CBSE 11TH MATHEMATICS - Online Test
Q2. If A, B and C are any three sets , then A – (B∩∩C ) is equal to
Answer : Option B
Explaination / Solution:
Q3. Ltx→0+x√16+x√√−4
Answer : Option D
Explaination / Solution:
Q4. Solve the system of inequalities
Answer : Option A
Explaination / Solution:
Hence solution set is
which implies no solution exists.
Q5. If a, b, c are in A. P. as well as in G.P.; then
Answer : Option B
Explaination / Solution:
a,b,c are in A.P, .......................(i)
a,b,c are in G.P, .............................(ii)
from (i) and (ii), we get
using a=c in (ii)
so, a=b=c
Q6. If a, b, c be any three positive numbers, then the least value of
Answer : Option A
Explaination / Solution:
A.M≥H.M⇒a+b+c3≥31a+1b+1c⇒(a+b+c)(1a+1b+1c)≥3×3≥9
Q7. The acute angle between the lines ax + by + c = 0 and ( a + b )x = ( a – b )y , a ≠ b , is
Answer : Option C
Explaination / Solution:
The slope of the first line is -(a/b) and slope of the second line is
Angle between the lines is =
Let m1 =-(a/b) and m2 =
Hence =[
On simplifying we get,
= 1
This implies = 450
Q8. Negation of the statement q∨∼(p∧r) is
Answer : Option C
Explaination / Solution:
∼(q∨∼(p∧r)) Since ∼(q∨r)≡∼q∧∼r
∼q∧(p∧r)
Q9. Which of the following lines is a normal to the circle
Answer : Option A
Explaination / Solution:
comparing the above equation with we get center as (1,2)
putting (1,2) in L.H.S of above line ,we get 1+2=3=R.H.S
=> it satisfies center satisfies the equation of line or the line passes through center
and the line passing through the centre is normal.
Q10.
cot = sin 2 n , n integer) if equals
Answer : Option B
Explaination / Solution:
sin2θ=cotθ⇒2sinθcosθ=sinθcosθ⇒cosθ(2sinθ−1sinθ)=0⇒−cosθ(1−2sin2θ)=0⇒−cosθ.cos2θ=0⇒cosθ=cosπ20rcos2θ=cosπ2⇒θ=π2,π4[∵θ≠nπ,nϵZ]
