CBSE
For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is
5/2
11/2
6
13/2
B.
11/2
Let p(x) be a function defined on R such that = 1, p'(x) p'(1-x),for all x∈[0,1] p(0) = 1 and p(1) = 41. Then equals
√41
21
41
42
B.
21
We have
p'(x) = p'(1-x), ∀ x ∈[0,1], p(0) = 1, p(1) = 41
⇒ p(x) = - p(1-x) + C
⇒ 1 = - 41+C
⇒ C = 42
therefore, p(x) + p(1-x) = 42
For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is
there is a regular polygon with r/R = 1/2
there is a regular polygon with
there is a regular polygon with r/R = 2/3
there is a regular polygon with
C.
there is a regular polygon with r/R = 2/3
Let S be a non empty subset of R. Consider the
following statement:
P: There is a rational number x∈S such that x > 0.
Which of the following statements is the negation of the statement P?
There is a rational number x∈S such that x ≤ 0.
There is no rational number x∈ S such that x≤0.
Every rational number x∈S satisfies x ≤ 0.
x∈S and x ≤ 0 ⇒ x is not rational.
C.
Every rational number x∈S satisfies x ≤ 0.
The equation of the tangent to the curve y = x +4/x^{2}, that is parallel to the x-axis, is
y= 0
y= 1
y= 2
y= 3
D.
y= 3
We have,
On differentiating w.r.t x, we get
since the tangent is parallel to X- axis, therefore
dy/dx = 0
⇒ x^{3} = 8
⇒ x = 2 abd y =3
A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a_{1} = a_{2} = ... = a_{10} = 150 and a_{10}, a_{11}, ...are in an A_{P} with common difference -2, then the time taken by him to count all notes is
24 min
34 min
125 min
135 min
B.
34 min
Let the first term of an AP be a and common difference be d and number of terms be n, then
t_{n} = a + (n-1)d and Sn = n/2 [ 2a + (n-1)d]
Number of notes that the person counts in 10 min = 10 x 150 = 1500
Since a10, a11, a12, .... are in AP in with common difference -2
Let n be the time has taken to count remaining 3000 notes than
n/2[2 x 148 + (n-1) x -2] = 3000
⇒ n2-149n +3000 = 0
⇒ (n-24)n-125) = 0
n = 24, 125
Then, the total time taken by the person to count all notes = 10 +24 = 34 min
n = 125 is discarded as putting n = 125
an = 148 + (125-1)(-2)
= 148 - 124 x 2 = 148-248 = -100
⇒ Number of notes cannot be negative.
If α and β are the roots of the equation x^{2} – x +1 =0, then α^{2009} + β^{2009} =
-2
-1
1
2
C.
1
The quadratic equation ax2 + bx +c = 0 has roots α and β,
Then α + β = - b/a, α β = c/a
Also, if ax^{2}+ bx +c = 0
Then,
We know that 1,ω, ω^{2} are cube roots of unity.
1+ω + ω^{2} = 0 (ω^{2} = 1)
and
Since α and β are roots of the equations
x^{2}-x+1 = 0
⇒ α + β = 1 , α β =1
x = ω^{2}, or -ω
α = -ω^{2}, then β =-ω
or α = -ω, then β =-ω^{2}, (where ω^{3} = 1)
Hence, α^{2009} + β^{2009} =(-ω)^{2009} + (-ω^{2})^{2009}
= - [(ω^{3})^{669}. ω^{2} + (ω^{3})^{1337}.ω]
= - [ω2 + ω]
= -(-1) = 1
Let cos (α + β) = 4/5 and let sin (α - β) = 5/13, where 0 ≤α,β ≤ π/4. Then tan 2α is equal to
25/16
56/33
19/12
20/7
B.
56/33
Cos (α + β) = 4/5
⇒ (α + β) ∈ 1st quadrant
and sin (α - β) = 5/13
⇒ (α - β) ∈ 1st quadrant
⇒ 2α =(α + β) + (α - β)
∴
Consider the following relations:
R = {(x, y)| x, y are real numbers and x = wy for some rational number w}; S = {(m/p, p/q)| m, n, p and q are integers such that n, q ≠ 0 and qm = pn}. Then
R is an equivalence relation but S is not an equivalence relation
neither R nor S is an equivalence relation
S is an equivalence relation but R is not an equivalence relation
R and S both are equivalence relations
D.
R and S both are equivalence relations
The Number of complex numbers z such that |z– 1| = |z + 1| = |z – i| equals
0
1
2
∞
B.
1
|z – 1| = |z + 1|
⇒ lies on y-axis (perpendicular bisector
of the line segment joining (0, 1) and (0,-1)].
|z + 1| = |z – 1|
⇒ lies on y = -x
hence (0 + oe) is the only solution.