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# JEE Mathematics 2010 Exam Questions

#### Multiple Choice Questions

1.

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

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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

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3.

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

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4.

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.

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5.

The equation of the tangent to the curve y = x +4/x2, 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
⇒ x3 = 8

⇒ x = 2 abd y =3

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6.

A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ...are in an AP 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
tn = 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.

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7.

If α and β are the roots of the equation x2 – x +1 =0, then α2009 + β2009 =

• -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 ax2+ bx +c = 0
Then,
We know that 1,ω, ω2 are cube roots of unity.
1+ω + ω2 = 0  (ω2 = 1)
and
Since α and β are roots of the equations
x2-x+1 = 0
⇒ α + β = 1 , α β =1

x = ω2, or -ω
α = -ω2, then β =-ω
or α = -ω, then β =-ω2, (where ω3 = 1)
Hence,  α2009 + β2009 =(-ω)2009 + (-ω2)2009
= - [(ω3)669. ω2 + (ω3)1337.ω]
= - [ω2 + ω]
= -(-1) = 1

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8.

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)

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9.

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

308 Views

10.

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.

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