Mathematics Test 107

Result

Mathematics Test 107

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Let A(1,3), B(a,b), C(4,7) are collinear points and x, then k CANNOT be-

A
4

B
5

C
6

D
9

Solution
Question 2
4 / -1

If a circle passes through point (1, 1) and cuts the circle x² + y² = 8 orthogonally, then the locus of its centre is -

A
x + y + 5 = 0

B
x + y = 5

C
x + y = 10

D
x + y + 10 = 0

Solution

Let circle be

x² + y² + 2gx + 2ƒy + c = 0

orthogonality condition

c = 8

circle passes through (1,1)

2 + 2g + 2ƒ + 8 = 0

g + ƒ + 5 = 0

–h – k + 5 = 0

x + y = 5
Question 3
4 / -1

If the two parabolas y² = 4x and y² = (x – k) have a common normal other than x-axis, then value of K cannot be -

A
1

B
2

C
3

D
4

Solution
Question 4
4 / -1

Let S = 0 be an ellipse whose vertices are the extremities of minor axis of the ellipse . If S = 0 passes through the foci of E, then its eccentricity is (considering the eccentricity of E as e)

A

B

C

D

Solution
Question 5
4 / -1

Value of

A
2

B
4

C
1

D
8

Solution
Question 6
4 / -1

The smallest positive root of the equation is equal to -

A
1/2 + 3π/4

B
1/2 + 7π/4

C
1/2 + 11π/4

D
1/2

Solution
Question 7
4 / -1

Let A_i, i = 1,2,3,......10 be vertices of regular decagon P. If area of polygon P is 5 sin (π/5) square units then the value of (A₁A₂)(A₁A₃)(A₁A₄)......(A₁A₁₀) is-

A
10

B
9

C
45sin (π/5)

D
45cos (π/5)

Solution
Question 8
4 / -1

If A = {1, 3, 5, 7, 9, 11},

B = {2, 4, 6, 8, 10, 12},

U is universal set, then A' ∪ ((A ∪ B) ∩ B') is-

A
U

B
A'

C
A ∩ B

D
A

Solution

(A ∪ B) ∩ B' = A

⇒ A' ∪ ((A ∪ B) ∩ B') = A' ∪ A = U
Question 9
4 / -1

Let I be the set of positve integers. R is a relation on the set I given by R = {(a,b) ∈ I × I | log₂ (a/b) is a non-negative integer}, then R is

A
neither symmetric nor transitive but reflexive.

B
reflexive, transitive but not symmetric

C
neither reflexive nor transitive but symmetric

D
equivalence relation.

Solution

a R b ⇔ a = 2^k.a it is true for k = 0

∴ reflexive

(2,1) ∈ R but (1,2) ∉ R ⇒ it is not symmetric

⇒ it is transistive.