Mathematics - Online Test
Q1. If the two lines of regression are at right angles , then ρ(X,Y) is equal to
Answer : Option C
Explaination / Solution:
Q2. The area bounded by the curves is equal to
Answer : Option A
Explaination / Solution:
Q3. A relation R in a set A is called universal relation, if
Answer : Option B
Explaination / Solution:
The relation R = A x A is called Universal relation.
Q4. The roots of the equation det. 
Answer : Option D
Explaination / Solution:
Expanding along C1
Q5. The triangle formed by the lines x + y = 1, 2x + 3y – 6 = 0 and 4x – y + 4 = 0 lies in
Answer : Option C
Explaination / Solution:
On solving line 1 and line 2 we get x = -3 andy =4. Hence the point of intersection is (-3,4)
On solving line 2 and line 3 we get x = (-3/7) and y = 16/7. Hence the point of intersection is (-3/7, 16/7)
On solving line 3 and line 1 we get x = -3/5 and y - 8/5.Hence the point of intersection is (-3/5,8/5)
All the above points lie in the second quadrant. Hence the triangle formed by these lines also lie in the second quadrant.
Q6. The negation of the proposition “if a quadrilateral is a square, then it is a rhombus “ is
Answer : Option C
Explaination / Solution:
rules of negation ∼(p→q)≡p∧∼q
Q7. is equal to
Answer : Option A
Explaination / Solution:
+ C+ C
Q8. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).
Answer : Option C
Explaination / Solution:
The scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7) is given by : (- 5 – 2 ) i.e. – 7 and (7 – 1 ) i.e. 6. Therefore, the scalar components are – 7 and 6 .,and vector components are - 
Q9. if then tan θ is equal to
Answer : Option C
Explaination / Solution:
Q10. Four distinct points and (0, 0) lie on a circle for
Answer : Option B
Explaination / Solution:
Because equation of circle will be x2 + y2 - x - y =0 which is a quadratic equation and by putting the given fourth point we will get two different values of lemda.
Proof: let equation of cirle be 
As (1,0) (0,1) and (0,0) lie on circle ,we get
Comparing (i) and (iii) ,we get h=
Comparing (ii) and (iii) we get k=
putting value of h and k in (iii) we get
hence we get equation of circle as 
now putting in above circle ,we get
which is quadratic and will give 2 values of
