CBSE
The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directives is x= –4, then the equation of the normal to it at (1,3/2) is
x + 2y = 4
2y – x = 2
4x – 2y = 1
4x + 2y =7
C.
4x – 2y = 1
Eccentricity of ellipse =1/2
Now, a/e = -4
⇒ a = 4 x (1/2) = 2
therefore, b^{2} = a2(1-e^{2})
= a^{2} (1-(1/4)) = 3
Equation of normal at (1, 3/2)
y-3/2 = 2(x – 1)
⇒ 2y – 3 = 4x – 4
⇒ 4x – 2y = 1
Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB.If ∠BPC = β , then tanβ is equal to
4/9
6/7
1/4
2/9
D.
2/9
Let ω be a complex number such that 2ω +1 = z where z = √-3. if
then k is equal to
1
-z
z
-1
B.
-z
Here ω is complex cube of unity
For three events A, B and C,
P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P(Exactly one of C or A occurs) = 1/4and P(All the three events occur simultaneously) = 1/16.Then the probability that at least one of the events occurs, is
3/16
7/32
7/16
7/64
C.
7/16
P(exactly one of A or B occurs)
= P(A) + P(B) – 2P(A ∩ B) =1/4
P(Exactly one of B or C occurs)
= P(B) + P(C) – 2P(B ∩ C) =1/4
P(Exactly one of C or A occurs)
= P(C) + P(A) – 2P(C ∩ A) =1/4
Adding all, we get
2ΣP(A) – 2ΣP(A ∩ B) =3/4
∴ΣP(A) – ΣP(A ∩ B) =3/8
Now, P(A ∩ B ∩ C) =1/16
(Given)
∴ P(A ∪ B ∪ C)
= ΣP(A) – ΣP(A∩B) + P(A ∩ B ∩ C)
If two different numbers are taken from the set {0, 1, 2, 3, ......., 10), then the probability that their sum, as well as absolute difference, are both multiple of 4, is
7/55
6/55
14/55
12/55
B.
6/55
Let A ≡ {0, 1,2 ,3 ,4 ......, 10 }
n(s) = 11C2 (where 'S' denotes sample space)
Let E be the given event
∴ E ≡ {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)} n(E) = 6
∴ n(E) = 6 = 6/55
Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point
A.
We have,
⇒ 5k^{2} + 13k – 46 = 0
or
5k^{2} + 13k + 66 = 0 (no real solution exist)
∴ k =–23/5
or k = 2
k is an integer, so k =2
As
Therefore (2, 1/2)
The radius of a circle, having minimum area, which touches the curve y = 4 – x^{2} and the lines, y = |x| is
D.
There are two circles satisfying the given
conditions. The circle shown is of least area.
Let radius of circle is 'r'
∴ co-ordinates of centre = (0, 4 – r)
∴ circle touches the line y = x in first quadrant
If S is the set of distinct values of 'b' for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is
a singleton
an empty set
an infinite set
a finite set containing two or more elements
A.
a singleton
but at a = 1 and b =1
First two equations are x +y+ z =1
and third equations is x + y +z = 0
⇒ There is no solution
therefore, b = {1}
⇒ it is a singleton set