CBSE
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?
3a^{2}−26a+55=0
3a^{2}−32a+84=0
3a^{2}−34a+91=0
3a^{2}−23a+44=0
B.
3a^{2}−32a+84=0
We know that, if x_{1}, x_{2}..... x_{n} are n observations, then their standard deviation is given by
If the line lies in the plane lx +my -z = 9, then l^{2} +m^{2} is equal to
26
18
5
2
D.
2
Since the line lies in the plane lx my-z =9, therefore we have 2l-m-3 = 0
[∴ normal will be perpendicular to the line]
⇒ 2l-m = 3
and 3l -2m +4 = 9
[∴ point (-3,-2-4) lies on the plane
⇒ 3l-2m = 5
On solving eqs (i) adn (ii), we get
l =1 and m=-1
therefore, l^{2} +m^{2} = 2
Let be three unit vectors such that then the angle between is
3π/4
π/2
2π/3
5π/6
D.
5π/6
The centres of those circles which touch the circle, x^{2}+y^{2}−8x−8y−4=0, externally and also touch the x-axis, lie on:
a circle
an ellipse which is not a circle
a hyperbola.
a parabola.
D.
a parabola.
If one of the diameters of the circle, given by the equation, x^{2}+y^{2}−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is:
5
10
B.
Given equation of a circle is x^{2} + y^{2} -4x +6y -12 = 0, whose centre is (2,-3) and radius
Now, according to given information, we have the following figure.
x^{2}+y^{2}-4x +6y-12 =0
Clearly, AO perpendicular to BC, as O is mid-point of the chord.
Now in ΔAOB,. we have
Let P be the point on the parabola, y^{2}=8x which is at a minimum distance from the centre C of the circle, x^{2}+(y+6)^{2}=1. Then the equation of the circle, passing through C and having its centre at P is:
x^{2}+y^{2}−4x+8y+12=0
x^{2}+y^{2}−x+4y−12=0
x^{2}+y^{2}− 4 x +2y−24=0
x^{2}+y^{2}−4x+9y+18=0
A.
x^{2}+y^{2}−4x+8y+12=0
Centre of circle x^{2} + (y+6)^{2} = 1 is C (0,6)
Let the coordinates of point P be (2t^{2}, 4t)
Now, let
Two sides of a rhombus are along the lines, x−y+1=0 and 7x−y−5=0. If its diagonals intersect at (−1, −2), then which one of the following is a vertex of this rhombus?
(−3, −9)
(−3, −8)
(1/3, -8/3)
(-10/3, -7/3)
C.
(1/3, -8/3)
As the given lines x-y +1 =0 and 7x-y-5 = 0 are not parallel, therefore they represent the adjacent sides of the rhombus.
On solving x-y+1 = 0 adn 7x - y -5 = 0. we get x =1 and y =2
Thus, one of the vertex is A(1,2)
Let the coordinate of point C be (x,y)
Then,
⇒ x+1 =- 2 and y =-4-2
⇒ x=-3 and y =-6
Hence, coordinates of C = (-3,-6)
Note that, vertices B and D will satisfy x-y +1 =0 and 7x - y-5 = 0, therefore the coordinate of vertex D is (1/3, -8/3)
The Boolean Expression (p∧~q)∨q∨(~p∧q) is equivalent to:
~p ∧ q
p ∧ q
p ∨ q
p ∨ ~ q
C.
p ∨ q
Consider, (p ∧~q) ∨ q ∨(~p ∧ q)
≡ [(p ∧~q) ∨ q] ∨ (~p ∧ q)
≡[(p ∨~q) ∧ t] ∨ (~p ∧ q)
≡((p ∨ q) ∨ (~p ∧ q)
≡(p ∨ q ∨ ~p) ∧ (p ∨ q ∨ q)
≡(q ∨ t) ∧ (p ∨ q)
≡ t ∧ (p ∨ q)
≡ p ∨ q
The distance of the point (1, −5, 9) from the plane x−y+z=5 measured along the line x=y=z is:
B.
the equation of the line passing through the point (1,5-9 and parallel to x =y=z is
Thus, any point on this line is of the form
(λ +1, λ-5 ,λ+9)
Now, if P (λ +1, λ-5, λ+9) is the point of intersection of line and plane, then
(λ+1) - (λ-5) +λ+9 = 5
λ +15 = 5
λ = -10
therefore coordinates of point P are (-9, -15,-1)
Hence, required distance
=
The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is:
4/3
C.
We have, 2b^{2}/a = 8 and 2b = ae
⇒ b^{2} = 4a and 2b = ae
Consider, 2b = ae
Consider, 2b = ae
⇒ 4b^{2} = a^{2}e^{2}
⇒ 4a^{2} (e^{2}-1)=a^{2}e^{2}
⇒ 4e^{2}-4 = e^{2}^{ [e>0]}