6 ĐỀ THI VÀ ĐÁP ÁN TIN HỌC TRONG KỸ THUẬT.
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...................................................
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2 /2
TRU'dNG BH SU' PHAM KY THUAT TPHCM
DE THI CUOI KY HOC KY I NAM HOC 2019-2020
KHOA CO KHI CHE TAO MAY
Mon: TIN HOC TRONG KY THUAT..................
BO MON CODIENTU'
M3 mnn hnc MF.TF1,14529.........................................
Chu’ky giam thi 1
Chu ky giam thi 2
Be so/Ma dl: 01........ Be thi co 08.......... trang.
Thai gian: 70.. phut.
CB cham thi thu nhat
CB cham thi thu hai
So cau dung:
So cau dung:
Diem va chir ky
Dilm va ch& ky
Dirac phep su dung tai lieu (KHONG su dung laptop).
Ho va ten:........................................... .........................
M as6SV :.....................................................................
So TT:.......................Phong thi:................................
PHIEU TRA LOfI
Hiro'ng dSn tra 161 cau hoi:
Chon cau tra lai dung:
No.
a
b
Bo chon:
c
d
Chon lai:
No.
1
21
2
22
3
23
4
24
5
25
6
26
7
27
8
28
9
29
10
30
11
31
12
32
13
33
14
34
15
35
16
36
17
37
18
38
19
39
20
40
S6 hieu: BM2/QT-PBBCL-RBTV
a
b
c
d
1/2
4
PHANI-M ATLAB (15 cau)
Cau 1: (0.25 diim)
Trong cac cau lenh Matlab sau, cau lenh nao la DUNG:
a. x_5 = 5
b. _x5 = 5
c. 5_x = 5
d. Tat ca deu dung..
Cau 2: (0.25 diem)
Cho biet ket qua cua phep toan sau.
» 0 /0
a. 0.
b. Inf.
c. NaN.
d. Tat ca deu sai.
Cau 3: (0.25 diem)
Lenh close all trong Matlab co chuc nang:
a. Dong toan bo cua so lam viec
b. Thoat Matlab
c. Dong toan bo cua so do hoa (Figure)
d. Tdt ca deu sai.
Cau 4: (0.25 diem)
Trong cua so Command cua Matlab, chung ta thuc hien lenh sau:
»
2*3A2
Ket qua la:
a. 16
b. 18
c. Cau lenh bao loi
d. Tat ca deu sai.
Cau 5: (0.25 diem)
Trong cua s6 Command cua Matlab, chung ta thuc hien lenh sau:
»
abs(3+4j)
Ket qua la:
a. 5
b. 7
c. Cau lenh bao 16i
d. Tdt ca deu sai.
Cau 6: (0.25 diim)
Trong Matlab, d l tlnh phep toan y = ln(x), ta su dung cau lenh sau:
a. y = ln(x)
b. y = log(x)
c. y = loge(x)
d. Tat ca deu sai.
Cau 7: (0.25 diem)
Trong cua so command cua Matlab, chung ta thuc hien lenh sau:
»
a=[l 2 3];
»
b= [l;l;l];
»
aAb
Ket qua in ra la:
a. [1 2 3]
b. 6
c. Cau lenh bao loi
d. Tat ca deu sai.
S6 hieu: BM2/QT-PDBCL-RDTV
2/8
Cau 8: (0.25 dilm )
Trong cua so command cua Matlab, chung ta thuc hien lenh sau:
»
a=[l 2 3];
»
b = a ';
»
a.*b
Ket qua in ra la:
a. 10
b. 14
c. Cau lenh bao loi
d. Tit ca dlu sai.
Cau 9: (0.25 diem)
Trong cua s6 Command cua Matlab, cho biet gia tri cua x khi chung ta thuc hien lenh sau:
»
x = 0:2.5:8
a. 0 2.5 5.0 7.5
c. Cau lenh bao loi
b. 0 2 4 8
d. Tat ca deu sai.
Cau 10: (0.25 diem)
Trong cua s6 command cua Matlab, cho bi6t gia trj cua y khi chung ta thuc hien lenh sau:
»
x = 0:2:6
»
y = 2*sin(x)
a. Vecto 1 hang 3 cot
c. Vecto 1 hang 1 cot
b. Vecot 1 hang 4 cot
d. Cau lenh bao 16i.
Cau 11: (0.25 diim)
Cho 2 vector x = [1 2 3] va y = [1 0 0], cau lenh nao sau day la SAI:
a. pie(x)
b. pie(x,y)
c. pie(y,x)
d. Khong cau lenh nao SAI.
Cau 12: (0.25 dilm)
Cho vecto x co gia trj nhu sau:
» x = [1:5]
B6 ve d6 thi sin2(x), cau lenh nao sau day la BUNG:
a. plot(sin(x)A2)
b. plot( 'sin(x)A2 ')
c. fplot( 'sin(x)A2 ')
d. Tat ca d6u sai.
Cau 13: (0.25 diem)
Trong cac cau lenh Matlab sau, cau lenh nao la BUNG:
a. » y = sym(x)
b. » y = syms(x)
c. » y = sym( 'x ')
d. » y = syms( 'x ')
Cau 14: (0.25 diem)
Cho cau lenh sau:
» x= 0:10;
» plot(x,sin(x),'dk')
S6 hieu: BM2/QT-PBBCL-RBTV
3/2
Khi chay chuang trinh, ducmg d6 thj co dang:
a. Net lien, mau xanh
c. Net dut, mau xanh
b. Net li€n, mau den
d. Net dut, mau den
Cau 15: (0.25 di£m)
Cho cau lenh sau:
» x = [1:5]
» pIot(x,xA2)
Khi chay chuong trinh, ket qua se la:
a. Duong d6 thi co mau xanh
c. Cau lenh bao loi
b. Duong do thi c6 mau do
d. Tat ca deu sai.
P H A N II - C (25 cau)
Can 16: (0.25 diem)
Cho a,b la bien s6 nguyen (int) va a = 5, b = 2, c = 2. Hay cho biet gia tri cua bieu thuc: a/b/c
a. 0.
b. 1.
c. 2.
d. TSt ca d6u sai.
Cau 17: (0.25 diem)
Cho a,b la bien s6 nguyen (int) va a = 4, b = 3. Hay cho bi6t gia tri cua bilu thuc: (++a/c)%b:
a. 0.
c. 2.
b. 1.
d. Tat ca deu sai.
Cau 18: (0.25 diem)
Cho a,b la bien so nguyen (int) va a = 4, b = 3. Hay cho bi6t gia tri cua biSu thuc: (a++*++a)%b:
a. 0.
b. 1.
c. 2.
d. Tat ca deu sai.
Cau 19: (0.25 diem)
Cho a,b,c la cac bi6n s6 nguyen va a = 4, b = 2, c = 3. Hay cho biet gia tri cua bieu thuc:
(a>5)&&(b— 2)||(!(c==4)):
a. Dung/True.
b. Sai/False.
Cau 20: (0.25 diem)
Cho a,b la bi6n so nguyen a = 1, b = 2. Hay cho biet gia tri cua bieu thuc: (a<3)||( a<0)&&(b>a)
a. Dung/True.
b. Sai/False.
Cau 21: (0.25 dilm)
Cho doan chuong trinh nhu sau:
char S[20] = "KIEM TRA";
int n = strlen(S);
Gia tri cua bien n la:
a. 8
c. 10
S6 hieu: BM2/QT-PDBCL-RDTV
b. 99
d. T
dtcadeusai.
Tdt
ca deu sai.
4/8
Cau 22: (0.25 diem)
Trong cac cau lenh sau khai bao chuoi S sau, cau lenh nao la DUNG:
a. CharS[10];
b. char [10] S;
c. char S[10];
d. T it ca diu dung.
Cau 23: (0.25 diem)
Cho mang M dugc khai bao nhu sau:
int M[3][3];
Trong cac cau lenh sau, cau lenh nao la DUNG:
a. M = 5;
b. M[][] = 5;
c. M[0][0] = 5;
d. Khong dap an nao dung.
Cau 24: (0.25 diem)
Trong cac cau lenh sau, cau lenh nao la DUNG:
a. int A[3] = {1,2,3};
b. int A[3] = {1,2};
d. T it ca deu dung.
c. int A[3] = {1};
D o a n c h u ffn g tr in h s a u d u n g ch o c a u 2 5 ,2 6 :
char S I[10] = "12345", S2[10] = "abcde";
for (int i = 3; i>=0; i—)
{
if ( i%2 == 0) Sl[i] = S2[i+1];
else
S2[i] = Sl[i+1];
}
Cau 25: (0.25 diem)
Cho biet gia tri cua chuSi SI sau khi kit thuc doan chuong trinh tren:
a. 51545
b. 52545
c. 53545
d. Tit ca diu SAI.
Cau 26: (0.25 diem)
Cho biet gia tri cua chuoi S2 sau khi ket thuc doan chuong trinh tren:
a. a3c4e
b. a4c5e
c. a5c5e
d. T it ca diu SAI.
D o a n c h u a n g trin h s a u d u n g ch o c a u 2 7 ,2 8 :
int x = 0, y = 0;
int M[3][3] = {{1,2,3},{1,1,1},{3,2,1}};
for (int i = 0;i<3;i++)
{
for (int j = 0y<3J++)
{
if (i = j)
else
x = x + M[i][j];
y=y + M[i][j];
}
}
So hieu: BM2/QT-PDBCL-RDTV
5/2
Cau 27: (0.25 diem)
Cho biet gia tri cua bien x sau khi ket thuc doan chiro'ng trinh tren:
a. 3.
b. 4.
c. 5.
d. Tat c&deu SAI.
Cau 28: (0.25 dilm)
Cho biet gia tr| cua bien y sau khi kit thuc doan chucmg trinh tren:
a. 11.
b. 13.
c. 15.
d. Tat ca deu SAI.
D o a n ch u ctn g tr in h s a u d u n g ch o c a u 2 9 ,3 0 ,3 1 :
int M[2][2] = {{1,2},{3,4}};
int x = 1;
for (int i = 0;i<2;i++)
{
for (int j = 0;j<2J++)
{
if(M [i][j]= x )
x++;
else
M[i][j]=x;
}
printf("%d ", x);
}
Cau 29: (0.25 diem)
Cho biet gia tri cua phan tur M [l][l] sau khi ket thuc doan chuong trinh tren:
a. 3.
b. 4.
c. 5.
d. Tdt ca deu SAI.
Cau 30: (0.25 diem)
Cho biet gia tri cua bien x sau khi ket thuc doan chuong trinh tren:
a. 2.
c. 4.
b. 3.
d. Tit ca deu SAI.
Cau 31: (0.25 diem)
Cho biet ket qua in ra man hinh sau khi chay doan chuong trinh tren:
a. 2 4
c. 4 5
b. 3 5
d. Tit ca deu SAI.
D o a n ch u ctn g tr in h s a u d u n g ch o c a u 3 2 ,3 3 ,3 4 :
int M[5] = {-1,3,2,4,7};
for (int i = 0;i<4;i++)
{
if (M[i]>M[i+l])
M[i] = M[i+1];
else
printf("%d",M[i]);
}
So hieu: BM2/QT-PDBCL-RDTV
6/8
Cau 32: (0.25 diem)
Cho bilt gia tri cua phln tur M[2] sau khi ket thuc doan chuong trinh tren:
a. -1
c. 2
b. 3
d. Tat ca
deuSAI.
Cau 33: (0.25 diem)
Cho bilt gia tri cua phln tu M[4] sau khi kit thuc doan chuong trinh tren:
a. 5
b. 6
c. 7
d. Tat ca
deuSAI.
Cau 34: (0.25 dilm)
Cho biet ket qua in ra man hinh sau khi chay doan chuong trinh tren:
a. -124
b. -12
c. -13
d. T it ca
deuSAI.
D o a n c h u ffn g tr in h s a u d u n g cho c a u 3 5 ,3 6 ,3 7 :
inta = 0, b = 0;
for (int i = 0; i<3; i++)
{
for (int j = 2; j>=0; j —)
{
if (j>i)
else
a++;
b = b - a;
}
printf("%d", a+b);
}
Cau 35: (0.25 diem)
Cho biet gia tri cua biln a sau khi ket thuc doan chuong trinh tren:
a. 0.
b. 1.
c. 2.
d. Tit ca dlu SAI.
Cau 36: (0.25 diem)
Cho biet gia trj cua bien b sau khi ket thuc doan chuong trinh tren:
a. -10.
b. -17.
c. -20.
d. Tit ca d6u SAI.
Cau 37: (0.25 dilm)
Cho biet ket qua in ra man hinh sau khi chay doan chuong trinh tren:
a. 0-5-11
b. 0-5-13
c. 0-5-14
d. Tat ca deu SAI.
S6 hieu: BM2/QT-PDBCL-RDTV
7/2
D o a n chiro 'ng tr in h s a u d u n g ch o c a u 3 8 ,3 9 ,4 0 :
int x = 13, y = 6, n = 0;
for (int i = 0; i<10; i++)
{
if ( x%y == 0) break;
else
{
x++;
y -;
}
n++;
}
Cau 38: (0.25 diem)
Cho biet gia trj cua bien x sau khi ket thuc doan chuong trinh tren:
a. 15.
b. 18.
c. 20.
d. Tat ca deu SAI.
Cau 39: (0.25 diem)
Cho biet gia trj cua bien y sau khi ket thuc doan chuong trinh tren:
a. 1.
b. 2.
c. 3.
d. Tat ca deu SAI.
Cau 40: (0.25 diem)
Cho biet gia trj cua bien n sau khi ket thuc doan chuong trinh tren:
a. 3.
b. 4.
c. 5.
d. Tat ca deu SAI.
Ghi chu:Can bo coi thi khong duoc giai thick de thi.
Cliuan dau ra cua hoc phan (ve kien thuc)
Noi dung kiem tra
[CBR 1.1]: Giai thich duoc, mo ta duoc hoat dong cua mot
Cau 1,2,3,4,5,6,7,8,9,10,11,12
chuong trinh lap trinh may tinh.
31,32,33,34,35,36,37,38,39,40
[CBR 1.2]: HiSu va giai thich duoc cac luu do giai thuat.
Cau 13,14,15,16.
[CBR 2.2]: Trinh bay duoc cac bai toan dieu khien duoi
Cau 17,18,19,20,21.
dang thuat toan va giai thuat.
[CBR 4.1]: Bi6t su dung cac phuong phap lap trinh de xay
dung chuong trinh.
Cau 22,23,24,25,26,27
28,29,30.
Ngay 15 thang 12 nam 2019
^Truong bo mon
S6 hieu: BM2/QT-PDBCL-RDTV
8/8
TRU'CSNG BH SU' PHAM KY THUAT TPHCM
D£ THI CUOl KY HOC KY I NAM HOC 2019-2020
KHOA CO KHI CHE TAO MAY
b O m o n CODIEN TU”
Chu ky giam thi 2
Chu ky giam thi 1
Mon: TIN HOC TRONG KY THUAT..................
M3 mnn hnr,- ME,TF1 ^4^7.9..........................................
B6 so/Ma de: 02........ Be thi co 08.......... trang.
n/y
Then gian: 70.. phut.
Dirffc phep sir dung tai lieu (KHONG su dung laptop).
CB cham thi thu nhat CB chtim thi thu hai
S6 cau dung:
So cau dung:
Diem va ch& ky
Diem va chu ky
Ho va ten:.....................................................................
M asoSV :.....................................................................
S6 TT:...................... Phong thi:................................
PHIEU TRA LOT
Hiring dan tra 161 cau hoi:
Chon cau tra lcri dung: N
a
b
Bo chon: ]^T
c
d
No.
1
21
2
22
3
23
4
24
5
25
6
26
7
27
8
28
9
29
10
30
11
31
12
32
13
33
14
34
15
35
16
36
17
37
18
38
19
39
20
40
S6 hleu: BM2/QT-PBBCL-RBTV
Chon lai:
a
b
c
d
1/8
PHANI-M ATLAB (15 cau)
C3u 1: (0.25 diem)
Trong cac cau lenh Matlab sau, cau lenh nao la DUNG:
a. » y = sym(x)
b. » y = syms(x)
c. » y = sym( 'x')
d. » y = syms( 'x' )
Cau 2: (0.25 diem)
Cho 2 vector x = [1 2 3] va y = [1 0 0], cau lenh nao sau day la SAI:
a. pie(x)
b. pie(x,y)
c. pie(y,x)
d. Khong cau lenh nao SAI.
Cau 3: (0.25 diem)
Cho cau lenh sau:
» x= 0:10;
» plot(x,sin(x),'dk')
Khi chay chuong trinh, duong do thj co dang:
a. Net lien, mau xanh
c. Net dut, mau xanh
b. Net li6n, mau den
d. Net dut, mau den
Cau 4: (0.25 diem)
Trong cua so Command cua Matlab, cho biet gia tri cua x khi chung ta thuc hien lenh sau:
»
x = 0:2.5:8
a. 0 2.5 5.0 7.5
c. Cau lenh bao loi
b. 0 2 4 8
d. Tat ca deu sai.
Cau 5: (0.25 diem)
Cho cau lenh sau:
» x = [1:5]
» pIot(x,xA2)
Khi chay chuong trinh, ket qua se la:
a. Duong do thj co mau xanh
c. Cau lenh bao loi
b. Duong do thi co mau do
d. Tat ca deu sai.
Cau 6: (0.25 diem)
Trong cua so command cua Matlab, cho biet gia tri cua y khi chung ta thuc hien lenh sau:
»
x = 0:2:6
»
y = 2*sin(x)
a. Vecto 1 hang 3 cot
c. Yecto 1 hang 1 cot
S6 hieu: BM2/QT-PDBCL-RDTV
b. Vecot 1 hang 4 cot
d. Cau lenh bao loi.
2/8
Cau 7: (0.25 diem)
Cho vecto x co gia tri nhu sau:
» x = [1:5]
Be ve do thi sin2(x), cau lenh nao sau day la DUNG:
b. plot( 'sin(x)A2 ')
a. plot(sin(x)A2)
c. fplot( 'sin(x)A2 ')
d. T it ca dlu sai.
Cau 8: (0.25 diem)
Trong cua so command cua Matlab, chung ta thuc hien lenh sau:
»
a=[l 2 3];
»
b = a' ;
»
a.*b
K it qua in ra la:
a. 10
b. 14
c. Cau lenh bao loi
d. Tat ca dlu sai.
Cau 9: (0.25 dilm)
Trong cua so Command cua Matlab, chung ta thuc hien lenh sau:
»
2*3A2
K it qua la:
a. 16
b. 18
c. Cau lenh bao loi
d. T it ca dlu sai.
Cau 10: (0.25 diem)
Trong Matlab, de tinh phep toan y = ln(x), ta su dung cau lenh sau:
a. y = ln(x)
b. y = log(x)
c. y = loge(x)
d. T it ca dlu sai.
Cau 11: (0.25 diem)
Cho biet ket qua cua phep toan sau:
» 0/0
a. 0.
b. Inf.
c. NaN.
d. Tat ca deu sai.
Cau 12: (0.25 dilm)
Trong cua so command cua Matlab, chung ta thuc hien lenh sau:
»
a=[l 2 3];
» b=[l;1; 1];
» aAb
K it qua in ra la:
a. [1 2 3]
b. 6
c. Cau lenh bao loi
d. Tat ca deu sai.
S6 hieu: BM2/QT-PDBCL-RDTV
3/8
*
Cau 13: (0.25 diem)
Trong cac cau lenh Matlab sau, cau lenh nao la DUNG:
a. x 5 = 5
b.
x5 = 5
c. 5_x = 5
d. Tat ca deu dung..
Cau 14: (0.25 diem)
Lenh close all trong Matlab co chuc nang:
a. Dong toan bo cua so lam viec
b. Thoat Matlab
c. Dong toan bo cua so d6 hoa (Figure)
d. Tat ca deu sai.
Cau 15: (0.25 diem)
Trong cua so Command cua Matlab, chung ta thuc hien lenh sau:
»
abs(3+4j)
Ket qua la:
a. 5
b. 7
c. Cau lenh bao loi
d. Tat ca d6u sai.
PHAN II - C (25 cau)
Cau 16: (0.25 diem)
Cho mang M duoc khai bao nhu sau:
int M[3][3];
Trong cac cau lenh sau, cau lenh nao la DUNG:
a. M = 5;
b. M[][] = 5;
c. M[0][0] = 5;
d. Khong dap an nao dung.
Cau 17: (0.25 diem)
Trong cac cau lenh sau khai bao chuoi S sau, cau lenh nao la DUNG:
a. CharS[10];
b. char [10] S;
c. charS[10];
d. TSt ca deu dung.
Cau 18: (0.25 diem)
Trong cac cau lenh sau, cau lenh nao la DUNG:
a. int A[3] = {1,2,3};
b. int A[3] = {1,2};
c. int A[3] = {1};
d. Tat ca deu dung.
Cau 19: (0.25 diem)
Cho doan chuong trinh nhu sau:
char S[20] = "KIEM TRA";
int n = strlen(S);
Gia tri cua bien n la:
a. 8
b. 9
c. 10
d. Tilt ca deu sai.
S6 hieu: BM2/QT-PDBCL-RDTV
4/8
Cau 20: (0.25 diem)
Cho a,b la biin so nguyen a = 1, b = 2. Hay cho biit gia tri cua biiu thuc: (a<3)||( a<0)&&(b>a)
a. Dung/True.
b. Sai/False.
Cau 21: (0.25 diem)
Cho a,b,c la cac biin so nguyen va a = 4, b = 2, c = 3. Hay cho biet gia tri cua bieu thuc:
(a>5)&&(b=2)| |(! (c— 4)):
a. Dung/True.
b. Sai/False.
Cau 22: (0.25 diem)
Cho a,b la biin so nguyen (int) va a = 4, b = 3. Hay cho biet gia tri cua bieu thuc: (++a/c)%b:
a. 0.
b. 1.
c. 2.
d. T it ca deu sai.
Cau 23: (0.25 diem)
Cho a,b la bien so nguyen (int) va a = 5, b = 2, c = 2. Hay cho biit gia tri cua bieu thuc: a/b/c
a. 0.
b. 1.
c. 2.
d. T it ca deu sai.
Cau 24: (0.25 diim)
Cho a,b la bien s6 nguyen (int) va a = 4, b = 3. Hay cho biet gia tri cua bieu thuc: (a++*++a)%b:
a. 0.
b. 1.
c. 2.
d. Tat ca deu sai.
D o a n c h u ffn g tr ln h s a u d u n g ch o c d u 2 5 ,2 6 ,2 7 :
int M[2][2] = {{1,2},{3,4}};
int x = 1;
for (int i = 0;i<2;i++)
{
for (int j = 0y<2y++)
{
if (M[i][j] == x)
x++;
else
M [i][j]=x;
}
printf("%d ", x);
}
Cau 25: (0.25 diim)
Cho biet gia tri cua phan tu M [l][l] sau khi ket thuc doan chuong trlnh tren:
a. 3.
b. 4.
c. 5.
d. Tat ca deu SAI.
Cau 26: (0.25 diem)
Cho biet gia tri cua biin x sau khi ket thuc doan chuong trinh tren:
a. 2.
c. 4.
S6 hieu: BM2/QT-PDBCL-RDTV
b. 3.
d. Tit ca diu SAI.
5/8
Cau 27: (0.25 diem)
Cho bilt kit qua in ra man hinh sau khi chay doan chucmg trinh tren:
a. 2 4
c. 4 5
b. 3 5
d. Tat ca deu SAI.
D o a n chuofng tr in h s a u d u n g c h o c a u 2 8 ,2 9 :
char Sl[10] = "12345", S2[10] = "abcde";
for (int i = 3; i>=0; i—)
{
if ( i%2 = 0) S1[i] = S2[i+1];
else
S2[i] = Sl[i+1];
}
Cau 28: (0.25 diem)
Cho biet gia tr| cua chuoi SI sau khi ket thuc doan chucmg trinh tren:
a. 51545
b. 52545
c. 53545
d. Tat ca deu SAI.
Cau 29: (0.25 dilm)
Cho biet gia trj cua chuSi S2 sau khi k it thuc doan chucmg trinh tren:
a. a3c4e
b. a4c5e
c. a5c5e
d. Tat ca deu SAI.
D o a n c h u a n g tr in h s a u d u n g c h o c a u 3 0 ,3 1 ,3 2 :
int x = 13, y = 6, n = 0;
for (int i = 0; i<10; i++)
{
if ( x%y == 0) break;
else
{
x++;
y~ ;
}
n++;
}
Cau 30: (0.25 diem)
Cho bilt gia tri cua biln x sau khi ket thuc doan chuong trinh tren:
a. 15.
b. 18.
c. 20.
d. Tat ca deu SAI.
Cau 31: (0.25 diim)
Cho biet gia tri cua bien y sau khi ket thuc doan chuang trinh tren:
a. 1.
b. 2.
c. 3.
d. T it ca deu SAI.
S6 hieu: BM2/QT-PBBCL-RBTV
6/8
Cau 32: (0.25 diem)
Cho biet gia tri cua bien n sau khi ket thuc doan chuang trinh tren:
a. 3.
b. 4.
c. 5.
d. Tat ca deu SAI.
D o a n c h u a n g tr in h s a u d u n g c h o c au 3 3 ,3 4 :
int x = 0, y = 0;
intM [3][3] = {{1,2,3}, {1,1,1}, {3,2,1}};
for (int i = 0;i<3;i++)
{
for (intj = OJOJ-H-)
{
if (i == j)
x = x + M[i][j];
else
y= y + M[i][j];
}
}
Cau 33: (0.25 dilm)
Cho biet gia tri cua bien x sau khi kit thuc doan chuang trinh tren:
a. 3.
b. 4.
c. 5.
d. Tat ca deu SAI.
Cau 34: (0.25 diem)
Cho bilt gia trj cua bien y sau khi ket thuc doan chuang trinh tren:
a. 11.
b. 13.
c. 15.
d. Tat ca deu SAI.
D o a n c h u a n g tr in h s a u d u n g c h o c au 3 5 ,3 6 ,3 7 :
int M[5] = {-1,3,2,4,7};
for (int i = 0;i<4;i++)
{
if (M[i]>M[i+l])
M[i] = M[i+1];
else
printf("%d",M[i]);
}
Cau 35: (0.25 diem)
Cho biet gia trj cua phan tu M[2] sau khi kit thuc doan chuang trinh tren:
a. -1
b. 3
c. 2
d. Tat ca dlu SAI.
Cau 36: (0.25 dilm)
Cho biet gia tri cua phan tu M[4] sau khi k it thuc doan chuang trinh tren:
a. 5
b. 6
c. 7
d. Tit ca diu SAI.
S6 hieu: BM2/QT-PDBCL-RDTV
7/8
-
.
-
Cau 37: (0.25 diem)
Cho biet ket qua in ra man hinh sau khi chay doan chuong trinh tren:
a. -124
b. -12
c. -13
d. Tat ca deu SAI.
D o a n c h u o n g tr in h s a u d u n g c h o c a u 3 8 ,3 9 ,4 0 :
int a = 0, b = 0;
for (int i = 0; i<3; i++)
{
for (int j = 2; j>=0; j —)
{
if (j>i)
else
a++;
b = b - a;
}
printf("%d", a+b);
}
Cau 38: (0.25 diem)
Cho biet gia trj cua bien a sau khi ket thuc doan chuong trinh tren:
a. 0.
c. 2.
b. 1.
d. Tat ca deu SAI.
Cau 39: (0.25 dilm)
Cho biet gia tri cua bien b sau khi ket thuc doan chuong trinh tren:
a. -10.
c. -20.
b. -17.
d. Tat ca deu SAI.
Cau 40: (0.25 diem)
Cho biet ket qua in ra man hinh sau khi chay doan chuong trinh tren:
a. 0-5-11
c. 0-5-14
b. 0-5-13
d. Tat ca deu SAI.
Ghi chit:Can bo coi thi khong duac giai thick de thi.
Chuan dau ra cua hoc phan (ve kien thuc)
[CBR 1.1]: Giai thich duoc, mo ta dugc hoat dong cua mot
chuong trinh lap trinh may tinh.
Noi dung kiem tra
Cau 1,2,3,4,5,6,7,8,9,10,11,12
31,32,33,34,35,36,37,38,39,40
[CBR 1.2]: Hilu va giai thich duoc cac luu do giai thuat.
Cau 13,14,15,16.
[CBR 2.2]: Trinh bay duoc cac bai toan dieu khien duoi
dang thuat toan va giai thuat.
Cau 17,18,19,20,21.
[CBR 4.1]: Biet su dung cac phuong phap lap trinh de xay
dung chuong trinh.
Cau 22,23,24,25,26,27
28,29,30.
Ngay 15 thang 12 nam 2019
P. Truong bo mon
So hieu: BM2/QT-PDBCL-RDTV
8/8
\
TRCONG BH SC PHAM KY THUAT TP.HCM
DE THI CUOI KY HOC KY II NAM HQC 2019-2020
KHOACO KHI CHE TAO MAY
Mon: TIN HQC TRONG KY THUAT.......................
BO MON CO BIENTC
Chu ky giam thi 1
Chii ky giam thi 2
Ma mon hoc: MEIF134529..........................................
P c
B6 s6/Ma d6: 01.
Thai gian: 60phut.
Bugc phep sir dung tai lieu (KHONG dung laptop).
SV lam bai true tiep tren de thi va nop lai de
Diem va chir ky
CB cham thi thu nhat
B6 thi co 08trang.
CB cham thi thu hai
Ho va ten:.....................................................................
M as6 S V :.....................................................................
So T T :...................... Phong t h i : ...............................
PHIEU TRA LOT
Hirtmg d in tra 151 cau hoi:
Chon cau tra ldi dung:
STT
a
Bo chon:
b
c
Chon lai:
d
STT
1
21
2
22
3
23
4
24
5
25
6
26
7
27
8
28
9
29
10
30
11
31
12
32
13
33
14
34
15
35
16
36
17
37
18
38
19
39
20
40
S6 hieu: BM3/QT-PBBCL-RBTV
a
j^T
b
c
d
Trang: 1/8
A
r
PHAN I - MATLAB (15 cau)
Cau 1: (0.25 diem)
Trong cac cau lenh Matlab sau, cau lenh nao la DUNG (khong bao loi):
a. x = 5
b. x = 5.0
c. x = 5i
d.Tat ca dlu dung.
Cau2: (0.25 diem)
Cho biet ket qua cua phep toan sau:
» 1 - 0*lnf
a.l.
b.Inf.
c. NaN.
d. Tit ca deu sai.
Cau3: (0.25 diem)
Ham ceil(x) trong Matlab duqc hilu la:
a.Lam tron len
b.Tinh |x|
c.Lam tron xuong
d.Tat ca dlu sai.
Cau4: (0.25 diem)
Trong cira so Command cua Matlab, chung ta thuc hien lenh sau, hay cho bilt k it qua:
» 3 A3/3\3A3
a.l
b.3
c.Cau lenh bao loi
d.Tat ca deu sai.
Cau5: (0.25 dilm)
Trong cua so Command cua Matlab, chung ta thuc hien lenh sau,hay cho bilt k it qua:
»abs(3+4i)
a.3
b.4
c.5
d.Cau lenh bao loi.
Cau6: (0.25 diem)
Trong Matlab, d i giai phuoug trinh sin(x) == 0, ta sir dung cau lenh nao sau day:
a.solve(‘sind(x)’)
b.solve(‘sind(x) = O’)
c.solve(‘sind(x) = O’)
d.Ca 3 cau lenh deu dung.
Cau7: (0.25 dilm)
Trong cua s6 command cua Matlab, chung ta thuc hien lenh sau:
»
a=[l 2; 3 4];
»
b = a ';
» x = b(2)
Ket qua cua x la:
a.l
b.2
c.3
d.Tat ca deu sai.
S6 hieu: BM3/QT-PDBCL-RDTV
Trang:2/8
Cau8: (0.25 diem)
Trong cua s6 Command cua Matlab, chung ta thuc hien lenh sau,hay cho biet ket qua:
»mod(7.5,2.5)
a.O
c.Cau lenh bao loi
b.2.5
d.Tat ca deu sai.
Cau9: (0.25 diem)
Trong cua s6 command cua Matlab, chung ta thuc hien lenh sau:
»
a=[l 2 3];
»
b = [1; 1 ;1);
» a .* b
Ket qua in ra la:
a.l 2 3
b.6
c.Cau lenh bao loi
d.Tat ca deu sai.
Cau 10: (0.25 diem)
Trong cua so command cua Matlab, cho biet gia tri cua y khi chung ta thuc hien lenh sau:
» x = 0:3:5
»
y = x(3)
a.2
b.3
c.4
d.Cau lenh bao loi.
Cau 11: (0.25 diem)
Cho ma tran A co gia trj nhu sau:
»
A = [1 2;3 4;5 6]
Trong cac cau lenh Matlab sau, cau lenh nao la DUNG (khong bao 16i):
a .» A .A2
b .» 2 .AA
c .» A .A(-l)
d. Tat ca deu dung.
Cau 12: (0.25 diem)
Cho vecto x co gia tri nhu sau:
» x = [1:5]
Be ve d6 thi sin(x)cos(x), ta su dung cau lenh nao trong cac cau lenh sau:
a.fplot('sin(x*cos(x)')
b.fplot(sin(x)*cos(x))
c.plot( sin(x).*cos(x))
d.plot(sin(x)*cos(x))
Cau 13: (0.25 diem)
Cho vecto x co gia trj nhu sau:
» x = [1:5]
D6 ve d6 thi y = xexcau lenh khai bao vecto y nao sau day la DUNG:
a . » y = x*eAx
c . » y = x.*e.Ax
S6 hieu: BM3/QT-PDBCL-RDTV
b . » y = x.*eAx
d. Khong cau lenh nao dung.
Trang: 3/8
f
Cau 14: (0.25 diem)
Cho vector x = [1 2 3], de ve d l thi hinh tron, ta su dung cau lenh:
a. pie(x)
b.plot(x)
c.bar(x)
d.stem(x)
Cau 15: (0.25 diem)
Cho cau lenh sau:
» x = [1:5]
» p lot(x,xA2)
Khi chay chuong trinh, k it qua se la:
a.Budng do thi co mau xanh
c.Cau lenh bao 16i
b.Buong d6 thi co mau do
d.Tat ca dlu sai.
PHAN II - C (25 cau)
Cau 16: (0.25 diem)
Cho a,b la bien so nguyen (int) va a = 1, b = 2, c = 3. Hay cho biet gia tri cua bilu thuc: a-b%c
a. 0.
b. 1.
c.-l.
d. T it ca dlu sai.
Caul7: (0.25 diem)
Cho a,b la biln s6 nguyen (int) va a = 2, b = 3. Hay cho bilt gia tri cua bilu thuc: (++a/++b)%a:
a. 0.
c. 2.
Caul8: (0.25 diem)
b. 1.
d. Tat ca dlu sai.
Cho a,b la bien s6 nguyena = 1, b = 2. Hay cho biet gia tri cua bieu thuc: !(a>b)&&(a
b. Sai/False.
Caul9: (0.25 diem)
Cho doan chuong trinh nhu sau, hay cho bilt gia tri cua biln n:
char S[20] = "12345";
cham = S[strlen(S)];
a. '4'
c. Cau lenh bao loi
b. '5'
d. Tit ca dlu sai.
Cau20: (0.25 diem)
Be sir dung ham strcpyO, can khai bao thu vien nao sau day:
a.stdio.h
b.conio.h
c.string.h
d. Khong dap an nao dung.
Cau21: (0.25 diem)
Trong cac cau lenh sau, cau lenh nao la BUNG:
a.char A[] = “ 123”;
b.char A[3] = 123;
c.char A[3] = “ 123”;
d. Tat ca dlu dung.
So hieu: BM3/QT-PBBCL-RBTV
Trang:4/8
D o a n chicffng trin h s au d u n g c h o c a u 2 2 ,2 3 :
int tong = 0, max = 0;
for(int i = 0; i<5; i++)
if(i%2==0)
tong = tong + i;
else
max
Cau 22: (0,25dilm)
Gia trj cua tong sau khi ket thuc doan chuong trinh tren la:
a.l
b.3
d. Tat ca deu sai.
Cau 23: (0,25diem)
Gia tri cua max sau khi ket thuc doan chuong trinh tren la:
a.2
b.3
c.4
d. Tit ca diu sai.
D o an c h ita n g trin h sau dung cho cau 24,25,26:
char S[20] = "11223";
int n = 0;
for (int i = 0;i
if ( S[i] == S[i+1])
n++;
else
S[i+1] = ‘O’;
printf("%c",S[i]);
}
Cau 24: (0,25diem)
K it qua in ra man hinh sau khi kit thuc doan chuong trinh tren la:
a.l 100
b.1110
c.1000
d. Tit ca diu sai.
Cau25: (0.25 diem)
Cho biet gia tri cua chuoi S sau khi ket thuc doan chuong trinh tren:
a.l 1223
b.11002
c.11000
d. Tat ca deu sai.
Cau26: (0.25 dilm)
Cho biet gia tri cua n sau khi k it thuc doan chuong trinh tren:
a.O
b.l
c.2
d. Tat ca deu sai.
S6 hieu: BM3/QT-PBBCL-RBTV
Trang: 5/8
D o a n c h ita n g trin h sau d u n g cho cau 27,28:
intx = 0, y = 0, i = 0;
int M[5] = {1,2,3,4,5};
while ( x < 5 && i<5)
{
{
if (M[i]%2 == 0)
x = x + M[i];
else
y= y + M[i];
}
Cau27: (0.25 dilm)
Cho biet gia tri ciia bien x sau khi ket thuc doan chuong trinh tren:
a.
4.
b. 5.
c.
6.
d.
Tit ca diu sai.
Cau28: (0.25 diem)
Cho biet gia tri cua bien y sau khi k it thuc doan chuong trinh tren:
a.
5.
b. 7.
c.
9.
d.
T it ca deusai.
D o a n c h u a n g trin h sau d un g cho cau 29,30,31:
charS[10] = "abc";
intx = 0;
for (int i = 0;i
for (int j = 0y
if (S[i] = S[j]}
x++;
else
S[i] = SO];
}
printf("%d", x);
}
Cau29: (0.25 diem)
Cho biet gia trj cua chuoi S sau khi ket thuc doan chuang trinh tren:
a. “abc”
b. “aaa”
c. “bbb”
d. “ccc”
Cau30: (0.25 diem)
Cho biet gia tri cua bien x sau khi ket thuc doan chuong trinh tren:
a.2.
c.4.
S6 hieu: BM3/QT-PDBCL-RDTV
b.3.
d. Tat ca deu sai.
Trang: 6/8
Cau31: (0.25 diem)
Cho biSt k it qua in ra man hinh sau khi chay doan chucmg trinh tren:
a. 123
c. 136
b. 135
d. Tat ca deu sai.
D oan c h u ffn g t r in h sau d un g cho cau 32,33,34:
int M[2][3] = {1,2,3,4,5,6};
for (int i = 0;i<2;i++)
{
for (intj = 0;j<3y++)
{
if(M[i][j]
M[i][j] = M[i][i];
M[i][j] = 0;
printf("%d",M[i][j]);
}
}
Cau32: (0.25 dilm)
Cho biet gia tri cua phan tu M[1][0] sau khi kit thuc doan chuong trinh tren:
a.O
b.l
c.2
d.
Tat ca diu sai.
Cau33: (0.25 diem)
Cho biet gia tri cua phan tu M[0][1] sau khi ket thuc doan chuong trinh tren:
a.O
b. 3
c.5
d.
Tit ca diu sai.
Cau34: (0.25 diem)
Cho biet ket qua in ra man hinh sau khi chay doan chuong trinh tren:
a.000000
b.000300
c.000500
d. Tat ca deu sai.
D oan chieang trin h sau d u n g cho cau 35,36,37:
int a = 0, b = 0;
while (a+b < 5)
{
if (a>b)
b++;
else
a=a+2;
printf("%d", a+b);
}
Cau35: (0.25 diem)
Cho biet gia trj cua bien a sau khi ket thuc doan chuong trinh tren:
a.2.
b.3.
c.4.
d. Tat ca diu sai.
S6 hieu: BM3/QT-PDBCL-RDTV
Trang: 7/8
