@wovfactorAPS said:i did by differentiation..
plz tell the eqn you got after diff..
@vijay_chandola said:If the roots of the equation ax3 + bx2 + cx + d = 0 are in Geometric Progression, then which of the following relations is true?(a) a*c^2 = b^2*d(b) a*c^3 = b^3*d (c) a^2*c = b*d^2 (d) a^3*c = b*d^3
(b) ?
1 2 4
b = -7a
c = 14a
d = -8a
option B
@vijay_chandola said:If the roots of the equation ax3 + bx2 + cx + d = 0 are in Geometric Progression, then which of the following relations is true?(a) a*c^2 = b^2*d(b) a*c^3 = b^3*d (c) a^2*c = b*d^2 (d) a^3*c = b*d^3
4th option
k/r+k+kr=-b/a
k^2/r+k^2+k^2r=c/a
k^3=-d/a
c/b=1/k
c^3/d^3=a/d
k/r+k+kr=-b/a
k^2/r+k^2+k^2r=c/a
k^3=-d/a
c/b=1/k
c^3/d^3=a/d
:D@Anivesh90 said:plz tell the eqn you got after diff..
assuming UQ=X
the eqn. i got is after
x^2/(x^2+1)=(2-x)^2/((2-x)^2+4)
reciprocating and substracting 1 gives
1/x^2=4/(2-x)^2
the eqn. i got is after
x^2/(x^2+1)=(2-x)^2/((2-x)^2+4)
reciprocating and substracting 1 gives
1/x^2=4/(2-x)^2
How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?
(a) 21 (b) 40 (c) 41 (d) 39
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
(c) 41 ?
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
Numbers are of the form = LCM(2,4,6,8)*k - 1
24k - 1 => k= 1 to 41 => 41 numbers ?
@vijay_chandola said:If the roots of the equation ax3 + bx2 + cx + d = 0 are in Geometric Progression, then which of the following relations is true?(a) a*c^2 = b^2*d(b) a*c^3 = b^3*d (c) a^2*c = b*d^2 (d) a^3*c = b*d^3
(b) a*c^3 = b^3*d
Let the roots be 1,2,4
Let the roots be 1,2,4
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
41?
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
LCM (2,4,6,8 ) - 1 = 23
So numbers of form 24k + 23 will satisfy the condition
k can take any value from 0 to 40 (since the limit is till 1000)
So answer should be 41 numbers
Please correct me if i am wrong
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
41
Nos will be of the form 24x-1
Nos will be of the form 24x-1
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
41...hcf and lcm model :P
@audiq7 said:the remainder obtained when 1989*1990*1991+1992^3 is divided by 3982?
1989*1990*1991+1992^3/(2*1991)
Remainder is 1
How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?
(a) 21 (b) 40 (c) 41 (d) 39
41 😃 ==>No.'s of the form = LCM(2,4,6,8)*k - 1
24k - 1 => k= 1 to 41 => 41 :)
"LCM first model" cncpt
(a) 21 (b) 40 (c) 41 (d) 39
41 😃 ==>No.'s of the form = LCM(2,4,6,8)*k - 1
24k - 1 => k= 1 to 41 => 41 :)
"LCM first model" cncpt
@vijay_chandola said:How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders 1, 3, 5, 7 respectively?(a) 21 (b) 40 (c) 41 (d) 39
getting 41
24k+23 ..form of the number 23-->983
24k+23 ..form of the number 23-->983
@ All: Correct. kuchh jyada hi easy ho gaya 😛 :splat:
If 'a' is one of the roots of x^5 – 1 = 0 and a ≠ 1, then what can be the value of a^15 + a^16 + a^17 +.......a^50?
(a) 1 (b) 5a (c) 35 (d) None of these
@vijay_chandola said:If the roots of the equation ax3 + bx2 + cx + d = 0 are in Geometric Progression, then which of the following relations is true?(a) a*c^2 = b^2*d(b) a*c^3 = b^3*d (c) a^2*c = b*d^2 (d) a^3*c = b*d^3
product of three roots = -(d/a) =>one of the roots must be -(d/a)^(1/3)
put this value in original eqn to get:
a(-d/a) + b(-d/a)^(2/3) +c(-d/a)^(1/3) +d =0
=>b(-d/a)^(2/3) = -c(-d/a)^(1/3)
cubing both sides we get
=>b^3*d^2/a^2 = c^3*d/a
=>b^3*d = c^3*a
hence option (b)
ATDH.
@vijay_chandola said:@ All: Correct. kuchh jyada hi easy ho gayaIf 'a' is one of the roots of x^5 – 1 = 0 and a ≠ 1, then what can be the value of a^15 + a^16 + a^17 +.......a^50?(a) 1 (b) 5a (c) 35 (d) None of these
1