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CBSE

Subject

Mathematics

Class

JEE Class 12

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

because space straight sigma subscript straight x superscript 2 space equals space 4 space and space straight sigma subscript straight y superscript 2 space equals 5
Also comma space straight x with minus on top space equals space 2 space and space straight y with minus on top space equals space 4 space
Now comma space Σx subscript straight i over 5 space equals space 2
rightwards double arrow space straight capital sigma space straight x subscript straight i space equals 10 space semicolon space space Σy subscript straight i space equals space 20
and space straight sigma subscript straight x superscript 2 space equals space open parentheses 1 half Σx subscript 1 superscript 2 close parentheses minus left parenthesis straight x with minus on top space right parenthesis squared space equals space 1 fifth space left parenthesis straight capital sigma space straight y subscript straight i superscript 2 right parenthesis minus 16
rightwards double arrow space Σy subscript straight i superscript 2 space equals space 105
space straight sigma subscript straight x superscript 2 space space equals space 1 over 10 left parenthesis straight capital sigma space straight x subscript straight i superscript 2 space plus space straight capital sigma space straight y subscript straight i superscript 2 right parenthesis space minus open parentheses fraction numerator begin display style straight x with minus on top end style plus begin display style straight y with minus on top end style over denominator 2 end fraction close parentheses squared
space equals space 1 over 10 left parenthesis 40 plus 105 right parenthesis minus 9 space equals space fraction numerator 145 minus 90 over denominator 10 end fraction space equals space 55 over 10 space equals space 11 over 2
202 Views

2.

Let p(x) be a function defined on R such that limit as straight x space rightwards arrow infinity of space fraction numerator straight f left parenthesis 3 straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction = 1, p'(x)  p'(1-x),for all x∈[0,1] p(0) = 1 and p(1) = 41. Then integral subscript 0 superscript straight x space straight p space left parenthesis straight x right parenthesis space dx 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

straight I space equals space integral subscript 0 superscript 1 space straight p left parenthesis straight x right parenthesis dx space equals space integral subscript 0 superscript 1 space straight p left parenthesis 1 minus straight x right parenthesis space dx
rightwards double arrow space 2 space straight I space equals space integral subscript 0 superscript 1 space left parenthesis straight p left parenthesis straight x right parenthesis space plus space straight p space left parenthesis 1 minus straight x right parenthesis space dx
space equals space integral subscript 0 superscript 1 space 42 space dx space equals space 42
rightwards double arrow space straight I space equals 21

224 Views

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 straight r over straight R space equals space fraction numerator 1 over denominator square root of 2 end fraction

  • there is a regular polygon with r/R = 2/3

  • there is a regular polygon with straight r over straight R space space equals space fraction numerator square root of 3 over denominator 2 end fraction


C.

there is a regular polygon with r/R = 2/3

190 Views

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.

141 Views

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, 
straight y space equals space straight x space plus space 4 over straight x squared
On differentiating w.r.t x, we get

dy over dx space equals space 1 space minus space 8 over straight x cubed
since the tangent is parallel to X- axis, therefore
dy/dx = 0
⇒ x3 = 8

⇒ x = 2 abd y =3

162 Views

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.

349 Views

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, straight x space equals space fraction numerator negative straight b plus-or-minus space square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction
We know that 1,ω, ω2 are cube roots of unity.
1+ω + ω2 = 0  (ω2 = 1)
and straight omega space equals space fraction numerator negative 1 space plus space straight i square root of 3 over denominator 2 end fraction comma space straight omega squared space equals space fraction numerator negative 1 minus straight i square root of 3 over denominator 2 end fraction
Since α and β are roots of the equations
x2-x+1 = 0
⇒ α + β = 1 , α β =1
rightwards double arrow space straight x space equals space fraction numerator 1 space plus-or-minus space square root of 3 straight i over denominator 2 end fraction
rightwards double arrow space straight x space equals space fraction numerator 1 plus space square root of 3 straight i over denominator 2 end fraction
rightwards double arrow space equals space fraction numerator 1 minus square root of 3 straight i over denominator 2 end fraction
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

157 Views

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)
space tan space 2 straight alpha space equals space fraction numerator tan space left parenthesis straight alpha space space plus space straight beta right parenthesis space plus space tan space left parenthesis straight alpha space minus straight beta right parenthesis over denominator 1 minus tan space left parenthesis straight alpha space plus straight beta right parenthesis space tan left parenthesis straight alpha minus straight beta right parenthesis end fraction
space equals space fraction numerator begin display style 3 over 4 end style plus begin display style 5 over 12 end style over denominator 1 minus begin display style 3 over 4 end style. begin display style 5 over 12 end style end fraction space equals space 56 over 33

145 Views

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.

149 Views

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