﻿ JEE Mathematics Solved Question Paper 2014 | Previous Year Papers | Zigya

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# JEE Mathematics Solved Question Paper 2014

#### Multiple Choice Questions

1.

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

D.

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
= -

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2.

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
X ⊂ Y
Therefore, X U Y = Y

380 Views

3.

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
⇒
∵ t = x - [x] = {X}
∴ 0≤ t≤1

Taking positive sign, we get

⇒
⇒ 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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4.
• π/2

• 1

• π

D.

π

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5.

If z is a complex number such that |z|≥2, then the minimum value of

• 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  is distance of z, which lies on circle |z| = 2 from
therefore, minimum  = Distance of  from (-2,0)

Hence, minimum value of   lies in the interval (1,2)

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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  = 4, then the value of |α- β| is

D.

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7.

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  =
[∵ 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

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8.

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

192 Views

9.

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

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)

199 Views

10.

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

D.

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

250 Views