Socie 1 / 4 There is only one correct answer
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2019 1st SOCIE MATH Problems and Solution-1
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SOCIE - 1 / 4 per each question. Mark your answer choice on the OMR answer sheet. ○ For each correct answer, you will get the points indicated next to each question number. ○ No penalty point is applied to an incorrect answer. 1. [3 points] When
for
find . ① ② ③ ④ ⑤ (sol) Since
and
, we have
.
Hence, . The answer is ③ 2. [3 points] When the minimum value of
for is , find . ① ② ③ ④ ⑤ (sol) Since
, the minimum value is . That is, . Hence, . The answer is ① 3. [3 points] When and
, find . ① ② ③ ④ ⑤ (sol) Since ln ln
and ln
ln , we have ln
ln log
ln ln
. Hence, . The answer is ② 4. [3 points] Compute
, where .
① ② ③ ④ ⑤ (sol) Since
,
for any positive integer . In particular,
.
The answer is ③ 5. [3 points] When are the solutions of find
.
① ② ③ ④ ⑤ (sol) Since , we have
. The answer is ④ 6. [3 points] When and
, find
① ② ③ ④ ⑤
(sol) From , we know and
2019 IUT Admission Test(SOCIE) Math Examination(A TYPE) SOCIE - 2 / 4
, which leads to ± . Since
. The answer is ⑤
7. [3 points] Let be the remainder when is
divided by . Find .
① ② ③ ④ ⑤
(sol) If we denote the quotient by , we can write
. Hence, . Taking derivative, we have
′ . Hence, ′
and and ′
we have
. Hence, . The answer is ① 8. [3 points] When sin cos for , find sin cos . ①
② ③ ④ ⑤ (sol) Since sin cos
sin cos sin cos . Hence, sin cos , which shows that sin cos sin cos . Hence, sin cos ± . Since sin cos for , sin cos . Hence, sin cos . The answer is ⑤ 9. [3 points] For
, we write
. Find . ① ② ③ ④ ⑤ (sol) For
,
.
Simple computation shows that
and
. In particular,
and . The answer is ① 10. [3 points] Compute log log log
log
.
① ② ③ ④ ⑤ (sol) We note that log
ln ln
, log ln ln
, log
ln ln
, log ln ln
. log log
log
log
ln
ln ×
ln ln
× . The answer is ⑤ 11. [4 points] Compute
, where .
① ② ③ ④ ⑤ (sol)
×
. The answer is ③ SOCIE - 3 / 4 12. [4 points] Compute sin . ①
②
③
④
⑤
(sol) sin
sin sin cos sin cos
. The answer is ④ 13. [4 points] Find the sum of all solutions of log
for . ①
② ③ ④ ⑤ (sol) Taking logarithm on log , we have ln
ln ln ln ln , which leads to ln ln ln ln .
Hence, we have ln ln ln ln , which shows that ln ln ln or ln ln ln . Hence, or
. The answer is ⑤ 14. [4 points] When
lim →
for some constants and , find
. ① ② ③ ④ ⑤ (sol) From lim
→ , we have . Since , lim
→ lim
→
. Since , we have and . Hence . The answer is ①
15. [4 points] Compute lim →
sin cos . ① ② ③ ④ ⑤ (sol) Note that sin
cos
sin
cos sin
and
lim →
sin . Hence, lim
→ sin cos
. The answer is ④ 16. [4 points] Compute
ln . ① ② ③ ④ ⑤ (sol) Setting
ln , we have
. Hence,
ln
. The answer is ② 17. [4 points] Compute
. ①
② ③ ④
⑤ (sol) Setting , we have . Hence, SOCIE - 4 / 4
The answer is ② 18. [4 points] Find the area of the region enclosed by two curves
. ①
② ③ ④ ⑤ (sol) From
, we obtain .
Hence, two curves meet at and . Since
≤
for ≤ ≤ , the area is the following.
× . The answer is ③ 19. [4 points] Let be a differentiable function and be the inverse function of . When
lim →
, find ′ . ① ② ③ ④ ⑤ (sol) From lim
→
, we have . Note that
.
.
lim
→
′ ⋅ ′
Hence, ′ . Since , it follows that ′ ′ and ′ ′
. If we put , ′ ′ ′
. The answer is ④
20. [4 points] When is the inverse function of , find
. ①
② ③ ④ ⑤ (sol) We put . Then, , and since is the inverse function of . Hence,
′ ′
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