Mathematics - Online Test
Q1. 
Answer : Option C
Explaination / Solution:
Q2. Let f (x) = then f (x) is
Answer : Option C
Explaination / Solution:
Hence an increasing function.
Q3.
The line which passes through the point ( 0 , 1 ) and perpendicular to the line x – 2y + 11 = 0 is
Answer : Option A
Explaination / Solution:
The line which is perpendicular to the given line is 2x + y + k = 0
Since it passes through (0,1)
2(0) + 1 + k = 0
This implies k = -1
Hence the equation of the required line is 2x + y - 1 = 0
Q4. ∼p∨∼q is logically equivalent to
Answer : Option B
Explaination / Solution:
The answer is rule of negation for ∼(p→∼q)≡∼p∨∼q
Q5. dx is not equal to
Answer : Option C
Explaination / Solution:
Q6. If and are any two points, then the vector joining P1 and P2 is the vector P1P2. Magnitude of the vector is
Answer : Option A
Explaination / Solution:
If and are any two points, then the vector joining P1 and P2 is the vector P1P2. then: = 
Q7. If E, F and G are events with P(G) ≠ 0 then P ((E ∪ F)|G) given by
Answer : Option C
Explaination / Solution:
P(EUF/G) ==
= +-
= P (E|G) + P (F|G) – P ((E ∩ F)|G)
Q8. If sin A + cos A = 1, then sin 2A is equal to
Answer : Option D
Explaination / Solution:
1 = 1 + Sin 2A
so, Sin 2A = 0
Hence A = 0
Q9.
For each n N , n (n + 1 ) ( 2n + 1 ) is divisible by :
Answer : Option A
Explaination / Solution:
When n = 1 the value is 6 . The subsequent substitution will give the value as a multiple of 6.
Q10. If f(x) = x | x | ∀x∈R, then
Answer : Option B
Explaination / Solution:
Also, f'(x)
,
Therefore f'(x) exists at all
Further, f'(0) =
