Mathematics - Online Test
Q1. If the coefficient of variation between x and y is 0.28, covariance between x and y is 7.6, and the variance of x is 9, then the S.D. of the y series is
Answer : Option C
Explaination / Solution:
Q2. Consider the differential equation 
Which of the following is a solution
to this differential equation for x > 0?
Which of the following is a solution
to this differential equation for x > 0?
Answer : Option C
Explaination / Solution:
Q3. If A = { 1, 2, 3}, then the relation R = {(1, 2), (2, 3), (1, 3)} in A is
Answer : Option B
Explaination / Solution:
A relation R on a non empty set A is said to be transitive if x Ry and y Rz ⇒xRz, for all x ∈ R. Here , (1, 2) and (2, 3) belongs to R implies that (1, 3) belongs to R.
Q4. DELETE IT 
Answer : Option B
Explaination / Solution:
Q5. The lines x + ( k – 1) y + 1 = 0 and are at right angles if
Answer : Option A
Explaination / Solution:
If the lines are at right angles to each other, then the product of their slopes = -1. Slope of any line = -(coefficient of x/coeffiecient of y)
Therefore slope of line 1 = -(1/(k-1)
Slope of line 2 = -2/k2
Therefore X = - 1
That is k2 (k-1) = -2
i.e; k3 - k2 +2 = 0
On factorizing we get (k+1)(k2 -2k - 2) = 0
This imlpies k+1 is a factor, hence k = -1
Hence they are at right angles if k = -1
Q6. The compound statement p→(∼p∨q) is false , then the truth values of p and q are respectively
Answer : Option B
Explaination / Solution:
since T->F is false.
Hence q=F.
So p=T and q=F
Q7.
is equal to
Answer : Option B
Explaination / Solution:
Q8. If a vector is expressed in component form as then the magnitude of the vectoris
Answer : Option D
Explaination / Solution:
Q9. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E)
Answer : Option A
Explaination / Solution:
We have ,
P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then ,
P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then ,
Q10. equals
Answer : Option A
Explaination / Solution:
we know that sin−1x+cos−1x=π2sin(sin−112+cos−112)=sin(π2)=1
