CBSE 12TH MATHEMATICS - Online Test
Q2. For the curve tangent is parallel to X – axis where
Answer : Option D
Explaination / Solution:
dydx=0⇒dydt=0⇒2t−1=0⇒t=12.
Q3. The area bounded by the curves y = - 1 and y = - + 1 is
Answer : Option A
Explaination / Solution:
The graph of modulus function is V-shaped graph. Therefore , from graph , the area is = √2 × √2 =2.
Q4. Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Answer : Option A
Explaination / Solution:
Objective function is Z = 3x + 5 y ……………………(1).
The given constraints are : x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 .
The corner points obtained by drawing the lines x+3y=3 and x+y=2 are (0,0), (3,0),(0,2) and (3/2,1/2)
Corner points | Z = 3x + 5 y |
A(3 , 0 ) | 9 |
B ( 0 ,2 ) | 10 |
C( 3/2 , 1/2 ) | 7………………..( Min. ) |
Here Z = 7 is minimum at ( 3/2 , ½ ) .
Q5. If R is a relation from a set A to a set B and S is a relation from B to C, then the relation S∘R.
Answer : Option C
Explaination / Solution:
Let R and S be two relations from sets A to B and B to C respectively.Then we can define a relation
from A to C such that this relation is called the composition of R and S.
Q6. Solution set of the equation 
Answer : Option D
Explaination / Solution:
expanding along R1
Q8. Which of the following is a vector?
Answer : Option A
Explaination / Solution:
Acceleration is a vector quantity , because acceleration is defined as the rate of change of velocity w.r.t. time.and velocity specifies direction
Q9. If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A ∩ B)
Answer : Option A
Explaination / Solution:
We have ,
P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4
P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4
Q10. If x then sin 2 θ is equal to
Answer : Option B
Explaination / Solution:
.
