Mathematics Solved MCQs
Mathematics Solved MCQs for ECATMathematics SOLVED MCQs for Engineering Collages Admission Test ECAT
1. 3
3 ?
(A)
3
(B)
3
(C)
3i
(D)
3i
(E)
None of these
Solution:
We can write 3
3i
Hence,
3i
3i 3i2 3( 1)
3
Answer is: 3
2. dxd bx ?
(A)
bx
(B)
bx
(C)
bx ln b
(D)
bx
ln
x
(E)
b x x ln x
Solution:
Remember: dxd ax ax ln a Hence, dxd bx bx ln a Answer
3. If A a b Then A 1 ?
c d
(A)
ad
bc
(B)
ad
bc
(C) |
1 |
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b |
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ad bc c |
d |
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(D) |
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b |
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ad bc c |
a |
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(E) |
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ad bc c |
a |
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Remember,
if |
A |
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b |
Then its inverse
is: A 1 |
1 |
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c |
d |
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ad bc c |
a |
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Hence,
A 1 |
1 |
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d |
b |
Answer |
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ad bc c |
a |
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4. If a
sin xdx
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Then a ? |
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2 |
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2 |
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(A)
0 |
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(B) 1 |
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(C) |
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(D) |
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2 |
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(E) |
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3 |
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Solution:
a
sin xdx
cos x |
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= cos a cos |
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cos a 0 cos a |
1 |
OR cos a |
1 |
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2 |
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Hence,
a |
Answer |
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3 |
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5. Find the values of x and y from the following equations: 3x 2y
4
x y 2
(A)
x
2 and y 4
(B)
x
4 and y 6
(C)
x
32 and y 23
(D)
x
85 and y 52
(E) x 43 and y 35 Solution:
Multiply eq:2
with 3. 3x 2 y 4
3x 3y 6
5y 2 5y 2
y 52
Now
put value of y in eq:2, x y 2
x 25 2 x 85
Hence, answer is (D).
6.
If |
f (
x) |
x 2 |
Then |
f 1 (x) ? |
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(A) 3x 2
(B) 2x 3
(C)
3
x 2
(D) Does not exist
(E)
None of these
Solution:
f (
x) |
x 2 |
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3 |
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Step.1 Replace |
f (
x) |
by
y |
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y |
x 2 |
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3 |
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Step.2 Replace |
y with
x , and x with y |
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x |
y 2 |
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3 |
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Step.3 Solve for y . |
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y 2 3x |
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y 3x 2 |
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Step.4
Replace |
y by |
f 1 (x) |
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f 1 (x) 3x 2
Answer
7. Matrix A has 4 rows and 3 columns, and Matrix B has 5 rows and 2 columns. The Matrix AB will have?
(A)
4 rows and 2 columns
(B)
5 rows and 3 columns
(C)
2 rows and 4 columns
(D)
3 rows and 5 columns
(E)
3 rows and 2 columns
Solution:
If
rows = m , and columns = n , then
m n n p (Columns of 1st are equal to rows of 2nd). Cancel n , hence result
is m
p
Hence, In the Matrix AB there are rows of 1st, and columns of 2nd. That is, 4 rows and 2 columns. (A) is answer
2 3
8. Given that the Matrix is
singular. Find the value of a ?
4 a
(A) 2
(B) 2
(C)
3
(D)
6
(E) 6
Solution: |
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Since
the Matrix is singular, therefore Matrix |
2 |
3 |
0 |
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Now,
2a 12 0 |
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4 |
a |
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2a 12 |
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a 6 Answer |
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9.
(1 i)4 |
? |
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(A) 2 |
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(B) 2i |
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(C) 4 |
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(D)
4i |
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(E) 6i |
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Solution: |
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We
can write (1 i)4 |
{(1 i)2 }2 |
= {(1 2i
1)}2 |
= (2i)2 = 4i2 = 4( 1) |
= 4
Answer |
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10.
(logx |
xy)(logxy
xy
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(A)
1 |
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(B)
x |
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