If the coefficients of x^{3} and x^{4} in the expansion of (1+ax+bx^{2})(1-2x)^{18} in powers of x are both zero, then (a,b) is equal to
D.
In expansion of (1+ax+bx^{2})(1-2x)^{18},
Coefficient of x3 = Coefficient of x3 in (1-2x)^{18}
+Coefficient of x^{2} in a(1-2x)^{18}
+Coefficient of x in b(1-2x)^{18}
= -
If X = {4^{n} - 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 = {4^{n} - 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
If a ε R and the equation - 3(x-[x]2 + 2(x-[x] +a^{2} = 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] +a^{2} = 0
Let t = x - [x], then equation is
-3t^{2} +2t+ a^{2} = 0
⇒
∵ t = x - [x] = {X}
∴ 0≤ t≤1
Taking positive sign, we get
⇒
⇒ 1+3a^{2} <4
⇒ a^{2}-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)
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)
Let α and β be the roots of equation px^{2} +qx r =0 p ≠0. If p,q and r are in AP and = 4, then the value of |α- β| is
D.
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
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
The angle between the lines whose direction cosines satisfy the equations l +m+n=0 and l^{2} = m^{2}+n^{2} 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} = l^{2}
⇒ m^{2} +n^{2} +2mn = m^{2} +n^{2}
[∵ l^{2} = m^{2} +n^{2}, 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)
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, ar^{2} be in GP (r>1)
On multiplying middle term by 2, a 2ar are in AP.
⇒ 4ar = a+ ar^{2}
⇒ r^{2}-4r +1 = 0
Severity: Notice
Message: Undefined variable: flex_subject_id
Filename: features/flexipad.php
Line Number: 348
Backtrace:
File: /var/www/html/public_html/application/views/features/flexipad.php
Line: 348
Function: _error_handler
File: /var/www/html/public_html/application/views/previous_year_papers/paper.php
Line: 163
Function: view
File: /var/www/html/public_html/application/controllers/Study.php
Line: 2085
Function: view
File: /var/www/html/public_html/index.php
Line: 315
Function: require_once