Mathematics - Online Test
Q1. The area enclosed between the curves y = ,x- axis and two ordinates x = 1 to x = 2 is [ in square units ]
Answer : Option C
Explaination / Solution:
Required area :
= = = = == sq units
Q2. The medians of a triangle are concurrent at the point called
Answer : Option C
Explaination / Solution:
The centroid is the point of concurrency of the medians of the triangle.it is a point of centre of gravity of triangle
Q3. is the matrix
Answer : Option B
Explaination / Solution:
It is an identity matrix with the order 2*2,
Q4. Let f be a real valued function defined on (0, 1) ∪(2, 4) such that f ‘ (x) = 0 for every x, then
Answer : Option D
Explaination / Solution:
f ‘ (x) =0 f (x) is constant in ( 0 , 1 ) and also in ( 2, 4 ). But this does not mean that f ( x) has the same value in both the intervals . However , if f ( c ) =f ( d ) , where c ( 0 , 1 ) and d ( 2, 4) then f ( x ) assumes the same value at all x ( 0 ,1 ) U (2, 4 ) and hence f is a constant function.
Q5. If A is any set , then
Answer : Option A
Explaination / Solution:
{tex}A \cup A' = \{ x \in U:x \in A\} \cup \{ x \in U:x \notin A\} = U{/tex}
Q6. If Z = then equals
Answer : Option C
Explaination / Solution:
Q7. If A, B be two square matrices such that |AB|=O, then
Answer : Option A
Explaination / Solution:
If A B be two square matrices such that AB=O, then, |AB|=O⇒|A||B|=O⇒either|A|=0,or|B|=0
Q8.
If the rth term in the expansion of contains then r =
Answer : Option A
Explaination / Solution:
Q9. Identify the solution set for .
Answer : Option B
Explaination / Solution:
Multiplying both sides byLCM ,we get
So solution set is
Q10. In linear programming feasible region (or solution region) for the problem is
Answer : Option B
Explaination / Solution:
In linear programming feasible region (or solution region) for the problem is given by the common region determined by all the constraints including the non – negative constraints x ⩾ 0, y ⩾ 0
