B) 1

๐‘’ ln(๐‘ฅ๐‘’+1 + ๐‘’๐‘ฅ+1)+ c

(C) 1

๐‘’ ln((๐‘ฅ + 1) ๐‘’ + ๐‘’๐‘ฅ+1)+ c

(D) 1

๐‘’ ln(๐‘ฅ ๐‘’ + ๐‘’๐‘ฅ)+ c

(E)

19. Suppose 2,500 items are produced by a machine and 2 % of the product are randomly

selected and tested. If 5 of the tested items have a defect, then what is the probability that an

item produced by the machine No defect?

(A) 0.80

(B) 0.90

(C) 0.85

(D) 095

(E)

AnS. C โ†’ ๐‘ โ‡’ ๐‘žโ‹€๐‘Ÿ P: I am L

๐‘žโˆงยฌ๐‘Ÿ

ยฌ๐‘ q: I can read

r- I can write

โˆซ (๐‘ฅ๐‘’โˆ’1 + ๐‘’๐‘ฅโˆ’1

๐‘ฅ๐‘’ + ๐‘’๐‘ฅ ) ๐‘‘๐‘ฅ

(๐‘ฅ๐‘’ + ๐‘’๐‘ฅ )1 = ๐‘’๐‘ฅ๐‘’โˆ’1 + ๐‘’๐‘ฅ

Multiply both sides by 1 ๐‘’โ„ ๐‘ค๐‘’ ๐‘”๐‘’๐‘ก

(๐‘ฅ๐‘’โˆ’1 + ๐‘’๐‘ฅโˆ’1) 1๐‘’โ„

โˆซ๐‘ฅ๐‘’โˆ’1 + ๐‘’๐‘ฅโˆ’1

๐‘ฅ๐‘’ + ๐‘’๐‘ฅ๐‘‘๐‘ฅ =

1

๐‘’โ„“๐‘› |๐‘ฅ๐‘’ + ๐‘’๐‘ฅ| + ๐‘

Ans. D

Total item 2500

Test 2% of 2500=50

Defect=5

Non defect=45

P(E)= 4550

=0.9

Ans. B

20. If ๐‘“(๐‘ฅ) = 1

๐‘ฅ, then what is the value of ๐‘“(๐‘›)(๐‘ฅ)?

(A) ๐‘“(๐‘›)(๐‘ฅ)= (โˆ’1) ๐‘› ๐‘›!

๐‘ฅ๐‘›

(B) ๐‘“(๐‘›)(๐‘ฅ)= (โˆ’1) ๐‘›+1 ( ๐‘›+1)!

๐‘ฅ๐‘›+1

(C) ๐‘“(๐‘›)(๐‘ฅ)= (โˆ’1) ๐‘› ๐‘›!

๐‘ฅ๐‘›+1

(D) ๐‘“(๐‘›)(๐‘ฅ)= (โˆ’1) ๐‘› ๐‘›!

๐‘ฅ๐‘›โˆ’1

21. If the region enclosed by the graphs of ๐‘“(๐‘ฅ)= ๐‘ฅ2 and ๐‘“(๐‘ฅ)= ๐‘ฅ3 from ๐‘ฅ =0 to ๐‘ฅ =1 rotates

about the ๐‘ฅ- axis, what is the volume of the solid of revolution?

(A) 2๐œ‹

35 cubic units

(B) 2๐œ‹

25 cubic units

(C) 2๐œ‹

5 cubic units

(D) 2๐œ‹

27 cubic units

(E)

(F)

๐‘“(๐‘ฅ) =1

๐‘ฅ , ๐‘“โ€ฒ(๐‘ฅ) =

โˆ’1

๐‘ฅ2 , ๐‘“โ€ฒโ€ฒ(๐‘ฅ) =

2

๐‘ฅ3

๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ฅ) =โˆ’6

๐‘ฅ4

โˆด ๐‘“๐‘›(๐‘ฅ) =(โˆ’1)๐‘›. ๐‘›:

๐‘ฅ๐‘›+1

Ans. C

๐‘‰ = ๐œ‹ โˆซ (๐‘ฅ2)21

0

โˆ’ (๐‘ฅ3)2๐‘‘๐‘ฅ

= ๐œ‹ โˆซ ๐‘ฅ4 โˆ’ ๐‘ฅ6๐‘‘๐‘ฅ1

0

= ๐œ‹ [๐‘ฅ5

5โˆ’

๐‘ฅ7

7]

= ๐œ‹ [1

5โˆ’

1

7]

=2๐œ‹

35

Ans. A

22. What is the limit of sequence 1, 2

22โˆ’12 ,

3

32โˆ’22 ,

4

42โˆ’32 ,โ€ฆ?

(A) 0

(B) 1

2

(C) 1

(D) 3

2

(E)

23. At the value of x does the function ๐‘“(๐‘ฅ) = 4๐‘ฅ2

3 - ๐‘ฅ4 attains maximum value?

(A) -1

(B) 0

(C) 4

3

(D) 1

2

22 โˆ’ 12 ,

3

32 โˆ’ 22 ,

4

42 โˆ’ 32 , โ€ฆ

โ‡’๐‘›

๐‘›2 โˆ’ (๐‘› โˆ’ 1)2

โ‡’๐‘›

๐‘›2 โˆ’ (๐‘›2 โˆ’ 2๐‘› + 1 =

๐‘›

๐‘›2 โˆ’ ๐‘›2 + 2๐‘› โˆ’ 1

โ‡’๐‘›

2๐‘› โˆ’ 1 =

1

2 โˆ’1๐‘›

, ๐‘› โ†’ โˆž

lim๐‘›โ†’โˆž

1

2 โˆ’1๐‘›

=1

2

Ans.B

๐‘“(๐‘ฅ) =4๐‘ฅ3

3โˆ’ ๐‘ฅ4

๐‘“โ€ฒ(๐‘ฅ) = 4๐‘ฅ2 โˆ’ 4๐‘ฅ3

= 4๐‘ฅ2(1 โˆ’ ๐‘ฅ)

๐‘“โ€ฒ(๐‘ฅ) = 0, ๐‘ฅ = 0,1 ๐‘Ž๐‘Ÿ๐‘’ ๐‘๐‘Ÿ๐‘–๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘๐‘œ๐‘–๐‘›๐‘ก

0 1

4๐‘ฅ2 โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ + + + + + + + + +

1 โˆ’ ๐‘ฅ + + + + + โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ โˆ’

(4๐‘ฅ)2(1 โˆ’ ๐‘ฅ) โˆ’ โˆ’ โˆ’ โˆ’0 + +0 โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ โˆ’ โˆ’

๐‘“โ€ฒ(๐‘ฅ)๐‘โ„Ž๐‘Ž๐‘›๐‘”๐‘’๐‘ ๐‘ ๐‘–๐‘”๐‘› ๐‘“๐‘Ÿ๐‘œ๐‘š โˆ’ ๐‘ฃ๐‘’ ๐‘ก๐‘œ + ๐‘ฃ๐‘’ โˆ’ ๐‘Ž๐‘ก ๐‘ฅ = 0

๐‘“ โ„Ž๐‘Ž๐‘  ๐‘Ž ๐‘™๐‘œ๐‘๐‘Ž๐‘™ ๐‘š๐‘–๐‘›๐‘–๐‘š๐‘ข๐‘š

๐‘“โ€ฒ(๐‘ฅ)๐‘โ„Ž๐‘Ž๐‘›๐‘”๐‘’๐‘  ๐‘ ๐‘–๐‘”๐‘› ๐‘“๐‘Ÿ๐‘œ๐‘š + ๐‘ฃ๐‘’ ๐‘ก๐‘œ โˆ’ ๐‘ฃ๐‘’ ๐‘Ž๐‘ก ๐‘ฅ = 1

๐‘“ โ„Ž๐‘Ž๐‘  ๐‘Ž๐‘™๐‘œ๐‘๐‘Ž๐‘™ ๐‘š๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š

Ans. D

24. If โ„Ž(๐‘ฅ) =โˆš1 + โˆš๐‘ฅ , then which of the following is equal to โ„Žโ€ฒ(๐‘ฅ)?

(A) 1

4โˆš๐‘ฅ+๐‘ฅโˆš๐‘ฅ

(B) 1

2โˆš1+โˆš๐‘ฅ

(C) ๐‘ฅ

2โˆš1+โˆš๐‘ฅ

(D) ๐‘ฅ

4โˆš๐‘ฅ+๐‘ฅโˆš๐‘ฅ

25. Fatuma can solve 90% of the problems given in a book and Mesfin can solve 70%. What is the

probability that at least one of them will solve the problem, selected at random from the book?

(A) 0.77

(B) 0.97

(C) 0.87

(D) 0.67

Fatuma answers 90% โ‡’ ๐‘(๐น) = 0.9

Mesfin answers 70%โ‡’ ๐‘(๐‘€) = 0.7

โˆด ๐‘ƒ(๐น๐‘ˆ๐‘€) = ๐‘ƒ(๐น) + ๐‘ƒ(๐‘€) โˆ’ ๐‘ƒ(๐น๐‘›๐‘€)

= ๐‘(๐น) + ๐‘(๐‘€) โˆ’ [๐‘ƒ(๐น) โˆ— ๐‘ƒ(๐‘€)]

= 0.9 + 0.7 โˆ’ [0.9 โˆ— 0.7]

= 1.6 โˆ’ 0.63

= 0.97

Ans. B

โ„Ž(๐‘ฅ) = โˆš1 + โˆš๐‘ฅ

Let โ„Ž(๐‘ฅ) = ๐‘“(๐‘”(๐‘ฅ))๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’

๐‘“(๐‘ฅ) = โˆš๐‘ฅ , ๐‘“โ€ฒ(๐‘ฅ) = ๐‘“โ€ฒ(๐‘ฅ) =1

2(๐‘ฅ

12โ„ ) =

1

2โˆš๐‘ฅ

๐‘”(๐‘ฅ) = 1 + โˆš๐‘ฅ =1

2โˆš๐‘ฅ

โ„Žโ€ฒ(๐‘ฅ) = ๐‘“โ€ฒ(๐‘”(๐‘ฅ)), ๐‘”(๐‘ฅ) =1

2โˆš1 + โˆš๐‘ฅ ,

1

2โˆš๐‘ฅ

=1

4โˆš๐‘ฅ(1 + โˆš๐‘ฅ

= 1

4โˆšx+xโˆšx

Ans. A

26. What is the polar form of 7โˆ’๐‘–

3โˆ’4๐‘– ?

(A) โˆš2(๐‘๐‘œ๐‘  ๐œ‹

2+ ๐‘– ๐‘ ๐‘–๐‘›

๐œ‹

2)

(B) 2(๐‘๐‘œ๐‘  ๐œ‹

4+ ๐‘– ๐‘ ๐‘–๐‘›

๐œ‹

4)

(C) 2(๐‘๐‘œ๐‘  ๐œ‹

2+ ๐‘– ๐‘ ๐‘–๐‘›

๐œ‹

2)

(D) โˆš2(๐‘๐‘œ๐‘  ๐œ‹

4+ ๐‘– ๐‘ ๐‘–๐‘›

๐œ‹

4)

27. ๐ผ๐‘“ ยฌ๐‘ โ‡’ ๐‘Ÿ is False and p q is True , then which of the following is True?

(A) ๐‘ v (ยฌq ^r)

(B) ยฌ ๐‘ โ‡’ (q v r)

(C) ยฌ ๐‘ ^(q โ‡’ r)

(D) P (ยฌq v r)

Polar form of

7 โˆ’ ๐‘–

3 โˆ’ 4๐‘– ,

3 + 4๐‘–

3 + 4๐‘–=

7(3 + 4๐‘–) โˆ’ ๐‘–(3 + 4๐‘–)

3(3 + 4๐‘–) โˆ’ 4๐‘–(3 + 4๐‘–)

=21 + 28๐‘– โˆ’ 3๐‘– + 4

9 + 12๐‘– โˆ’ 12๐‘– + 16

=25 + 25๐‘–

25

= 1 + ๐‘– โ‡’ ๐‘ฅ = 1, ๐‘ฆ = 1

๐‘ก๐‘Ž๐‘› ๐œƒ =๐‘ฆ

๐‘ฅ= 01, ๐œƒ = ๐œ‹

4โ„ , ๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš2 =

โˆด ๐‘Ÿ(cos ๐œƒ + ๐‘– ๐‘ ๐‘–๐‘› ๐œƒ)

โˆš2 (๐‘๐‘œ๐‘  ๐œ‹4โ„ + ๐‘– ๐‘ ๐‘–๐‘› ๐œ‹

4โ„ )

Ans. D

ยฌpโ‡’ ๐‘Ÿ ๐‘–๐‘  ๐‘“๐‘Ž๐‘™๐‘ ๐‘’

ยฌ๐‘ is T and F is false

๐‘ƒ ๐‘–๐‘  ๐น

๐‘ โ‡” ๐‘ž ๐‘–๐‘  ๐‘‡๐‘Ÿ๐‘ข๐‘’

โˆด ๐‘ž ๐‘–๐‘  ๐น๐‘Ž๐‘™๐‘ ๐‘’

Truth value of p, q and r respectively

F,F,F

Ans.C ยฌ ๐‘ โˆง (๐‘ž โ‡’ ๐‘Ÿ)

T โˆง (๐น โ‡’ ๐น)

T โˆง ๐‘‡ = ๐‘‡

28. Which one of the following is the set of critical numbers of ๐‘“(๐‘ฅ) =3

8 ๐‘ฅ8/3- 6๐‘ฅ2/3?

(A) {โˆ’2, 2}

(B) {โˆ’1, 1}

(C) {โˆ’2, 0, 2}

(D) {โˆ’1, 0, 1}

29. The center of a circle on the line y= 2x and the line x=1 is tangent to the circle at (1,6). How

long is the radius of the circle?

(A) 5

(B) 4

(C) 2

(D) 3

30. If ๐‘Ž1 =2, ๐‘Ž2=6, ๐‘Ž3=10, ๐‘Ž4= 14,โ€ฆ. then ๐‘›=1

100๐‘Ž๐‘› =

(A) 20,020

(B) 20,000

(C) 20,200

(D) 22,200

๐‘“(๐‘ฅ) =๐‘ฅ

8 ๐‘ฅ

83โ„ โˆ’ 6๐‘ฅ

23โ„

๐‘“โ€ฒ(๐‘ฅ)3

8,8

3๐‘ฅ

53โ„ โˆ’ 6.

2

3 ๐‘ฅ

โˆ’13โ„

=๐‘ฅ5/3 โˆ’4

๐‘ฅ1

3โ„

๐‘“โ€ฒ(๐‘ฅ) =๐‘ฅ

53

+13โ„ โˆ’ 4

๐‘ฅ1

3โ„=

๐‘ฅ2 โˆ’ 4

๐‘ฅ1

3โ„

๐‘“โ€ฒ(๐‘ฅ) = 0, ๐‘ฅ2 โˆ’ 4 = 0, ๐‘ฅ = ยฑ 2, 0

Ans.C

Tangent at (1,6) and center at (h, k) here K=6, and y= 2x is on the center .

Hence, 2h=6โ‡’ โ„Ž = 3

Distance b/n (1, 6) to (3,6) is r

๐‘Ÿ = โˆš(1 โˆ’ 3)2 + (6 โˆ’ 6)2 = 42

Ans. C

31. Let ๐‘“(๐‘ฅ) = 2๐‘’๐‘ฅ-๐‘˜ sin ๐‘ฅ+1. If the equation of the tangent line to the graph of ๐‘“ at (0, 3) is ๐‘ฆ=

5x+3, then what is the value of ๐‘˜?

(A) -3

(B) 3

(C) -5

(D) 2

32. Among 2000 students who took a regional exam, the percentile of certain studentโ€™s score is 90.

Which of the following is correct about the studentโ€™s score?

(A) The students have answered 90% of the questions correctly.

(B) The score of the student is as good as that of 90% of the students.

(C) The studentโ€™s score is the same as the top 10% of the score.

(D) The studentโ€™s score is greater than or equal to that of 1800 students.

๐‘Ž1=2 ๐‘Ž2 = 6, ๐‘Ž3=10 โˆด ๐‘‘ = ๐‘Ž3โˆ’ ๐‘Ž2= ๐‘Ž2โˆ’ ๐‘Ž1= 4

Sn=โˆ‘ ๐ด๐‘›๐‘›๐‘˜=0 =

๐‘›

2(2๐ด1 + (๐‘› โˆ’ 1)๐‘‘)

โˆ‘ ๐ด๐‘›= 100

2 (2.2 + (100 โˆ’ 1)4)

100

๐‘›=1

= 50(4 + 99.4)

= 50(4 + 100) = 20,000

Ans. B

๐‘“(๐‘ฅ) = 2๐‘’๐‘ฅ โˆ’ k sin x + 1

Tangent at (0,3) the line is y=5x+3

๐‘“โ€ฒ(๐‘ฅ) = 2๐‘’๐‘ฅ โˆ’ ๐‘˜ cos ๐‘ฅ

๐‘“โ€ฒ(0) = ๐‘ ๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐‘ก๐‘Ž๐‘›๐‘”๐‘’๐‘›๐‘ก ๐‘™๐‘–๐‘›๐‘’

โ‡’ 5 = 2 โˆ’ ๐‘˜ โ‡’ 5 โˆ’ 2 = โˆ’๐‘˜ โ‡’ ๐‘˜ = โˆ’3

Ans. D

Ans. D

33. What is the value of lim๐‘›โ†’โˆž

(๐‘ฅ3โˆ’8

๐‘ฅ2โˆ’4)?

(A) 3

(B) 2

(C) -2

(D) The limit does not exist.

34. What is the greatest lower bound of sequence {(โˆ’1)๐‘› 1

๐‘›+1}

๐‘›=0

โˆž?

(A) -1

(B) 0

(C) -1/2

(D) ยฝ

35. What is the value of lim๐‘›โ†’โˆž

(3๐‘›+ 2โ€ฒโ€ฒ

6โ€ฒโ€ฒ )?

(A) 1

(B) โˆž

(C) 5

6

(D) 0

๐’๐’Š๐’Ž๐’™โ†’๐Ÿ ๐’™๐Ÿ‘ โˆ’ ๐Ÿ–

๐’™๐Ÿโˆ’๐Ÿ’= ๐’๐’Š๐’Ž๐’™โ†’๐Ÿ

(๐’™ โˆ’ ๐Ÿ)(๐’™๐Ÿ + ๐Ÿ๐’™ + ๐Ÿ’

(๐’™ โˆ’ ๐Ÿ)(๐’™ + ๐Ÿ)

=4+4+4

4=

12

4= 3

Ans. A

Ans. B

lim๐‘›โ†’โˆž

(3๐‘› + 2๐‘›

6๐‘›)

= lim๐‘›โ†’โˆž

3๐‘›

6๐‘›+ lim

๐‘›โ†’โˆž

2๐‘›

6๐‘›

= lim๐‘›โ†’โˆž

(1

2)

๐‘›+ lim

๐‘›โ†’โˆž (

1

3)

๐‘›

= lim ๐‘›โ†’โˆž

1

2๐‘›+ lim

๐‘›โ†’โˆž

1

3๐‘›

= 0+0

= 0

Ans. D

36. What is the value of limโ„Žโ†’0

8 (1

2+ โ„Ž) โˆ’ 8 (

1

2)

2

?

โ„Ž

(A) 8

(B) 4

(C) 0

(D) The limit does not exist

37. Which one of the following is the multiplicative inverse of ๐‘ง = 3+4๐‘–

4โˆ’5๐‘–

(A) 8

25โˆ’

31

25๐‘–

(B) 8

25โˆ’

25

31๐‘–

(C) 8

25+

25

31๐‘–

(D) 8

25+

31

25๐‘–

38. If A is a 3x3 matrix and det (A)= 5, then how much is det (2A A) ?

(A) 20

(B) 100

(C) 50

(D) 200

8 (

12 + h)

2

+ 8 (12)

2

h

lim โ„Žโ†’0

8h2 + 8h + 2 โˆ’ 2

h

lim 8โ„Žโ†’0

โ„Ž + 8 = 8

Ans. A

๐‘ง =3 + 4๐‘–

4 โˆ’ 5๐‘–

โˆด 1

๐‘ง=

13 + 4๐‘–4 โˆ’ 5๐‘–

= 4 โˆ’ 5๐‘–

3 + 4๐‘– ,

3 โˆ’ 4๐‘–

3 โˆ’ 4๐‘–=

โˆ’8 โˆ’ 3 + ๐‘–

2.5

Ans, B

39. Which one of the following is true about the pair of line: 3x +9y-24=0 and 4x+12y+32=0 ?

(A) parallel and distinct line

(B) intersecting line

(C) perpendicular line

(D) representing the same lines

40. If F(x)=โˆซ ๐‘’โˆ’1๐‘ฅ

0 dt , then what is the value of Fโ€™ (x) ?

(A) โˆ’๐‘’โˆ’1-1

(B) ๐‘’โˆ’๐‘ฅ

(C) ๐‘’โˆ’๐‘ฅ+1

โˆ’๐‘ฅ+1

(D) โˆ’๐‘’โˆ’๐‘ฅ

โ„“1 โˆถ ๐‘ฆ = โˆ’13โ„ ๐‘ฅ + 8

3โ„

โ„“2 โˆถ ๐‘ฆ = โˆ’13โ„ ๐‘ฅ โˆ’ 8

3โ„

Having the same slope

Ans. A

det (A) =5

det(2 ๐ด ๐‘‡๐ด)

Since T+ is 3x3 max

det (2 ๐ด ๐‘‡๐ด) = 23๐‘ฅ5๐‘ฅ5 = 200

Ans.D

๐น(๐‘ฅ) = โˆซ ๐‘’โˆ’๐‘ก

๐‘ฅ

0

๐‘‘๐‘ก

๐นโ€ฒ(๐‘ฅ) = โˆ’[๐‘’โˆ’๐‘ก]๐‘ฅ

= ๐น(๐‘ฅ) โˆ’ ๐น(0)

= โˆ’๐‘’โˆ’๐‘ฅ + 1, ๐‘กโ„Ž๐‘’๐‘› ๐นโ€ฒ(๐‘ฅ) = (โˆ’๐‘’โˆ’๐‘ฅ + 1)1 = ๐‘’โˆ’๐‘ฅ

Ans.B

41. If there are two children in a family, what is the probability that there is at least one girl in

the family?

(A) 1

2

(B) 1

4

(C) 2

3

(D) 3

4

42. Every day a person saves 5 cents more than the amount he saved on the previous day. His

target is to save the total amount of 3225 cents by the end of 30 days. How much must be

the starting saving to meet the target?

(A) 50 cents

(B) 25 cents

(C) 35 cents

(D) 60 cents

43. What is the value of โˆซ ln ๐‘ฅ

2 1

๐‘ฅ2 dx ?

(A) - ๐‘™๐‘› 2

2 +

1

2

(๐ต) ๐‘™๐‘› 2

2 -

1

2

A. - ๐‘™๐‘› 2

2 -

1

2

(๐ท) ๐‘™๐‘› 2

2 +

1

2

๐‘‡โ„Ž๐‘’๐‘ ๐‘’ ๐‘’๐‘ฅ๐‘–๐‘ ๐‘ก๐‘  2 ๐‘โ„Ž๐‘–๐‘™๐‘‘๐‘Ÿ๐‘’๐‘›, ๐‘ ๐‘ก๐‘Ž๐‘Ÿ๐‘ก <๐บ<๐บ

๐ต๐‘<๐บ

๐ต = BB, BG,GB,GG

๐‘(๐บ)=3

4

Ans. D

๐‘†30 = 3225 ๐‘‘ = 5 , ๐‘› = 30 , ๐ด1 =?

๐‘†๐‘› =๐‘›

2(2๐ด1 + (๐‘› โˆ’ 1)๐‘‘) โ‡’ ๐‘†30 =

30

2(2๐ด1 + (30 โˆ’ 1)5)

3225 = 15(๐ด1 + 29๐‘ฅ5)

215 = 2๐ด1 + 145

70 = 2๐ด1 โ‡’ ๐ด1 = 35 ๐‘๐‘’๐‘›๐‘ก๐‘ 

Ans. C

44. If the circle passing through the point (-1, 0) touches the y-axis at (0,2), then what is the

equation of the circle ?

(A) ๐‘ฅ2 + ๐‘ฆ2 + 5๐‘ฅ + 4๐‘ฆ + 4 = 0

(B) ๐‘ฅ2 + ๐‘ฆ2 โˆ’ 5๐‘ฅ + 4๐‘ฆ + 4 = 0

(C) ๐‘ฅ2 + ๐‘ฆ2 + 5๐‘ฅ โˆ’ 4๐‘ฆ + 4 = 0

(D) ๐‘ฅ2 + ๐‘ฆ2 โˆ’ 5๐‘ฅ โˆ’ 4๐‘ฆ + 4 = 0

45. Let ศค be a complex number and ๐‘ค = 3 + 4๐‘– . if ๐‘ง2+1

๐‘ง+๐‘– = |๐‘ค|ศค โˆ’

1

๐‘– ๏ฟฝฬ…๏ฟฝ , then what is the value of ศค ?

(A) -4- 2i

(B) โˆ’4 โˆ’ 2๐‘–

(C) โˆ’1 + ๐‘–

(D) โˆ’2๐‘–

โˆซโ„“๐‘› ๐‘ฅ

๐‘ฅ2

2

1

๐‘‘๐‘ฅ

Let u=โ„“๐‘› ๐‘ฅ ๐‘‘๐‘ฃ =1

๐‘ฅ2 ๐‘‘๐‘ฅ

๐‘‘๐‘› =1

2 ๐‘‘๐‘ฅ ๐‘‰ =

โˆ’1

๐‘ฅ

โˆซโ„“๐‘› ๐‘ฅ

๐‘ฅ2

2

1

= [โˆ’โ„“๐‘› ๐‘ฅ

๐‘ฅโˆ’

1

2]

1

2

=โˆ’โ„“๐‘› 2

2+

1

2

Ans. A

let (h, k) be the center and touches at (0,12) hence (h,2) be the centers at p(-1, 0)

(๐‘ฅ โˆ’ โ„Ž)2 + (๐‘ฆ โˆ’ ๐‘˜)2 = ๐‘Ÿ2

(โˆ’1 โˆ’ โ„Ž)2 + (โˆ’2)2 = (โ„Ž โˆ’ 0)2 + (2 โˆ’ 2)2

โ„Ž2 โˆ’ 2โ„Ž + 1 + 4 = โ„Ž2

โˆ’2โ„Ž = โˆ’5

โ„Ž = 52โ„

โˆด (๐‘ฅ โˆ’ 52โ„ )2 + (๐‘ฆ โˆ’ 2)2 = โ„Ž2 =

25

4

โˆด ๐‘ฅ2 + ๐‘ฆ2 + 5๐‘ฅ โˆ’ 4๐‘ฆ + 4 = 0

Ans. C

46. What is the value of the left-hand side limit, lim๐‘ฅโ†’0โˆ’

sin ๐‘ฅ

๐‘ฅ2 โˆš4๐‘ฅ2 + 9๐‘ฅ2 ?

(A) Does not exist

(B) 3

(C) 2

(D) -3

47. If โˆ’1 1 23 2 ๐‘ฅ2 4 1

= โˆ’๐‘ฅ 3 22 2 31 โˆ’1 โˆ’1

, then what is the value of x

(๐ด) โˆ’2

5

(๐ต) 2

5

(๐ถ) 2

3

(๐ท) โˆ’2

3

lim๐‘ฅโ†’0ฬ…

sin ๐‘ฅ

๐‘ฅ2 โˆš4๐‘ฅ3 + 9๐‘ฅ2

โ‡’ lim๐‘ฅโ†’0ฬ…

sin ๐‘ฅ

๐‘ฅ โˆš

4๐‘ฅ3 + 9๐‘ฅ2

๐‘ฅ2

โ‡’ lim๐‘ฅโ†’0ฬ…

sin ๐‘ฅ

๐‘ฅ , lim

๐‘ฅโ†’0ฬ…โˆš4๐‘ฅ + 9

1,-3

=-3

Ans. D

๐‘ง2 + 1 = (๐‘ง + ๐‘–)(๐‘ง โˆ’ ๐‘–)

[๐‘ค] = โˆš32 + 42 = 5

๐‘ค ฬ…ฬ… ฬ… = 3 โˆ’ 4๐‘–

โˆด๐‘ง2 + 1

๐‘ง + ๐‘–= |๐‘ค|๐‘ง โˆ’

1

๐‘– ๏ฟฝฬ…๏ฟฝ

(๐‘ง + ๐‘–)(๐‘ง โˆ’ ๐‘–)

2 + ๐‘–= 5๐‘ง + ๐‘–(3 โˆ’ 4๐‘–)

๐‘ง โˆ’ ๐‘– = 5๐‘ง + 3๐‘– + 4

๐‘ง โˆ’ 5๐‘ง = 3๐‘– + ๐‘– + 4

โˆ’4๐‘ง = 4๐‘– + 4

๐‘ง =4๐‘– + 4

โˆ’4

๐‘ง = โˆ’๐‘– โˆ’ 1

Ans. B

48. Consider the following system of equations: {

๐‘ฅ โˆ’ 2๐‘ฆ + ๐‘ง = 1โˆ’๐‘ฅ + ๐‘ฆ + ๐‘ง = 33๐‘ฅ โˆ’ 5๐‘ฆ + ๐‘ง = ๐‘˜

How much must be the value of k so that the system has a solution?

(A) 7

(B) 1

(C) -1

(D) 0

+ - + + - +

๐ด = |โˆ’1 1 23 2 ๐‘ฅ2 4 1

| = ๐ต |โˆ’๐‘ฅ 3 22 2 31 โˆ’1 โˆ’2

|

โ‡’ โˆ’1 |2 ๐‘ฅ4 1

| โˆ’ |3 ๐‘ฅ2 1

| + 2 |3 32 4

| = โˆ’1(2 โˆ’ 4๐‘ฅ) โˆ’ 1(3 โˆ’ 2๐‘ฅ) + 2(12 โˆ’ 4)

โ‡’ โˆ’2 + 4๐‘ฅ โˆ’ 3 + 2๐‘ฅ + 16

โ‡’ 6๐‘ฅ + 11

โ‡’ โˆ’๐‘ฅ | 2 3

โˆ’1 โˆ’ 2| โˆ’ 3 |

2 31 โˆ’ 2

| + 2 |2 2

1 โˆ’ 1| = โˆ’๐‘ฅ(โˆ’4 + 3) โˆ’ 3(โˆ’4 โˆ’ 3) + 2(โˆ’2 โˆ’ 2)

โ‡’ โˆ’๐‘ฅ (โˆ’1) โˆ’ 3(โˆ’7) + 2(โˆ’4)

โ‡’ ๐‘ฅ + 21 โˆ’ 8

โ‡’ ๐‘ฅ โˆ’ 13

โˆด 6๐‘ฅ + 11 = ๐‘ฅ + 13

6๐‘ฅ โˆ’ ๐‘ฅ = 13 โˆ’ 11

5๐‘ฅ = 2

๐‘ฅ = 25โ„

Ans. B

Using Augment matrix and by applying elementary raw operations to write in row echelon from then

solve for k. so that k = -1 .

Ans. C

49. What is the simplified form of ๐‘Žโˆ’1 ๐‘โˆ’1

๐‘Žโˆ’3โˆ’ ๐‘โˆ’3 ?

(๐ด) ๐‘Ž2๐‘2

๐‘2 โˆ’ ๐‘Ž2

(๐ต) ๐‘Ž3 โˆ’ ๐‘3

๐‘Ž โˆ’ ๐‘

(๐ถ) ๐‘Ž3 โˆ’ ๐‘3

๐‘Ž๐‘

(๐ท) ๐‘Ž2 ๐‘2

๐‘3 โˆ’ ๐‘Ž3

50. If ๐‘“(๐‘ฅ) = ๐‘Ž๐‘ฅ3 +๐‘

๐‘ฅ +5 has local minimum at (2, -3) , what are the values of a and b?

(A) ๐‘Ž = +1

4 , b=12

(B) ๐‘Ž =1

4 , b=12

(C) ๐‘Ž = โˆ’1

4 , b=-12

(D) ๐‘Ž =1

4 , b=-12

๐‘“(๐‘ฅ) = ๐‘Ž๐‘ฅ3 +๐‘

๐‘ฅ+ 5 โ„Ž๐‘Ž๐‘  ๐‘š๐‘–๐‘›๐‘–๐‘š๐‘ข๐‘š ๐‘Ž๐‘ก (2, โˆ’3)

๐‘“(2) = โˆ’3

16๐‘Ž + ๐‘ = โˆ’16 โ€ฆ โ€ฆ (๐‘ฅ)

๐‘†๐‘–๐‘›๐‘๐‘’ โˆ’ 2 ๐‘–๐‘  ๐‘Ž ๐‘๐‘Ÿ๐‘–๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ

๐‘“โ€ฒ(๐‘ฅ) = 0, ๐‘ฅ = 2

๐‘“โ€ฒ(๐‘ฅ) = 3๐‘Ž๐‘ฅ2 โˆ’๐‘

๐‘ฅ2

๐‘“โ€ฒ(2) = 0

48๐‘Ž โˆ’ ๐‘ = 0 โˆ’ โˆ’ โˆ’ (๐‘ฅ๐‘ฅ)

๐‘“๐‘Ÿ๐‘œ๐‘š (โˆ—)๐‘Ž๐‘›๐‘‘ (โˆ—โˆ—)๐‘ค๐‘’ ๐‘”๐‘’๐‘ก

๐‘Ž =โˆ’1

4 ๐‘Ž๐‘›๐‘‘ ๐‘ = โˆ’12

Ans. C

๐‘Žโˆ’1 ๐‘โˆ’1

๐‘Žโˆ’3 โˆ’ ๐‘โˆ’3=

1๐‘Ž๐‘

1๐‘Ž3 โˆ’

1๐‘3

=๐‘Ž2๐‘2

๐‘3 โˆ’ ๐‘Ž3

Ans. D

51. Which one of the following is equal to (๐‘ฅ) = โˆš(๐‘ฅ + 4)2 , for every ๐‘ฅ ๐œ– ?

(A) ๐‘”(๐‘ฅ) = |๐‘ฅ + 4|

(๐‘) ๐‘”(๐‘ฅ) = ๐‘ฅ + 2

(๐ถ) ๐‘” (๐‘ฅ) = ๐‘ฅ + 4

(๐ท) ๐‘”(๐‘ฅ) = |๐‘ฅ| +4

52. Ship A and B depart from the point at the same time on the course N600 E and N400 E,

respectively. If the speed of ship A is 20 km per hour and the speed of ship B is 30 km per hour,

what is the distance between the two ships just after 30 minutes of their departure?

[๐‘ฆ๐‘œ๐‘ข ๐‘š๐‘Ž๐‘ฆ ๐‘ก๐‘Ž๐‘˜๐‘’: cos(400) = 0.77, cos(200) = 0.94, sin(200) = 0.34 ]

(A) โˆš43 km

(B) โˆš40 km

(C) โˆš50 km

(D) โˆš53 km

53. Which the plane is rotated 450 about the point (1,-2), then what would be the image of the point

(2,4)?

(A) (1 โˆ’ โˆš2

2, โˆ’2 +

โˆš2

2)

(B) (1 โˆ’ โˆš 2

5

2, โˆ’2 +

โˆš27

2)

๐‘“(๐‘ฅ)โˆš(๐‘ฅ + 4)2, ๐‘ ๐‘–๐‘›๐‘๐‘’ ๐‘กโ„Ž๐‘’ ๐‘–๐‘›๐‘‘๐‘’๐‘ฅ ๐‘–๐‘  ๐‘’๐‘ฃ๐‘’๐‘›

๐‘“(๐‘ฅ) = |๐‘ฅ + 4|

Ans. A

Speed of ship A= 20 km/hr

Speed of shiep B= 30 km/hr

Both goes with in 30 min=0.5hr

The distance covered by sheep A= 10 km and, B= 15km using the relation S= Vt and angle b/n two

ships is ๐œƒ = 200 using cosine low the distance b/n them is ๐‘‘2 = 102 + 152 โˆ’ 2 โˆ— 10 โˆ— 15 cos 20

๐‘‘ = โˆš43 ๐‘˜๐‘š

Ans. A

(C) (โˆš 2

5

2,

โˆš27

2)

(D) (โˆš2

2,

โˆš2

2)

54. If u= 2 ๐‘—โ†’ โˆ’ ๐‘˜โ†’ and v=๐‘–โ†’ โˆ’ 8๐‘—โ†’ + 3๐‘˜โ†’, then what is the unit vector in the

direction of 5u+v?

(A) ๐‘–โ†’ + 2๐‘—โ†’ โˆ’ 2๐‘˜โ†’

(B) 2

3๐‘–โ†’ +

1

3๐‘—โ†’ โˆ’

2

3๐‘˜โ†’

(C) 1

3๐‘–โ†’ +

2

3๐‘—โ†’ +

2

3๐‘˜โ†’

(D) 1

3๐‘–โ†’ +

2

3๐‘—โ†’ โˆ’

2

3๐‘˜โ†’

๐‘†๐‘ข๐‘๐‘ ๐‘ก๐‘–๐‘ก๐‘ข๐‘ก๐‘’ (๐‘ฅ0, ๐‘ฆ0) = (1, โˆ’2), ๐œƒ = 450 ๐‘Ž๐‘›๐‘‘ (๐‘ฅ, ๐‘ฆ) = (2,4)

๐‘ฅ1 = ๐‘ฅ0 + (๐‘ฅ โˆ’ ๐‘ฅ0)๐‘๐‘œ๐‘ ๐œƒ โˆ’ (๐‘ฆ โˆ’ ๐‘ฆ0) ๐‘ ๐‘–๐‘› ๐œƒ

= 1 + (2, โˆ’1) ๐‘๐‘œ๐‘  45โ€” (4 โˆ’ (โˆ’2)) ๐‘ ๐‘–๐‘› 450

= 1 +โˆš2

2โˆ’

6โˆš2

2= 1 โˆ’

5โˆš2

2

๐‘ฆโ€ฒ = ๐‘ฆ00 + (๐‘ฅ โˆ’ ๐‘ฅ0) ๐‘ ๐‘–๐‘› ๐œƒ + (๐‘ฆ โˆ’ ๐‘ฆ0) ๐‘๐‘œ๐‘ ๐œƒ

= โˆ’2 +โˆš2

2+

6โˆš2

2= โˆ’2 +

7โˆš2

2

Ans. B

๐‘ข = 2๐‘— โˆ’ ๐‘˜ ๐‘Ž๐‘›๐‘‘ ๐‘ฃ = ๐‘– โˆ’ 8๐‘— + 3๐‘˜

5๐‘ข = 10๐‘— โˆ’ 5๐‘˜

5๐‘ข + ๐‘ฃ = ๐‘– + 2๐‘— โˆ’ 2๐‘˜

|5๐‘ข + ๐‘ฃ| = โˆš12 + 22 + (22) = โˆš9 =3

The unit rector in the direction of 5u+v is

=5u+v

|5u+v|

1๐‘–

3+

2

3๐‘— โˆ’

2

3๐‘˜

Ans. D

55. The diagram below is a representation of a 25m vertical observation tower TB and two cars K

and L on a road. The angle of depression from T to car L is300. The angle of elevation from car

K to the of the tower is 600. B, K and L lie in a straight line lie on the same horizontal plane as

the base of the tower.

T

300

600

B K L

What is the distance between the two cars?

(A) 50โˆš3

2 m

(B) 50โˆš3 m

(C) 50โˆš3

3 m

(D) 50+โˆš3 m

25 m

30

25 B=60

๐œƒ = 60

x y

tan ๐œƒ = 25

๐‘ฅ โ‡’ ๐‘ฅ =

25

tan 60=

25

โˆš3=

25โˆš3

3

tan ๐œƒ = ๐‘ฅ + ๐‘ฆ

25 โ‡’ ๐‘ฅ + ๐‘ฆ = 25 tan ๐ต = 25โˆš3

Now we need to calculate y to determine the distance b/n the two ships

๐‘ฆ = 25โˆš3 โˆ’ ๐‘ฅ = 25โˆš3 โˆ’25โˆš3

3

๐‘ฆ = 50โˆš3

3m

Ans. C

56. Which of the following is a correct assertion that can be proved by the principle of

mathematical induction?

(A) 1

๐‘›+1 โ‰ค 1, for each real number ๐‘› โ‰ฅ 1.

(B) ๐‘˜! โ‰ฅ 2๐‘˜, for each integer ๐‘˜ โ‰ฅ 4.

(C) 2๐‘ โˆ’ 1, is prime for each prime integer ๐‘.

(D) ๐‘š! โ‰ค 4๐‘š, for each positive integer ๐‘š.

57. If the between the vectors ๐ดโ†’=(2,-1,1) and ๐ตโ†’=(1,1, ๐‘Ž) is ๐œ‹

3, then what is the value of ๐‘Ž ?

(A) 2

(B) -1

(C) 1

(D) -2

Ans. B

๐ด(2, โˆ’1,1) ๐‘Ž๐‘›๐‘‘ ๏ฟฝโƒ—โƒ—๏ฟฝ = (1,1, ๐œƒ), ๐œƒ =๐œ‹

3

๐ด. ๐ต = |๐ด| |๐ต| cos ๐œƒ

(2)(1) + (โˆ’1)(1) + (1)(๐‘Ž) = โˆš22 + 12 + 12, โˆš12 + 12 + ๐‘Ž2 cos 600

1 + ๐‘Ž = โˆš6 , โˆš2 + ๐‘Ž2.1

2

2 + 2๐‘Ž = โˆš12 + 6๐‘Ž2

(2 + 2๐‘Ž)2 = 12 + 6๐‘Ž2

4๐‘Ž2 + 8๐‘Ž + 4 = 12 + 6๐‘Ž2

โˆ’2๐‘Ž2 + 8๐‘Ž โˆ’ 8 = 0

๐‘Ž2 โˆ’ 4๐‘Ž + 4 = 0

(๐‘Ž โˆ’ 2)2 = 0

๐‘Ž โˆ’ 2 = 0

๐‘Ž = 2

Ans. A

58. Let L be the line by the vector equation (๐‘ฅ, ๐‘ฆ) = (1,1) + ๐‘ก(โˆš3.1 ),๐‘ก๐œ–.. What is the equation of

the image of L after being rotated 150 about (1,1) and then translated by vector u= (โˆ’1,1)

(A) ๐‘ฅ โˆ’ ๐‘ฆ = 2

(B) โ€“ ๐‘ฅ + ๐‘ฆ = 2

(C) โˆš3 ๐‘ฅ โˆ’ ๐‘ฆ = 2

(D) โ€“+โˆš3 y=1

59. What is the solution set of ๐‘ ๐‘–๐‘›2 ๐‘ฅ โˆ’ ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ = 0 over [0,2๐œ‹]?

(A) {0, ๐œ‹,5๐œ‹

4, 2๐œ‹}

(B) {0,๐œ‹

4, ๐œ‹, ,2๐œ‹}

(C) {0,๐œ‹

4, ๐œ‹,

5๐œ‹

4}

(D) {0,๐œ‹

4, ๐œ‹,

5๐œ‹

4, 2๐œ‹}

L:(๐‘ฅ, ๐‘ฆ) = (1,1) + ๐‘ก(โˆš3 ,1) , ๐‘ก๐ธ๐‘…

๐‘ฅ = 1 + โˆš3 ๐‘ก , ๐‘ฆ = 1 + ๐‘ก

โ‡’๐‘ฅ โˆ’ 1

โˆš3= ๐‘ฆ โˆ’ 1

๐‘ฅ โˆ’ 1 = โˆš3 ๐‘ฆ โˆ’ โˆš3

๐‘ฆ =1

โˆš3 ๐‘ฅ +

โˆš3 โˆ’ 1

โˆš3

Here, slope m=1

โˆš3 , tan ๐œƒ =

1

โˆš3

๐œƒ = 300

(1,1) ๐‘‡๐‘Ÿ(โˆ’1,1), (0,2)

Slope of image of โ„“ = 450 , tan 450 = 1

โˆด ๐‘‡๐‘Ž๐‘˜๐‘–๐‘›๐‘” ๐‘๐‘œ๐‘–๐‘›๐‘ก (0,2)๐‘Ž๐‘›๐‘‘ ๐‘ ๐‘™๐‘œ๐‘๐‘’ ๐‘š = 1

๐‘ฆ โˆ’ 2 = 1(๐‘ฅ โˆ’ 0)

๐‘ฆ โˆ’ 2 = ๐‘ฅ

๐‘ฆ = ๐‘ฅ + 2 โ‡’ ๐‘ฆ โˆ’ ๐‘ฅ = 2

Ans. B

60. Consider a rectangle ABCD with base vertices A= (0,3) and B= (4,0) and the other vertices, C

and D, in the first-quadrant of the coordinate plane. If its height BC is half of the length of its

base, then which of the following indicates the coordinates of vertex C?

(A) (4,5/2)

(B) (11/2,2)

(C) (5/2,2)

(D) (6,3/2)

61. โˆ€๐‘› โˆˆN, 3๐‘› โˆ’ 2 prime that can be proved or disproved by which one of the following

mathematical proofs?

(A) Disprove by counter example

(B) Proof by exhaustion

(C) Direct proof

(D) Proof by contradiction

๐ด๐ต๐ถ๐ท ๐‘–๐‘  ๐‘Ÿ๐‘’๐‘๐‘ก๐‘Ž๐‘›๐‘”๐‘™๐‘’

๐ด๐ต = โˆš32 + 42 = 5

๐ต๐ถ =1

2 ๐ด๐ต =

5

2

(๐‘ฅ โˆ’ 4)2 + ๐‘ฆ2 =25

4โˆ’ ๐‘–๐‘  ๐‘Ž ๐‘๐‘–๐‘Ÿ๐‘๐‘™๐‘’ ๐‘ค๐‘–๐‘กโ„Ž ๐‘๐‘’๐‘›๐‘ก๐‘’๐‘  ๐‘(4,0)๐‘Ž๐‘›๐‘‘ ๐‘Ÿ =

5

2

โˆด ๐‘ ๐‘–๐‘  ๐‘Ž ๐‘๐‘œ๐‘–๐‘›๐‘ก ๐‘œ๐‘› ๐‘กโ„Ž๐‘’ ๐ถ๐‘–๐‘Ÿ๐‘๐‘™๐‘’

๐ถ(6, 32โ„ )

Ans. D

๐‘ ๐‘–๐‘›2๐‘ฅ โˆ’ sin ๐‘ฅ cos ๐‘ฅ = 0

sin ๐‘ฅ (sin ๐‘ฅ โˆ’ cos ๐‘ฅ) = 0

sin ๐‘ฅ = 0 or sin ๐‘ฅ โˆ’ cos ๐‘ฅ = 0

sin ๐‘ฅ = 0, ๐‘ฅ = 0 , ๐œ‹, 2๐œ‹

sin ๐‘ฅ โˆ’ cos ๐‘ฅ = 0 โ‡’ sin ๐‘ฅ = cos ๐‘ฅ, ๐‘ฅ =๐œ‹

4 ,

5๐œ‹

4

๐‘ . ๐‘  = {0,๐œ‹

4 , ๐œ‹,

5๐œ‹

4, 2๐œ‹}

Ans. D

62. If the point (๐‘Ž, 0, 3) is on the sphere centered at (1, 2, 3) with radius 2, what is the value of ๐‘Ž?

(A) 0

(B) 2

(C) 1

(D) -3

(E)

63. What is the image of the circle ๐‘ฅ2+๐‘ฆ2 + 4๐‘ฅ โˆ’ 6๐‘ฆ + 11 = 0, when the origin is shifted to the

point (1,1) after translation of axes ?

(A) ๐‘ฅ2+๐‘ฆ2 โˆ’ 4๐‘ฅ โˆ’ 2๐‘ฆ + 3 = 0

(B) ๐‘ฅ2+๐‘ฆ2 โˆ’ 4๐‘ฅ โˆ’ 6๐‘ฆ + 3 = 0

(C) ๐‘ฅ2+๐‘ฆ2 โˆ’ 6๐‘ฅ + 8๐‘ฆ โˆ’ 23 = 0

(D) ๐‘ฅ2+๐‘ฆ2 โˆ’ 6๐‘ฅ โˆ’ 8๐‘ฆ + 23 = 0

Let ๐‘ฅ0 , ๐‘ฆ0 ๐‘Ž๐‘›๐‘‘ ๐‘ง0 be the center of the sphere

โˆด (๐‘ฅ โˆ’ ๐‘ฅ0)2 + (๐‘ฆ โˆ’ ๐‘ฆ0)2 + (๐‘ง โˆ’ ๐‘ง0)2 = ๐‘Ÿ2, ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ฅ, ๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘ง ๐‘Ž๐‘Ÿ๐‘’ ๐‘๐‘œ๐‘–๐‘›๐‘ก๐‘  ๐‘œ๐‘› ๐‘กโ„Ž๐‘’ ๐‘ ๐‘โ„Ž๐‘’๐‘Ÿ๐‘’.

(๐‘Ž โˆ’ 1)2 + (0 โˆ’ 2)2 + (3 โˆ’ 3)2 = 22

(๐‘Ž โˆ’ 1)2 + 4 + 0 = 4

(๐‘Ž โˆ’ 1)2 = 0 โ‡’ ๐‘Ž = 1

Ans.C

โˆ€๐‘› โˆˆ ๐‘, 3๐‘› โˆ’ 2 ๐‘–๐‘  ๐‘๐‘Ÿ๐‘–๐‘š๐‘’

๐ฟ๐‘’๐‘ก ๐‘› = 1

3๐‘› โˆ’ 2 = 31 โˆ’ 2 = 1 โˆ’ ๐‘๐‘’๐‘–๐‘กโ„Ž๐‘–๐‘›๐‘” ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘›๐‘œ๐‘Ÿ ๐‘๐‘œ๐‘š๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’

Disprove by a counter example.

Ans. A

Give a circle with equation

๐‘ฅ2 + ๐‘ฆ2 โˆ’ 4๐‘ฅ โˆ’ 6๐‘ฆ + 11 = 0, ๐‘๐‘œ๐‘ค ๐‘–๐‘“ ๐‘ค๐‘’ ๐‘ โ„Ž๐‘–๐‘“๐‘ก

(1.1)units the image becomes

(๐‘ฅ โˆ’ 1)2 + (๐‘ฆ โˆ’ 1)2 โˆ’ 4(๐‘ฅ โˆ’ 1) โˆ’ 6(๐‘ฆ โˆ’ 1) + 11 = 0

๐‘ฅ2 โˆ’ 2๐‘ฅ + 1 + ๐‘ฆ2 โˆ’ 2๐‘ฆ + 1 โˆ’ 4๐‘ฅ + 4 โˆ’ 6๐‘ฆ + 6 + 11 = 0

๐‘ฅ2 + ๐‘ฆ2 โˆ’ 2๐‘ฅ โˆ’ 4๐‘ฅ โˆ’ 2๐‘ฆ โˆ’ 6๐‘ฆ + 1 + 1 + 4 + 6 + 11 = 0

๐‘ฅ2 + ๐‘ฆ2 โˆ’ 6๐‘ฅ โˆ’ 8๐‘ฆ + 23 = 0

Ans. D

64. What is the value of cot 2700+2cos 900+4๐‘ ๐‘’๐‘2 + 1800 ?

(A) 4

(B) 8

(C) -2

(D) 7

65. Which one of the following is a valid assertion that can be proved by the principle of

mathematical induction?

(A) 3๐‘› + 25 < 3๐‘›, for every integer ๐‘› โ‰ฅ 3.

(B) 2๐‘› > ๐‘› + 20, for every integer ๐‘› โ‰ฅ 4.

(C) ๐‘›3 โˆ’ ๐‘› ๐‘–๐‘  ๐‘‘๐‘–๐‘ฃ๐‘–๐‘ ๐‘–๐‘๐‘™๐‘’ ๐‘๐‘ฆ 6, for every integer ๐‘› โ‰ฅ 1.

(D) ๐‘›2 โ‰ค 2๐‘›, for every integer ๐‘› โ‰ฅ 1.

THE END

๐‘‡โ„Ž๐‘’ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’ ๐‘œ๐‘“ cot 2700 + 2 ๐‘๐‘œ๐‘  900 + 4 ๐‘ ๐‘’๐‘2 1800

cot 2700 =1

tan 270=

๐‘๐‘œ๐‘  2700

๐‘ ๐‘–๐‘›2700=

0

โˆ’1= 0

2 cos 900 = ๐‘ฅ โˆ— 0 = 0

4 ๐‘ ๐‘’๐‘2 1800 = 4(โˆ’1)2 = 4

โˆด 0 + 0 + 4 = 4

Ans. A

Ans. C , sine ๐‘›3 โˆ’ ๐‘› ๐‘“๐‘œ๐‘Ÿ ๐‘› โ‰ฅ 1 ๐‘–๐‘  ๐‘Ž๐‘™๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘‘๐‘–๐‘ฃ๐‘–๐‘ ๐‘–๐‘๐‘™๐‘’ ๐‘๐‘ฆ 6

top related