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in this question, x^y means x raised the power of y . how many intiger x satisfied the equation(x^2-x-1)^(x+2)= 1?
Read Solution (Total 11)
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- there will be 4 solutions 0,-1,2,-2.
- 10 years agoHelpfull: Yes(35) No(0)
- (x^2-x-1)^(x+2) can only be one when
x^2-x-1=1 or x+2=0(anything raised to power 0 is 1)
two solutions for x^2-x-1=1 (x=-1,2)
and one solution for x+2=0 (x=-2)
so total possible solution are 3. - 10 years agoHelpfull: Yes(14) No(19)
- (X^2-X-1)^(X+2)=1
(4^2-4-1)^(4+2)=1
(16-4-1)^(6)=1
(11)^6=1
SO,THAT 11^6 IS UNIT DIGITS 1
ANSWER IS 4 - 10 years agoHelpfull: Yes(12) No(9)
- ans)=>3
1 , -1 and-2
2-x-1=0
x=1 or -1
x+2=0
x=-2
- 10 years agoHelpfull: Yes(3) No(5)
- 3
(X^2-x-1)^x+2=(1)^0
(X^2-x-1)=1
X+2=0
then x=-1,2,-2
So 3 solutions - 10 years agoHelpfull: Yes(2) No(4)
- x would be 2
- 10 years agoHelpfull: Yes(0) No(3)
- x+2=0;
x=-2;
x^2-x-1=0;
x=-1,2;
total three value;
- 9 years agoHelpfull: Yes(0) No(1)
- Ans is 4
let x^2-x-1=a and x+2=b
acc to this question (x^2-x-1)^(x+2)= 1 i.e., a^b = 1 when
1.) a=1
2.) b=0, or
3.) a=-1 and b can be any even integer(2,4,6,....)
1.) x^2-x-1 = 1 => x^2-x-2=0 => (x+1)(x-2) => x = (-1, 2)
2.) x+2=0 => x = -2
3.) x^2-x-1 = -1 => x^2-x = 0 => x = (0, -1)
therefore x = (-2, -1, 0, -1) - 9 years agoHelpfull: Yes(0) No(1)
- Ans is 4.
As solutions 0,-1,2,-2 - 8 years agoHelpfull: Yes(0) No(0)
- Ans is 5 (not in the options)
let x^2-x-1=a and x+2=b
acc to this question (x^2-x-1)^(x+2)= 1 i.e., a^b = 1 when
1.) a=1
2.) b=0, or
3.) a=-1 and b can be any even integer(2,4,6,....)
1.) x^2-x-1 = 1 => x^2-x-2=0 => (x+1)(x-2) => x = (-1, 2)
2.) x+2=0 => x = -2
3.) x^2-x-1 = -1 => x^2-x = 0 => x = (0, -1)
therefore x = (-2, -1, 0, 1,2) - 7 years agoHelpfull: Yes(0) No(0)
- (X^2-X-1)^(x+2)=1
(x+2)log(x^2-x-1)=0
x=-2 is one solution. Other possibility is
log(x^2-x-1)=0
x^2-x-1=0. From here we get two integral solutions x=-1,2
total 3 solution till now
now taking log on (x^2-x-1) we assumed (x^2-x-1)>0 but (x^2-x-1) may be negative
thus we solve the inequality:
(x^2-x-1) - 7 years agoHelpfull: Yes(0) No(0)
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