Mathematics Test 11

Result

Mathematics Test 11

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Question 1
4 / -1

The fundamental period of the function: f(x) = x + a – [x + b] + sin πx + cos 2πx + sin 3πx + cos 4πx + …. + sin (2n – 1) πx + cos 2 nπx for every a, b ∈R is:
(where [.] denotes the greatest integer function)

A
2

B
4

C
1

D
0

Solution

f(x) = x + b – [x + b] + sin πx + cos 2πx +…..+ cos 2πnx + a – b
= {x + b} + sin πx + cos 2πx + ….. + a – b
↓
T 1, 2, ……….
L.C.M. = 2
Thus period = 2
Question 2
4 / -1

If A is a square matrix of order 3, then correct statement is -

A
det (–A) = – det A

B
det (–A) = 0

C
det (A+ I) = I + det A

D
det 2A = 2 det A

Solution

Options 2, 3 and 4 are incorrect.
For option 2, there is not information given about A to conclude that the determinant of it's negative would be 0
For option 3, we cannot add I to a determinant (which is a scaler)
For option 4, det 2A = 2³ det A.
Similarly, in option A,
det (-A) = (-1)³ det A
= - det A
Hence option A is correct.