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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2014 Exam Questions

Multiple Choice Questions

1.

The angle between the lines whose direction cosines satisfy the equations l +m+n=0 and l2 = m2+n2 is

  • π/3

  • π/4

  • π/6

  • π/2


A.

π/3

We know that angle between two lines is 
cos space straight theta space equals space fraction numerator straight a subscript 1 straight a subscript 2 space plus straight b subscript 1 straight b subscript 2 space plus straight c subscript 1 straight c subscript 2 over denominator square root of straight a subscript 1 superscript 2 plus straight b subscript 1 superscript 2 plus straight c subscript 1 superscript 2 end root square root of straight a subscript 2 superscript 2 plus straight b subscript 2 superscript 2 plus straight c subscript 2 superscript 2 end root end fraction
l +m +n= 0
⇒ l = - (m+n)
⇒ (m+n)2 = l2
⇒ m2 +n2 +2mn = m2 +n2
[∵ l2 = m2 +n2, given]
⇒ 2mn = 0
when m = 0 ⇒ l =-n
Hence, (l, m, n) is (1,0-1)
When n =0, then l =-m
Hence, (l,m,n) is (1,0-1)
Cos space straight theta space equals space fraction numerator 1 plus 0 plus 0 over denominator square root of 2 space end root space straight x square root of 2 end fraction space equals space 1 half rightwards double arrow straight theta space equals space straight pi over 3

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2.

If the coefficients of x3 and x4 in the expansion of (1+ax+bx2)(1-2x)18 in powers of x are both zero, then (a,b) is equal to

  • open parentheses 16 comma 251 over 3 close parentheses
  • open parentheses 14 comma 251 over 3 close parentheses
  • open parentheses 14 comma 272 over 3 close parentheses
  • open parentheses 16 comma 272 over 3 close parentheses

D.

open parentheses 16 comma 272 over 3 close parentheses

In expansion of (1+ax+bx2)(1-2x)18,
Coefficient of x3 = Coefficient of x3 in (1-2x)18
+Coefficient of x2 in a(1-2x)18
+Coefficient of x in b(1-2x)18
 = - negative straight C presuperscript 18 subscript 3.2 cubed space plus space straight a straight C presuperscript 18 subscript 2.2 squared minus straight b to the power of 18 straight C subscript 1.2 space equals space 0
rightwards double arrow fraction numerator 18 space straight x 17 space straight x space 16 over denominator 3 space straight x 2 end fraction.8 space plus space straight a. fraction numerator 18 space straight x 17 over denominator 2 end fraction 2 squared minus straight b.18.2 space equals 0
rightwards double arrow space 17 straight a minus straight b space equals space fraction numerator 34 space straight x space 16 over denominator 3 end fraction space space... space left parenthesis straight i right parenthesis
Similarly space coefficient space of space straight x to the power of 4
straight C presuperscript 18 subscript 4.2 to the power of 4 space plus space straight a straight C presuperscript 18 subscript 3.2 cubed minus straight b to the power of 18 straight C subscript 2.2 squared space equals space 0
therefore space 32 straight a minus 3 straight b equals space 240
On space solving space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
straight a space equals space 16 space straight b space equals space 272 over 3

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3. limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style left parenthesis end style begin display style straight pi end style begin display style space end style begin display style begin display style cos end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction space is space equal space to
  • π/2

  • 1

  • π


D.

π

limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style left parenthesis end style begin display style straight pi end style begin display style space end style begin display style begin display style cos end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction
space equals space limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style straight pi end style begin display style left parenthesis end style begin display style 1 end style begin display style minus end style begin display style begin display style sin end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction space
equals space limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style left parenthesis end style begin display style straight pi end style begin display style minus end style begin display style begin display style πsin end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction
equals limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style left parenthesis end style begin display style straight pi end style begin display style space end style begin display style begin display style sin end style squared end style begin display style space end style begin display style straight pi end style begin display style right parenthesis end style over denominator straight x squared end fraction space space space space space left square bracket because space sin space left parenthesis straight pi space minus space straight theta space equals space sin space straight theta right square bracket
equals space limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style begin display style πsin end style squared end style begin display style straight x end style over denominator straight pi space sin squared straight x end fraction space straight x space left parenthesis straight pi right parenthesis open parentheses fraction numerator begin display style sin squared end style begin display style straight x end style over denominator straight x squared end fraction close parentheses space equals space straight pi space open square brackets because space limit as straight theta rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style straight theta end style over denominator straight theta end fraction space equals 1 space close square brackets
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4.

If a ε R and the equation - 3(x-[x]2 + 2(x-[x] +a2 = 0(where,[x] denotes the greatest integer ≤ x) has no  integral solution, then all possible value of  lie in the interval

  • (-1,0) ∪ (0,1)

  • (1,2)

  • (-2,-1)

  • (-∞,-2) ∪ (2, ∞)


A.

(-1,0) ∪ (0,1)

Given a ε R and equation is
-3{x-[x]}2 + 2{x-[x] +a2 = 0
Let t = x - [x], then equation is 
-3t2 +2t+ a2 = 0
⇒ straight t space equals fraction numerator 1 space plus-or-minus square root of 1 plus 3 straight a squared end root over denominator 3 end fraction space less or equal than space 1
∵ t = x - [x] = {X}
∴ 0≤ t≤1
0 space less or equal than fraction numerator 1 space plus-or-minus square root of 1 plus 3 straight a squared end root over denominator 3 end fraction space less or equal than space 1
Taking positive sign, we get
0 space less or equal than fraction numerator 1 space plus-or-minus square root of 1 plus 3 straight a squared end root over denominator 3 end fraction space less than space 1 space space space space space space space left square bracket because space open curly brackets straight x close curly brackets space greater than 0
⇒ square root of 1 plus 3 straight a squared end root space less than thin space 2 space
⇒ 1+3a2 <4 
⇒ a2-1 <0
⇒(a+1)(a-1) <0

a ε (-1,1)
For no integral solution of a, we consider the interval (-1,0) ∪ (0,1)

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5.

If PS is the median of the triangle with vertices P(2,1), Q(6,-1) and R (7,3), then equation of the line passing through (1,-1) and parallel to PS is

  • 4x-7y - 11 =0

  • 2x+9y+7=0

  • 4x+7y+3 = 0

  • 2x-9y-11  =0


B.

2x+9y+7=0

Coordinate of S  = open parentheses fraction numerator 7 plus 6 over denominator 2 end fraction comma fraction numerator 3 minus 1 over denominator 2 end fraction close parentheses space equals space open parentheses 13 over 2 comma 1 close parentheses
[∵ S is mid-point of line QR]
A slope of the line PS is -2/9.
Required equation passes through (1,-1) and parallel to PS is

straight y space plus space 1 space equals space fraction numerator negative 2 over denominator 9 end fraction left parenthesis straight x minus 1 right parenthesis
rightwards double arrow space 2 straight x space plus space 9 straight y space plus 7 space equals space 0

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6.

Let α and β be the roots of equation px2 +qx r =0 p ≠0. If p,q and r are in AP and 1 over straight alpha space plus space 1 over straight beta = 4, then the value of |α- β| is

  • fraction numerator square root of 61 over denominator 9 end fraction
  • fraction numerator 2 square root of 17 over denominator 9 end fraction
  • fraction numerator square root of 34 over denominator 9 end fraction
  • fraction numerator 2 square root of 13 over denominator 9 end fraction

D.

fraction numerator 2 square root of 13 over denominator 9 end fraction
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7.

If (10)9 +2 (11)2(10)7 + .....+10 (11)9 = K(10)9, then k is equal to

  • 121/10

  • 441/100

  • 100

  • 110


C.

100

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8.

If z is a complex number such that |z|≥2, then the minimum value of open vertical bar straight z space plus space 1 half close vertical bar

  • is equal to 5/2

  • lies in the interval (1,2)

  • is strictly greater than 5/2

  • is strictly greater than 3/2 but less than 5/2


B.

lies in the interval (1,2)

|z|  ≥2 is the region on or outside circle whose centre is (0,0) and the radius is 2.
Minimum open vertical bar straight z plus 1 half close vertical bar is distance of z, which lies on circle |z| = 2 from open parentheses negative 1 half comma 0 close parentheses
therefore, minimum open vertical bar straight z plus 1 half close vertical bar = Distance of open parentheses negative 1 half comma 0 close parentheses from (-2,0)

space equals space square root of open parentheses negative 2 space plus space 1 half close parentheses squared plus 0 end root space equals space 3 over 2 space equals space square root of open parentheses fraction numerator negative 1 over denominator 2 end fraction plus 2 close parentheses squared space plus 0 end root space equals space 3 over 2
Hence, minimum value of open vertical bar straight z plus 1 half close vertical bar  lies in the interval (1,2)

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9.

Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is

  • square root of 2 plus square root of 3
  • 3 plus square root of 2
  • 2 minus square root of 3
  • 2 plus square root of 3

D.

2 plus square root of 3

Let a, ar, ar2 be in GP (r>1)
On multiplying middle term by 2, a 2ar are in  AP.
⇒ 4ar = a+ ar2
⇒ r2-4r +1 = 0
straight r space equals space fraction numerator 4 plus-or-minus square root of 16 minus 4 end root over denominator 2 end fraction space equals space 2 plus-or-minus 3
straight r space equals space 2 plus square root of 3 space left square bracket space therefore space AP space is space increasing right square bracket


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10.

If X = {4n - 3n-1 : n ε N} and Y = {9(n-1):n εN}; where N is the set of natural numbers,then X U Y is equal to

  • N

  • Y-X

  • X

  • Y


D.

Y

We have X = {4n - 3n-1 : n ε N} 
X = {0,9,54,243,.....} [put n = 1,2,3....]
Y = {9(n-1):n ε N}
Y = {0,9,18,27,......}[Put n = 1,2,3....]
It is clear that 
⊂ Y
Therefore, X U Y = Y

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