ethiopian university entrance examination (euee
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 −3
2y
A (y) = (96- 3
2 𝑦) 𝑦 = 96−
3
2𝑦2 = ; 0≤ 𝑦 ≤ 64
A (y)= 96y -3
2𝑦2 is contain [0,64] and diff. on (0,64)
A’(y)= 0 ⟹ 96-3y=0
⟹y=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 𝑚2
Ans. C
2. If 𝑓(𝑥)= 𝑥3 -2 𝑥+1, then which one of the following is NOT true?
(A) 𝑓(𝑐)= 1
4 for some ∁ ∈ [−1, 0].
(B) 𝑓(𝑐)= 1
2 for some ∁ ∈ [0, 1].
(C) 𝑓(𝑐)= 0 for some ∁ ∈ [−2, 0].
(D) 𝑓(𝑐)= 3 for some ∁ ∈ [1, 2].
3. If ) 𝑓(𝑥) = In (2tan 𝑥), then what is the value of 𝑓′(0)?
(A) 1
(B) -2ln 2
(C) ln 2
2
(D) ln2
4. If 𝑓(𝑥) = 𝛼𝑥 − 𝑏 and 𝑓−1 (𝑥 + 1)=1
2𝑥+2, for each 𝑥 𝜖, then what must be the value of
𝛼 and 𝑏?
(A) 𝛼 = ½, 𝑏= -2
(B) 𝛼 =2, 𝑏 =3
Let C∈R such that
𝑓(𝑥) = 𝑥2 − 2𝑥 + 1
Ans. A is using intermediate value theorem
If C ∈ [−1,0]
𝑓(−1) = 2 𝑎𝑛𝑑 𝑓(0) = 1
Hence, 1
4 is not included b/n that values
𝑓(𝑥) = ln(2tan 𝑥)
𝑓(𝑥) = 𝑔(ℎ(𝑥))
Where, g(x) - ln 𝑥, 𝑔′(𝑥) =1
𝑥
h(x) = 2tan 𝑥 , h′(x) = 2tan 𝑥 , 𝑡𝑎 𝑛 2 𝑠𝑒𝑐2𝑥
𝑓′(𝑥) = 1
2tan 𝑥 . ln 2 𝑠𝑒𝑐2 𝑥
𝑓′(0) = 1
2tan 0. ln 2 𝑠𝑒𝑐2 𝑥
= 𝐥𝐧 2
Ans. D
(C) 𝛼 =1, 𝑏=1
(D) 𝛼 =2, 𝑏=2
(E)
5. If 2(2𝑥 𝑥−5 −3
) -1 = (3 2
−5 −4), then what is the value of 𝑥?
(A) 2
(B) -2
(C) 1
2
(D) - 1
2
6. What is the area of the region bounded by the lines 𝑥 = 0, 𝑥 = 2 and 𝑦 = 1 and the
curve 𝑦 = 𝑒2𝑥?
(A) 𝑒2
2 -
1
2
(B) 𝑒4
2 -
5
2
(C) 𝑒3
3 -
1
3
(D) - 𝑒4
2 +
5
2
𝑓(𝑥) = 𝑎𝑥 − 𝑏
𝑓−1(𝑥) =𝑥 + 𝑏
𝑎
𝑓−1(𝑥 + 1) =𝑥 + 1 + 𝑏
𝑏 , 𝑏𝑢𝑡 𝑓−1(𝑥 + 1) =
1
2𝑥 + 2 =
𝑥 + 4
2
𝑥 + 1 + 𝑏
𝑎=
𝑥 + 4
2
∴ 𝑎 = 2 𝑎𝑛𝑑 1 + 𝑏 = 4 𝑏 = 3
Ans. B
2(2𝑥 𝑥 −5 −3
)−1
= (3 2−5 −4
) , 𝐴−1 =1
2 ( 3 2
−5 −4)
Let 𝐴−1 =1
2( 1 2
−5 −4 ) 𝐴−1 =
1
2 (3 2
−5 −4)
−1
A=( 4 2−5 −3
) , 2x = 4
X= 2
Ans, A
7. If 𝑓 if the greatest integer function and g is the absolute value function, the what is the
value of (𝑓 𝑜 g) (3
2) + (g 𝑜 𝑓) (−
4
3) ?
(A) 1
(B) 2
(C) -1
(D) 3
𝒚 = 𝒆𝒙𝟐
The area b/n the line x=0, x=2 and y=1 is
𝑒4
y = 1
y = 0 x = 2
A ∫ 𝑒2𝑥2
0 𝑑𝑥 , let u = 2𝑥 , 𝑥 = 0, u =
𝑥 = 2, 𝑢 = 4
𝑑𝑛 = 2 𝑑𝑥
1
2 𝑑𝑛 = 𝑑𝑥
A1
2 ∫ 𝑒44
0 𝑑𝑦
= 1
2[𝑒4]4
=1
2 [𝑒4 − 𝑒0] =
1
2 [𝑒4 − 1]
𝑒4
2−
1
2 Area of rectangle
𝑒4
2−
1
2− 2 =
𝑒4
2−
5
2
Ans, B
8. Which one of following is true about the graph of 𝑓(𝑥)𝑥2+5𝑥+6
𝑥2−4 +3?
(A) A horizontal asymptote of the graph of the graph is 𝑦 =4.
(B) The vertical asymptotes of the graph are 𝑥 = -2 and 𝑥 =2.
(C) The graph has a hole at 𝑥 =2.
(D) The graph has 𝑦- intercept at (0, −3
2)
9. What is the solution of ∫ 4𝑥 (ℓ𝑛 𝑥 +1
𝑥2)d𝑥 ?
(A) 4𝑥2(ℓ𝑛 𝑥 + 1)-2𝑥2+c
(B) 𝑥2(2ℓ𝑛 𝑥 − 1)+4 𝑙𝑛 𝑥 +c
(C) 𝑥2(2ℓ𝑛 𝑥 + 1)+4 𝑙𝑛 𝑥 +c
(D) 𝑥2(4ℓ𝑛 𝑥 − 1)+2 𝑙𝑛 𝑥 +c
Give –f is the greatest integer function and g is the absolute value function , then (fog) (3/2)+ (gof)(-
4/3)= 1+2=3
(𝑓𝑜𝑔) (3/2) = 𝑓(𝑔 (3/2))
= 𝑓(3/2) = 1
(𝑔𝑜𝑓) (−4/3) = 𝑔(𝑓(−4/3))
= 𝑔(−2)
= 2
Ans. D
𝒇(𝒙) 𝒙𝟐+𝟓𝒙+𝟔
𝒙𝟐−𝟒 +3
=𝒙𝟐+𝟓𝒙+𝟔+𝟑(𝒙𝟐−𝟒)
𝒙𝟐−𝟒 =
𝟔
−𝟒+ 𝟑
=𝒙𝟐+𝟓𝒙+𝟔+𝟑𝒙 𝟐−𝟏𝟐
𝒙𝟐−𝟒 =
−𝟑
𝟐+ 𝟑 −
𝟑
𝟐
= 𝟒𝒙𝟐+𝟓𝒙−𝟔
𝒙𝟐−𝟒
Hence 𝒎 = 𝒏, ⇒ 𝒚 = 𝟒 is a H.A
𝑿 = 𝟐 is a V.A
Its y intercept is ( 𝟎,𝟑
𝟐)
Makes a hole at 𝒙 = −𝟐
Ans. A
10. The earth’s orbit has a semi-major axis 𝛼 ≈ 149.6 Gm (gigameters) and an eccentricity
of 𝑒 ≈ 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
∫ 4𝑥 (ℓ𝑛 𝑥 +1
𝑥2 ) 𝑑𝑥
= ∫ 4𝑥 ℓ𝑛 𝑥 +4
𝑥 𝑑𝑥
= ∫ 4𝑥 ℓ𝑛 𝑥 𝑑𝑥 + ∫4
𝑥 𝑑𝑥
= 4 [∫ 𝑥 ℓ𝑛 𝑥 𝑑𝑥 + ∫1
𝑥 𝑑𝑥]
= 4 [𝑥2ℓ𝑛 𝑥
2−
𝑥2
4+ ℓ𝑛 𝑥] + 𝑐
=2𝑥2ℓ𝑛 𝑥 − 𝑥2 + 4 ℓ𝑛 𝑥 + 𝑐
= 𝑥2(2ℓ𝑛 𝑥 − 1) + 4 ℓ𝑛 𝑥 + 𝑐
Ans. B
𝑎 ≈ 149 𝐺𝑀
𝑒 ≈ 0.017
𝑒𝑐
𝑎
⇒ 𝑐 = 𝑒𝑎
= 0.017 x149.6 G.M
= 2.5432 G.M
From the relation
𝑎2 = 𝑏2+𝑐2
𝑏2 = 𝑎2-𝑐2
=(149.6)2 − (2.5432)2
𝑏2 = 22,373.69 G.M2
𝑏 = 149.58𝑚
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 ∑ 5∞𝑘=0
2𝑘 + 5𝑘
10 𝑘
(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
𝑑𝑣
𝑑𝑡= (𝐴𝑏ℎ)
𝑑ℎ
𝑑𝑡 ,
𝑑𝑣
𝑑𝑡= 0.12𝑚2/𝑠𝑒𝑐
𝑑𝑣
𝑑𝑡= 𝐴𝑏
𝑑ℎ
𝑑𝑡
∴𝑑ℎ
𝑑𝑡=
𝑑𝑣
𝑑𝑡 ,
1
𝐴𝑏
=0.12𝑚3/𝑠𝑒𝑐
6𝑚2= 0.02𝑚/𝑠𝑒𝑐
Ans. C
Sum of ∑ 5 (2𝑘+5𝑘
10𝑘 )∞𝐿=0
⇒ 5 [∑2𝑘
10𝑘
∞
𝑘=0
+ ∑5𝑘
10𝑘
∞
𝑘=0
]
⇒ 5 [5
4+ 2]
⇒ 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 𝑓1 10 𝑓2
If the median of the data is 26, what are the values of 𝑓1𝑎𝑛𝑑 𝑓2 ?
(A) 𝑓1 = 20, 𝑓2 = 7
(B) 𝑓1 = 12, 𝑓2 = 15
(C) 𝑓1 = 15, 𝑓2 = 12
(D) 𝑓1 = 7, 𝑓2 = 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, �̅� = 5
New data, 4, 5, 5, 6, �̅� = 5 mean stays
S.D (data) =2
3= 0.67
S.D (new data)= 2
4= 0.5 It decreases
Ans. B
Given 𝑛 = 50, 𝐿𝐶𝐵 = 19.5 𝐶 + 𝑏 = 11
𝑓𝑏 = 𝑓1 , 𝑚𝑒𝑑𝑖𝑎𝑛 = 26
Median = LCB + (𝑛
2 −𝐶𝑓𝑏
𝑓𝑏) 𝑖
26 = 19.5 + (25 − 13
𝑓1) 11
6.5 =12.11
𝑓1
𝑓1 =132
6.5= 20 𝑓2 = 50 − 5 + 8 + 20 + 10 = 7
Ans.A
15. For real numbers x and y, which one of the following is NOT true?
(A) (∀𝑥)(∀𝑦) (𝑦2 + 𝑥2 ≥ −1)
(B) (∃𝑥)(∀𝑦) (𝑦 ≥ 𝑥2 + 1)
(C) (∀𝑥)(∃𝑦) (𝑦 ≥ 𝑥2 + 1)
(D) (∃𝑥)(∃𝑦) (𝑦 ≥ 𝑥2 + 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 don’t change my oil regularly, my engine will die. My engine died, Thus, I didn’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 can’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
𝑮𝟐 = −𝟏𝟐⁄ , 𝑮𝟓 = 𝟏
𝟏𝟔⁄
𝑮𝟐
𝑮𝟓
=𝒓𝑮𝟏
𝒓𝟒 𝑮𝟏
, 𝒓𝟑 =−𝟏
𝟐⁄𝟏
𝟏𝟔⁄⁄ =
−𝟏
𝟖
∴ 𝑮𝟏 = 𝟏 𝒓 =−𝟏
𝟐
𝑺𝟖 =𝑮𝟏(𝟏−𝒓𝟖)
𝟏 − 𝒓= 𝟖𝟓
𝟏𝟐𝟖⁄
Ans. C
18. What is the value of ∫𝑥𝑒−1 + 𝑒𝑥−1
𝑥𝑒 + 𝑒𝑥 d𝑥?
(A) 1
𝑒 ln(𝑥𝑒 + 𝑒𝑥+1)+ c
(B) 1
𝑒 ln(𝑥𝑒+1 + 𝑒𝑥+1)+ c
(C) 1
𝑒 ln((𝑥 + 1) 𝑒 + 𝑒𝑥+1)+ c
(D) 1
𝑒 ln(𝑥 𝑒 + 𝑒𝑥)+ 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 → 𝑝 ⇒ 𝑞⋀𝑟 P: I am L
𝑞∧¬𝑟
¬𝑝 q: I can read
r- I can write
∫ (𝑥𝑒−1 + 𝑒𝑥−1
𝑥𝑒 + 𝑒𝑥 ) 𝑑𝑥
(𝑥𝑒 + 𝑒𝑥 )1 = 𝑒𝑥𝑒−1 + 𝑒𝑥
Multiply both sides by 1 𝑒⁄ 𝑤𝑒 𝑔𝑒𝑡
(𝑥𝑒−1 + 𝑒𝑥−1) 1𝑒⁄
∫𝑥𝑒−1 + 𝑒𝑥−1
𝑥𝑒 + 𝑒𝑥𝑑𝑥 =
1
𝑒ℓ𝑛 |𝑥𝑒 + 𝑒𝑥| + 𝑐
Ans. D
Total item 2500
Test 2% of 2500=50
Defect=5
Non defect=45
P(E)= 4550
=0.9
Ans. B
20. If 𝑓(𝑥) = 1
𝑥, then what is the value of 𝑓(𝑛)(𝑥)?
(A) 𝑓(𝑛)(𝑥)= (−1) 𝑛 𝑛!
𝑥𝑛
(B) 𝑓(𝑛)(𝑥)= (−1) 𝑛+1 ( 𝑛+1)!
𝑥𝑛+1
(C) 𝑓(𝑛)(𝑥)= (−1) 𝑛 𝑛!
𝑥𝑛+1
(D) 𝑓(𝑛)(𝑥)= (−1) 𝑛 𝑛!
𝑥𝑛−1
21. If the region enclosed by the graphs of 𝑓(𝑥)= 𝑥2 and 𝑓(𝑥)= 𝑥3 from 𝑥 =0 to 𝑥 =1 rotates
about the 𝑥- axis, what is the volume of the solid of revolution?
(A) 2𝜋
35 cubic units
(B) 2𝜋
25 cubic units
(C) 2𝜋
5 cubic units
(D) 2𝜋
27 cubic units
(E)
(F)
𝑓(𝑥) =1
𝑥 , 𝑓′(𝑥) =
−1
𝑥2 , 𝑓′′(𝑥) =
2
𝑥3
𝑓′′′(𝑥) =−6
𝑥4
∴ 𝑓𝑛(𝑥) =(−1)𝑛. 𝑛:
𝑥𝑛+1
Ans. C
𝑉 = 𝜋 ∫ (𝑥2)21
0
− (𝑥3)2𝑑𝑥
= 𝜋 ∫ 𝑥4 − 𝑥6𝑑𝑥1
0
= 𝜋 [𝑥5
5−
𝑥7
7]
= 𝜋 [1
5−
1
7]
=2𝜋
35
Ans. A
22. What is the limit of sequence 1, 2
22−12 ,
3
32−22 ,
4
42−32 ,…?
(A) 0
(B) 1
2
(C) 1
(D) 3
2
(E)
23. At the value of x does the function 𝑓(𝑥) = 4𝑥2
3 - 𝑥4 attains maximum value?
(A) -1
(B) 0
(C) 4
3
(D) 1
2
22 − 12 ,
3
32 − 22 ,
4
42 − 32 , …
⇒𝑛
𝑛2 − (𝑛 − 1)2
⇒𝑛
𝑛2 − (𝑛2 − 2𝑛 + 1 =
𝑛
𝑛2 − 𝑛2 + 2𝑛 − 1
⇒𝑛
2𝑛 − 1 =
1
2 −1𝑛
, 𝑛 → ∞
lim𝑛→∞
1
2 −1𝑛
=1
2
Ans.B
𝑓(𝑥) =4𝑥3
3− 𝑥4
𝑓′(𝑥) = 4𝑥2 − 4𝑥3
= 4𝑥2(1 − 𝑥)
𝑓′(𝑥) = 0, 𝑥 = 0,1 𝑎𝑟𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡
0 1
4𝑥2 − − − − − − + + + + + + + + +
1 − 𝑥 + + + + + − − − − − − −
(4𝑥)2(1 − 𝑥) − − − −0 + +0 − − − − − − −
𝑓′(𝑥)𝑐ℎ𝑎𝑛𝑔𝑒𝑠𝑠𝑖𝑔𝑛 𝑓𝑟𝑜𝑚 − 𝑣𝑒 𝑡𝑜 + 𝑣𝑒 − 𝑎𝑡 𝑥 = 0
𝑓 ℎ𝑎𝑠 𝑎 𝑙𝑜𝑐𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑓′(𝑥)𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑠𝑖𝑔𝑛 𝑓𝑟𝑜𝑚 + 𝑣𝑒 𝑡𝑜 − 𝑣𝑒 𝑎𝑡 𝑥 = 1
𝑓 ℎ𝑎𝑠 𝑎𝑙𝑜𝑐𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
Ans. D
24. If ℎ(𝑥) =√1 + √𝑥 , then which of the following is equal to ℎ′(𝑥)?
(A) 1
4√𝑥+𝑥√𝑥
(B) 1
2√1+√𝑥
(C) 𝑥
2√1+√𝑥
(D) 𝑥
4√𝑥+𝑥√𝑥
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% ⇒ 𝑝(𝐹) = 0.9
Mesfin answers 70%⇒ 𝑝(𝑀) = 0.7
∴ 𝑃(𝐹𝑈𝑀) = 𝑃(𝐹) + 𝑃(𝑀) − 𝑃(𝐹𝑛𝑀)
= 𝑝(𝐹) + 𝑝(𝑀) − [𝑃(𝐹) ∗ 𝑃(𝑀)]
= 0.9 + 0.7 − [0.9 ∗ 0.7]
= 1.6 − 0.63
= 0.97
Ans. B
ℎ(𝑥) = √1 + √𝑥
Let ℎ(𝑥) = 𝑓(𝑔(𝑥))𝑤ℎ𝑒𝑟𝑒
𝑓(𝑥) = √𝑥 , 𝑓′(𝑥) = 𝑓′(𝑥) =1
2(𝑥
12⁄ ) =
1
2√𝑥
𝑔(𝑥) = 1 + √𝑥 =1
2√𝑥
ℎ′(𝑥) = 𝑓′(𝑔(𝑥)), 𝑔(𝑥) =1
2√1 + √𝑥 ,
1
2√𝑥
=1
4√𝑥(1 + √𝑥
= 1
4√x+x√x
Ans. A
26. What is the polar form of 7−𝑖
3−4𝑖 ?
(A) √2(𝑐𝑜𝑠 𝜋
2+ 𝑖 𝑠𝑖𝑛
𝜋
2)
(B) 2(𝑐𝑜𝑠 𝜋
4+ 𝑖 𝑠𝑖𝑛
𝜋
4)
(C) 2(𝑐𝑜𝑠 𝜋
2+ 𝑖 𝑠𝑖𝑛
𝜋
2)
(D) √2(𝑐𝑜𝑠 𝜋
4+ 𝑖 𝑠𝑖𝑛
𝜋
4)
27. 𝐼𝑓 ¬𝑝 ⇒ 𝑟 is False and p q is True , then which of the following is True?
(A) 𝑝 v (¬q ^r)
(B) ¬ 𝑝 ⇒ (q v r)
(C) ¬ 𝑝 ^(q ⇒ r)
(D) P (¬q v r)
Polar form of
7 − 𝑖
3 − 4𝑖 ,
3 + 4𝑖
3 + 4𝑖=
7(3 + 4𝑖) − 𝑖(3 + 4𝑖)
3(3 + 4𝑖) − 4𝑖(3 + 4𝑖)
=21 + 28𝑖 − 3𝑖 + 4
9 + 12𝑖 − 12𝑖 + 16
=25 + 25𝑖
25
= 1 + 𝑖 ⇒ 𝑥 = 1, 𝑦 = 1
𝑡𝑎𝑛 𝜃 =𝑦
𝑥= 01, 𝜃 = 𝜋
4⁄ , 𝑟 = √𝑥2 + 𝑦2 = √2 =
∴ 𝑟(cos 𝜃 + 𝑖 𝑠𝑖𝑛 𝜃)
√2 (𝑐𝑜𝑠 𝜋4⁄ + 𝑖 𝑠𝑖𝑛 𝜋
4⁄ )
Ans. D
¬p⇒ 𝑟 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒
¬𝑝 is T and F is false
𝑃 𝑖𝑠 𝐹
𝑝 ⇔ 𝑞 𝑖𝑠 𝑇𝑟𝑢𝑒
∴ 𝑞 𝑖𝑠 𝐹𝑎𝑙𝑠𝑒
Truth value of p, q and r respectively
F,F,F
Ans.C ¬ 𝑝 ∧ (𝑞 ⇒ 𝑟)
T ∧ (𝐹 ⇒ 𝐹)
T ∧ 𝑇 = 𝑇
28. Which one of the following is the set of critical numbers of 𝑓(𝑥) =3
8 𝑥8/3- 6𝑥2/3?
(A) {−2, 2}
(B) {−1, 1}
(C) {−2, 0, 2}
(D) {−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 𝑎1 =2, 𝑎2=6, 𝑎3=10, 𝑎4= 14,…. then 𝑛=1
100𝑎𝑛 =
(A) 20,020
(B) 20,000
(C) 20,200
(D) 22,200
𝑓(𝑥) =𝑥
8 𝑥
83⁄ − 6𝑥
23⁄
𝑓′(𝑥)3
8,8
3𝑥
53⁄ − 6.
2
3 𝑥
−13⁄
=𝑥5/3 −4
𝑥1
3⁄
𝑓′(𝑥) =𝑥
53
+13⁄ − 4
𝑥1
3⁄=
𝑥2 − 4
𝑥1
3⁄
𝑓′(𝑥) = 0, 𝑥2 − 4 = 0, 𝑥 = ± 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=6⇒ ℎ = 3
Distance b/n (1, 6) to (3,6) is r
𝑟 = √(1 − 3)2 + (6 − 6)2 = 42
Ans. C
31. Let 𝑓(𝑥) = 2𝑒𝑥-𝑘 sin 𝑥+1. If the equation of the tangent line to the graph of 𝑓 at (0, 3) is 𝑦=
5x+3, then what is the value of 𝑘?
(A) -3
(B) 3
(C) -5
(D) 2
32. Among 2000 students who took a regional exam, the percentile of certain student’s score is 90.
Which of the following is correct about the student’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 student’s score is the same as the top 10% of the score.
(D) The student’s score is greater than or equal to that of 1800 students.
𝑎1=2 𝑎2 = 6, 𝑎3=10 ∴ 𝑑 = 𝑎3− 𝑎2= 𝑎2− 𝑎1= 4
Sn=∑ 𝐴𝑛𝑛𝑘=0 =
𝑛
2(2𝐴1 + (𝑛 − 1)𝑑)
∑ 𝐴𝑛= 100
2 (2.2 + (100 − 1)4)
100
𝑛=1
= 50(4 + 99.4)
= 50(4 + 100) = 20,000
Ans. B
𝑓(𝑥) = 2𝑒𝑥 − k sin x + 1
Tangent at (0,3) the line is y=5x+3
𝑓′(𝑥) = 2𝑒𝑥 − 𝑘 cos 𝑥
𝑓′(0) = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒
⇒ 5 = 2 − 𝑘 ⇒ 5 − 2 = −𝑘 ⇒ 𝑘 = −3
Ans. D
Ans. D
33. What is the value of lim𝑛→∞
(𝑥3−8
𝑥2−4)?
(A) 3
(B) 2
(C) -2
(D) The limit does not exist.
34. What is the greatest lower bound of sequence {(−1)𝑛 1
𝑛+1}
𝑛=0
∞?
(A) -1
(B) 0
(C) -1/2
(D) ½
35. What is the value of lim𝑛→∞
(3𝑛+ 2′′
6′′ )?
(A) 1
(B) ∞
(C) 5
6
(D) 0
𝒍𝒊𝒎𝒙→𝟐 𝒙𝟑 − 𝟖
𝒙𝟐−𝟒= 𝒍𝒊𝒎𝒙→𝟐
(𝒙 − 𝟐)(𝒙𝟐 + 𝟐𝒙 + 𝟒
(𝒙 − 𝟐)(𝒙 + 𝟐)
=4+4+4
4=
12
4= 3
Ans. A
Ans. B
lim𝑛→∞
(3𝑛 + 2𝑛
6𝑛)
= lim𝑛→∞
3𝑛
6𝑛+ lim
𝑛→∞
2𝑛
6𝑛
= lim𝑛→∞
(1
2)
𝑛+ lim
𝑛→∞ (
1
3)
𝑛
= lim 𝑛→∞
1
2𝑛+ lim
𝑛→∞
1
3𝑛
= 0+0
= 0
Ans. D
36. What is the value of limℎ→0
8 (1
2+ ℎ) − 8 (
1
2)
2
?
ℎ
(A) 8
(B) 4
(C) 0
(D) The limit does not exist
37. Which one of the following is the multiplicative inverse of 𝑧 = 3+4𝑖
4−5𝑖
(A) 8
25−
31
25𝑖
(B) 8
25−
25
31𝑖
(C) 8
25+
25
31𝑖
(D) 8
25+
31
25𝑖
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 ℎ→0
8h2 + 8h + 2 − 2
h
lim 8ℎ→0
ℎ + 8 = 8
Ans. A
𝑧 =3 + 4𝑖
4 − 5𝑖
∴ 1
𝑧=
13 + 4𝑖4 − 5𝑖
= 4 − 5𝑖
3 + 4𝑖 ,
3 − 4𝑖
3 − 4𝑖=
−8 − 3 + 𝑖
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)=∫ 𝑒−1𝑥
0 dt , then what is the value of F’ (x) ?
(A) −𝑒−1-1
(B) 𝑒−𝑥
(C) 𝑒−𝑥+1
−𝑥+1
(D) −𝑒−𝑥
ℓ1 ∶ 𝑦 = −13⁄ 𝑥 + 8
3⁄
ℓ2 ∶ 𝑦 = −13⁄ 𝑥 − 8
3⁄
Having the same slope
Ans. A
det (A) =5
det(2 𝐴 𝑇𝐴)
Since T+ is 3x3 max
det (2 𝐴 𝑇𝐴) = 23𝑥5𝑥5 = 200
Ans.D
𝐹(𝑥) = ∫ 𝑒−𝑡
𝑥
0
𝑑𝑡
𝐹′(𝑥) = −[𝑒−𝑡]𝑥
= 𝐹(𝑥) − 𝐹(0)
= −𝑒−𝑥 + 1, 𝑡ℎ𝑒𝑛 𝐹′(𝑥) = (−𝑒−𝑥 + 1)1 = 𝑒−𝑥
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 ∫ ln 𝑥
2 1
𝑥2 dx ?
(A) - 𝑙𝑛 2
2 +
1
2
(𝐵) 𝑙𝑛 2
2 -
1
2
A. - 𝑙𝑛 2
2 -
1
2
(𝐷) 𝑙𝑛 2
2 +
1
2
𝑇ℎ𝑒𝑠𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 2 𝑐ℎ𝑖𝑙𝑑𝑟𝑒𝑛, 𝑠𝑡𝑎𝑟𝑡 <𝐺<𝐺
𝐵𝑏<𝐺
𝐵 = BB, BG,GB,GG
𝑝(𝐺)=3
4
Ans. D
𝑆30 = 3225 𝑑 = 5 , 𝑛 = 30 , 𝐴1 =?
𝑆𝑛 =𝑛
2(2𝐴1 + (𝑛 − 1)𝑑) ⇒ 𝑆30 =
30
2(2𝐴1 + (30 − 1)5)
3225 = 15(𝐴1 + 29𝑥5)
215 = 2𝐴1 + 145
70 = 2𝐴1 ⇒ 𝐴1 = 35 𝑐𝑒𝑛𝑡𝑠
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) 𝑥2 + 𝑦2 + 5𝑥 + 4𝑦 + 4 = 0
(B) 𝑥2 + 𝑦2 − 5𝑥 + 4𝑦 + 4 = 0
(C) 𝑥2 + 𝑦2 + 5𝑥 − 4𝑦 + 4 = 0
(D) 𝑥2 + 𝑦2 − 5𝑥 − 4𝑦 + 4 = 0
45. Let Ȥ be a complex number and 𝑤 = 3 + 4𝑖 . if 𝑧2+1
𝑧+𝑖 = |𝑤|Ȥ −
1
𝑖 �̅� , then what is the value of Ȥ ?
(A) -4- 2i
(B) −4 − 2𝑖
(C) −1 + 𝑖
(D) −2𝑖
∫ℓ𝑛 𝑥
𝑥2
2
1
𝑑𝑥
Let u=ℓ𝑛 𝑥 𝑑𝑣 =1
𝑥2 𝑑𝑥
𝑑𝑛 =1
2 𝑑𝑥 𝑉 =
−1
𝑥
∫ℓ𝑛 𝑥
𝑥2
2
1
= [−ℓ𝑛 𝑥
𝑥−
1
2]
1
2
=−ℓ𝑛 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)
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2
(−1 − ℎ)2 + (−2)2 = (ℎ − 0)2 + (2 − 2)2
ℎ2 − 2ℎ + 1 + 4 = ℎ2
−2ℎ = −5
ℎ = 52⁄
∴ (𝑥 − 52⁄ )2 + (𝑦 − 2)2 = ℎ2 =
25
4
∴ 𝑥2 + 𝑦2 + 5𝑥 − 4𝑦 + 4 = 0
Ans. C
46. What is the value of the left-hand side limit, lim𝑥→0−
sin 𝑥
𝑥2 √4𝑥2 + 9𝑥2 ?
(A) Does not exist
(B) 3
(C) 2
(D) -3
47. If −1 1 23 2 𝑥2 4 1
= −𝑥 3 22 2 31 −1 −1
, then what is the value of x
(𝐴) −2
5
(𝐵) 2
5
(𝐶) 2
3
(𝐷) −2
3
lim𝑥→0̅
sin 𝑥
𝑥2 √4𝑥3 + 9𝑥2
⇒ lim𝑥→0̅
sin 𝑥
𝑥 √
4𝑥3 + 9𝑥2
𝑥2
⇒ lim𝑥→0̅
sin 𝑥
𝑥 , lim
𝑥→0̅√4𝑥 + 9
1,-3
=-3
Ans. D
𝑧2 + 1 = (𝑧 + 𝑖)(𝑧 − 𝑖)
[𝑤] = √32 + 42 = 5
𝑤 ̅̅ ̅ = 3 − 4𝑖
∴𝑧2 + 1
𝑧 + 𝑖= |𝑤|𝑧 −
1
𝑖 �̅�
(𝑧 + 𝑖)(𝑧 − 𝑖)
2 + 𝑖= 5𝑧 + 𝑖(3 − 4𝑖)
𝑧 − 𝑖 = 5𝑧 + 3𝑖 + 4
𝑧 − 5𝑧 = 3𝑖 + 𝑖 + 4
−4𝑧 = 4𝑖 + 4
𝑧 =4𝑖 + 4
−4
𝑧 = −𝑖 − 1
Ans. B
48. Consider the following system of equations: {
𝑥 − 2𝑦 + 𝑧 = 1−𝑥 + 𝑦 + 𝑧 = 33𝑥 − 5𝑦 + 𝑧 = 𝑘
How much must be the value of k so that the system has a solution?
(A) 7
(B) 1
(C) -1
(D) 0
+ - + + - +
𝐴 = |−1 1 23 2 𝑥2 4 1
| = 𝐵 |−𝑥 3 22 2 31 −1 −2
|
⇒ −1 |2 𝑥4 1
| − |3 𝑥2 1
| + 2 |3 32 4
| = −1(2 − 4𝑥) − 1(3 − 2𝑥) + 2(12 − 4)
⇒ −2 + 4𝑥 − 3 + 2𝑥 + 16
⇒ 6𝑥 + 11
⇒ −𝑥 | 2 3
−1 − 2| − 3 |
2 31 − 2
| + 2 |2 2
1 − 1| = −𝑥(−4 + 3) − 3(−4 − 3) + 2(−2 − 2)
⇒ −𝑥 (−1) − 3(−7) + 2(−4)
⇒ 𝑥 + 21 − 8
⇒ 𝑥 − 13
∴ 6𝑥 + 11 = 𝑥 + 13
6𝑥 − 𝑥 = 13 − 11
5𝑥 = 2
𝑥 = 25⁄
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 𝑎−1 𝑏−1
𝑎−3− 𝑏−3 ?
(𝐴) 𝑎2𝑏2
𝑏2 − 𝑎2
(𝐵) 𝑎3 − 𝑏3
𝑎 − 𝑏
(𝐶) 𝑎3 − 𝑏3
𝑎𝑏
(𝐷) 𝑎2 𝑏2
𝑏3 − 𝑎3
50. If 𝑓(𝑥) = 𝑎𝑥3 +𝑏
𝑥 +5 has local minimum at (2, -3) , what are the values of a and b?
(A) 𝑎 = +1
4 , b=12
(B) 𝑎 =1
4 , b=12
(C) 𝑎 = −1
4 , b=-12
(D) 𝑎 =1
4 , b=-12
𝑓(𝑥) = 𝑎𝑥3 +𝑏
𝑥+ 5 ℎ𝑎𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑡 (2, −3)
𝑓(2) = −3
16𝑎 + 𝑏 = −16 … … (𝑥)
𝑆𝑖𝑛𝑐𝑒 − 2 𝑖𝑠 𝑎 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
𝑓′(𝑥) = 0, 𝑥 = 2
𝑓′(𝑥) = 3𝑎𝑥2 −𝑏
𝑥2
𝑓′(2) = 0
48𝑎 − 𝑏 = 0 − − − (𝑥𝑥)
𝑓𝑟𝑜𝑚 (∗)𝑎𝑛𝑑 (∗∗)𝑤𝑒 𝑔𝑒𝑡
𝑎 =−1
4 𝑎𝑛𝑑 𝑏 = −12
Ans. C
𝑎−1 𝑏−1
𝑎−3 − 𝑏−3=
1𝑎𝑏
1𝑎3 −
1𝑏3
=𝑎2𝑏2
𝑏3 − 𝑎3
Ans. D
51. Which one of the following is equal to (𝑥) = √(𝑥 + 4)2 , for every 𝑥 𝜖 ?
(A) 𝑔(𝑥) = |𝑥 + 4|
(𝑏) 𝑔(𝑥) = 𝑥 + 2
(𝐶) 𝑔 (𝑥) = 𝑥 + 4
(𝐷) 𝑔(𝑥) = |𝑥| +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?
[𝑦𝑜𝑢 𝑚𝑎𝑦 𝑡𝑎𝑘𝑒: cos(400) = 0.77, cos(200) = 0.94, sin(200) = 0.34 ]
(A) √43 km
(B) √40 km
(C) √50 km
(D) √53 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 − √2
2, −2 +
√2
2)
(B) (1 − √ 2
5
2, −2 +
√27
2)
𝑓(𝑥)√(𝑥 + 4)2, 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑖𝑛𝑑𝑒𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛
𝑓(𝑥) = |𝑥 + 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 𝜃 = 200 using cosine low the distance b/n them is 𝑑2 = 102 + 152 − 2 ∗ 10 ∗ 15 cos 20
𝑑 = √43 𝑘𝑚
Ans. A
(C) (√ 2
5
2,
√27
2)
(D) (√2
2,
√2
2)
54. If u= 2 𝑗→ − 𝑘→ and v=𝑖→ − 8𝑗→ + 3𝑘→, then what is the unit vector in the
direction of 5u+v?
(A) 𝑖→ + 2𝑗→ − 2𝑘→
(B) 2
3𝑖→ +
1
3𝑗→ −
2
3𝑘→
(C) 1
3𝑖→ +
2
3𝑗→ +
2
3𝑘→
(D) 1
3𝑖→ +
2
3𝑗→ −
2
3𝑘→
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 (𝑥0, 𝑦0) = (1, −2), 𝜃 = 450 𝑎𝑛𝑑 (𝑥, 𝑦) = (2,4)
𝑥1 = 𝑥0 + (𝑥 − 𝑥0)𝑐𝑜𝑠𝜃 − (𝑦 − 𝑦0) 𝑠𝑖𝑛 𝜃
= 1 + (2, −1) 𝑐𝑜𝑠 45— (4 − (−2)) 𝑠𝑖𝑛 450
= 1 +√2
2−
6√2
2= 1 −
5√2
2
𝑦′ = 𝑦00 + (𝑥 − 𝑥0) 𝑠𝑖𝑛 𝜃 + (𝑦 − 𝑦0) 𝑐𝑜𝑠𝜃
= −2 +√2
2+
6√2
2= −2 +
7√2
2
Ans. B
𝑢 = 2𝑗 − 𝑘 𝑎𝑛𝑑 𝑣 = 𝑖 − 8𝑗 + 3𝑘
5𝑢 = 10𝑗 − 5𝑘
5𝑢 + 𝑣 = 𝑖 + 2𝑗 − 2𝑘
|5𝑢 + 𝑣| = √12 + 22 + (22) = √9 =3
The unit rector in the direction of 5u+v is
=5u+v
|5u+v|
1𝑖
3+
2
3𝑗 −
2
3𝑘
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) 50√3
2 m
(B) 50√3 m
(C) 50√3
3 m
(D) 50+√3 m
25 m
30
25 B=60
𝜃 = 60
x y
tan 𝜃 = 25
𝑥 ⇒ 𝑥 =
25
tan 60=
25
√3=
25√3
3
tan 𝜃 = 𝑥 + 𝑦
25 ⇒ 𝑥 + 𝑦 = 25 tan 𝐵 = 25√3
Now we need to calculate y to determine the distance b/n the two ships
𝑦 = 25√3 − 𝑥 = 25√3 −25√3
3
𝑦 = 50√3
3m
Ans. C
56. Which of the following is a correct assertion that can be proved by the principle of
mathematical induction?
(A) 1
𝑛+1 ≤ 1, for each real number 𝑛 ≥ 1.
(B) 𝑘! ≥ 2𝑘, for each integer 𝑘 ≥ 4.
(C) 2𝑝 − 1, is prime for each prime integer 𝑝.
(D) 𝑚! ≤ 4𝑚, for each positive integer 𝑚.
57. If the between the vectors 𝐴→=(2,-1,1) and 𝐵→=(1,1, 𝑎) is 𝜋
3, then what is the value of 𝑎 ?
(A) 2
(B) -1
(C) 1
(D) -2
Ans. B
𝐴(2, −1,1) 𝑎𝑛𝑑 �⃗⃗� = (1,1, 𝜃), 𝜃 =𝜋
3
𝐴. 𝐵 = |𝐴| |𝐵| cos 𝜃
(2)(1) + (−1)(1) + (1)(𝑎) = √22 + 12 + 12, √12 + 12 + 𝑎2 cos 600
1 + 𝑎 = √6 , √2 + 𝑎2.1
2
2 + 2𝑎 = √12 + 6𝑎2
(2 + 2𝑎)2 = 12 + 6𝑎2
4𝑎2 + 8𝑎 + 4 = 12 + 6𝑎2
−2𝑎2 + 8𝑎 − 8 = 0
𝑎2 − 4𝑎 + 4 = 0
(𝑎 − 2)2 = 0
𝑎 − 2 = 0
𝑎 = 2
Ans. A
58. Let L be the line by the vector equation (𝑥, 𝑦) = (1,1) + 𝑡(√3.1 ),𝑡𝜖.. What is the equation of
the image of L after being rotated 150 about (1,1) and then translated by vector u= (−1,1)
(A) 𝑥 − 𝑦 = 2
(B) – 𝑥 + 𝑦 = 2
(C) √3 𝑥 − 𝑦 = 2
(D) –+√3 y=1
59. What is the solution set of 𝑠𝑖𝑛2 𝑥 − 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 = 0 over [0,2𝜋]?
(A) {0, 𝜋,5𝜋
4, 2𝜋}
(B) {0,𝜋
4, 𝜋, ,2𝜋}
(C) {0,𝜋
4, 𝜋,
5𝜋
4}
(D) {0,𝜋
4, 𝜋,
5𝜋
4, 2𝜋}
L:(𝑥, 𝑦) = (1,1) + 𝑡(√3 ,1) , 𝑡𝐸𝑅
𝑥 = 1 + √3 𝑡 , 𝑦 = 1 + 𝑡
⇒𝑥 − 1
√3= 𝑦 − 1
𝑥 − 1 = √3 𝑦 − √3
𝑦 =1
√3 𝑥 +
√3 − 1
√3
Here, slope m=1
√3 , tan 𝜃 =
1
√3
𝜃 = 300
(1,1) 𝑇𝑟(−1,1), (0,2)
Slope of image of ℓ = 450 , tan 450 = 1
∴ 𝑇𝑎𝑘𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 (0,2)𝑎𝑛𝑑 𝑠𝑙𝑜𝑝𝑒 𝑚 = 1
𝑦 − 2 = 1(𝑥 − 0)
𝑦 − 2 = 𝑥
𝑦 = 𝑥 + 2 ⇒ 𝑦 − 𝑥 = 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. ∀𝑛 ∈N, 3𝑛 − 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
𝐴𝐵𝐶𝐷 𝑖𝑠 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
𝐴𝐵 = √32 + 42 = 5
𝐵𝐶 =1
2 𝐴𝐵 =
5
2
(𝑥 − 4)2 + 𝑦2 =25
4− 𝑖𝑠 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 𝑤𝑖𝑡ℎ 𝑐𝑒𝑛𝑡𝑒𝑠 𝑐(4,0)𝑎𝑛𝑑 𝑟 =
5
2
∴ 𝑐 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝐶𝑖𝑟𝑐𝑙𝑒
𝐶(6, 32⁄ )
Ans. D
𝑠𝑖𝑛2𝑥 − sin 𝑥 cos 𝑥 = 0
sin 𝑥 (sin 𝑥 − cos 𝑥) = 0
sin 𝑥 = 0 or sin 𝑥 − cos 𝑥 = 0
sin 𝑥 = 0, 𝑥 = 0 , 𝜋, 2𝜋
sin 𝑥 − cos 𝑥 = 0 ⇒ sin 𝑥 = cos 𝑥, 𝑥 =𝜋
4 ,
5𝜋
4
𝑠. 𝑠 = {0,𝜋
4 , 𝜋,
5𝜋
4, 2𝜋}
Ans. D
62. If the point (𝑎, 0, 3) is on the sphere centered at (1, 2, 3) with radius 2, what is the value of 𝑎?
(A) 0
(B) 2
(C) 1
(D) -3
(E)
63. What is the image of the circle 𝑥2+𝑦2 + 4𝑥 − 6𝑦 + 11 = 0, when the origin is shifted to the
point (1,1) after translation of axes ?
(A) 𝑥2+𝑦2 − 4𝑥 − 2𝑦 + 3 = 0
(B) 𝑥2+𝑦2 − 4𝑥 − 6𝑦 + 3 = 0
(C) 𝑥2+𝑦2 − 6𝑥 + 8𝑦 − 23 = 0
(D) 𝑥2+𝑦2 − 6𝑥 − 8𝑦 + 23 = 0
Let 𝑥0 , 𝑦0 𝑎𝑛𝑑 𝑧0 be the center of the sphere
∴ (𝑥 − 𝑥0)2 + (𝑦 − 𝑦0)2 + (𝑧 − 𝑧0)2 = 𝑟2, 𝑤ℎ𝑒𝑟𝑒 𝑥, 𝑦 𝑎𝑛𝑑 𝑧 𝑎𝑟𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒.
(𝑎 − 1)2 + (0 − 2)2 + (3 − 3)2 = 22
(𝑎 − 1)2 + 4 + 0 = 4
(𝑎 − 1)2 = 0 ⇒ 𝑎 = 1
Ans.C
∀𝑛 ∈ 𝑁, 3𝑛 − 2 𝑖𝑠 𝑝𝑟𝑖𝑚𝑒
𝐿𝑒𝑡 𝑛 = 1
3𝑛 − 2 = 31 − 2 = 1 − 𝑁𝑒𝑖𝑡ℎ𝑖𝑛𝑔 𝑝𝑟𝑖𝑚𝑒 𝑛𝑜𝑟 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒
Disprove by a counter example.
Ans. A
Give a circle with equation
𝑥2 + 𝑦2 − 4𝑥 − 6𝑦 + 11 = 0, 𝑁𝑜𝑤 𝑖𝑓 𝑤𝑒 𝑠ℎ𝑖𝑓𝑡
(1.1)units the image becomes
(𝑥 − 1)2 + (𝑦 − 1)2 − 4(𝑥 − 1) − 6(𝑦 − 1) + 11 = 0
𝑥2 − 2𝑥 + 1 + 𝑦2 − 2𝑦 + 1 − 4𝑥 + 4 − 6𝑦 + 6 + 11 = 0
𝑥2 + 𝑦2 − 2𝑥 − 4𝑥 − 2𝑦 − 6𝑦 + 1 + 1 + 4 + 6 + 11 = 0
𝑥2 + 𝑦2 − 6𝑥 − 8𝑦 + 23 = 0
Ans. D
64. What is the value of cot 2700+2cos 900+4𝑠𝑒𝑐2 + 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) 3𝑛 + 25 < 3𝑛, for every integer 𝑛 ≥ 3.
(B) 2𝑛 > 𝑛 + 20, for every integer 𝑛 ≥ 4.
(C) 𝑛3 − 𝑛 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 6, for every integer 𝑛 ≥ 1.
(D) 𝑛2 ≤ 2𝑛, for every integer 𝑛 ≥ 1.
THE END
𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 cot 2700 + 2 𝑐𝑜𝑠 900 + 4 𝑠𝑒𝑐2 1800
cot 2700 =1
tan 270=
𝑐𝑜𝑠 2700
𝑠𝑖𝑛2700=
0
−1= 0
2 cos 900 = 𝑥 ∗ 0 = 0
4 𝑠𝑒𝑐2 1800 = 4(−1)2 = 4
∴ 0 + 0 + 4 = 4
Ans. A
Ans. C , sine 𝑛3 − 𝑛 𝑓𝑜𝑟 𝑛 ≥ 1 𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 6
