@vbhvgupta said:two equation have a common root which is positive. The other root of the equation satisfy x^2-9x+18=0 The product of the sum of the root of the two equation is 40. Find the common root.
2
@vbhvgupta said:The root of the equation X^3+PX^2+QX+R=0 are consecutive positive integer. Which of the following can never be the value of Q
1. 47
2. 11
3. 27
4. 107
EDIT:
27
@vbhvgupta said:The root of the equation X^3+PX^2+QX+R=0 are consecutive positive integer. Which of the following can never be the value of Q1. 472. 113. 274. 107
27 ?
x + x + 1 + x + 2 = -p
x(x + 1) + (x + 1)(x + 2) + x(x + 2) = q
x(x + 1)(x + 2) = -r
for x = 1...q = 11
x = 3....q = 47
x = 5....q = 107
@vbhvgupta said:The root of the equation X^3+PX^2+QX+R=0 are consecutive positive integer. Which of the following can never be the value of Q1. 472. 113. 274. 107
Let roots be, a-1, a and a+1
a(a-1)(a+1) = a^3 - a = -R
3a = -P
3a^2 - 1 = Q
Also, a^2 - 1 = 3R/P
3a^2 - 3 = 9R/P
Q = 3a^2 - 1 = 2 + 9(R/P) = 2 + 9k
So, 27 can never be value of Q
@vbhvgupta said:The root of the equation X^3+PX^2+QX+R=0 are consecutive positive integer. Which of the following can never be the value of Q1. 472. 113. 274. 107
roots be x-1, x, x+1
Q = x(x-1) + x(x+1) + (x2 - 1) = 3x^2 - 1 = 3k - 1
it cannot be a multiple of 3.
so 27.
f(n) is a function for all non-negative integers n, such that f(f(f(n))) + f(n) = 2n + 4 and f(0) = 1.
Find f(2013).
Find f(2013).
@vbhvgupta
@vbhvgupta said:pr^3 = q^3S
Pl share the method. Also confirm if it is geometric mean or progression.
@VJ12 said:@vbhvgupta Pl share the method. Also confirm if it is geometric mean or progression.
now in office...will post in the evening
@vbhvgupta said:the condition for the roots of the equation PX^3+3QX^2+3RX+S=0 to be in geometric mean is?
@vbhvgupta
Im getting it if it is geometric progression but answer i think is
(PR^3)=Q^3S
Method I used:
Let the roots be x,y,z.
If they are in G.P, y^2=zx and
x+y+z= -3Q/P---1
xy+yz+zx= 3R/P---2
xyz= -S/P---3
Substitute y^2=zx in (2) then,
y(x+y+z)= 3R/P
y( -3Q/P)= 3R/P
y= -R/Q; y^3= -R^3/Q^3---4
We know xyz= -S/P
Substituting y^2=xz;
y^3= -S/P---5
Since 4=5;
-R^3/Q^3=-S/P
PR^3=Q^3S.
Im getting it if it is geometric progression but answer i think is
(PR^3)=Q^3S
Method I used:
Let the roots be x,y,z.
If they are in G.P, y^2=zx and
x+y+z= -3Q/P---1
xy+yz+zx= 3R/P---2
xyz= -S/P---3
Substitute y^2=zx in (2) then,
y(x+y+z)= 3R/P
y( -3Q/P)= 3R/P
y= -R/Q; y^3= -R^3/Q^3---4
We know xyz= -S/P
Substituting y^2=xz;
y^3= -S/P---5
Since 4=5;
-R^3/Q^3=-S/P
PR^3=Q^3S.
@VJ12 said:@vbhvguptaIm getting it if it is geometric progression but answer i think is (-PR^3)=Q^3SMethod I used:Let the roots be x,y,z.If they are in G.P, y^2=zx andx+y+z= -3Q/P---1xy+yz+zx= 3R/P---2xyz= S/P---3Substitute y^2=zx in (2) then,y(x+y+z)= 3R/Py( -3Q/P)= 3R/Py= -R/Q; y^3= -R^3/Q^3---4We know xyz= S/PSubstituting y^2=xz;y^3= S/P---5Since 4=5;-R^3/Q^3=S/P-PR^3=Q^3S.Pl let me know if I made any calculation mistakes
let the roots be a/r , a, ar
a(1 + r + 1/r ) = -3Q/P
a^2(1 + r + 1/r ) =3R/P
dividing , a = -R/Q
also , a^3 = -S/P
putting the value of a
(-R/Q)^3 = -S/P
PR^3 = SQ^3
@techgeek2050 said:f(n) is a function for all non-negative integers n, such that f(f(f(n))) + f(n) = 2n + 4 and f(0) = 1.Find f(2013).
2014
Thanks bhai..
Product of roots in a cubic equation= -d/a, missed that.
edited the original solution as well.
Find the remainder when 22012 is divided by 2012.
OPTIONS 1)16 2)64 3)1024 4)503
pls post the approach...ty in advance
OPTIONS 1)16 2)64 3)1024 4)503
pls post the approach...ty in advance
All possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?
OPTIONS 1)632100 2)622110 3)633110 4)642100
OPTIONS 1)632100 2)622110 3)633110 4)642100
@vbhvgupta said:The root of the equation X^3+PX^2+QX+R=0 are consecutive positive integer. Which of the following can never be the value of Q1. 472. 113. 274. 107
it will be 27
a=-p/3
3a^2-1=q
put a=-p/3
using options 27
@saket.soni05 said:All possible 6 digit numbers, in each of which the digits occur in non - increasing order (from left to right) are written in increasing order. What is the 500th number in this sequence ?OPTIONS 1)632100 2)622110 3)633110 4)642100
642100