Latest TCS Aptitude Question SOLUTION: For how many integer x satisfy the equation :(x^2-x-1)^x+2 = 1?

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For how many integer x satisfy the equation :(x^2-x-1)^x+2 = 1?

Asked In TCS purva (9 years ago)

Unsolved Read Solution (25)

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sumit kumar  (9 years ago)

Ans is 4
when x=0,-1,2,-2
when x=-2 then
power x+2= -2+2= 0

Is this solution Helpfull? Yes (16) | No (12)  

RAKESH  (9 years ago)

only one integral soln , x=1

(x^2-x-1)^x+2 = 1
=> (x^2-x-1)^x = -1

=> x = 1 only

Is this solution Helpfull? Yes (11) | No (13)  

Saumya  (9 years ago)

We have (x^2-x-1)^x+2 = 1
so putting both side log we get
(x+2)log(x^2-x-1)=log(1)
=> (x+2)log(x^2-x-1)=0
=>either (x+2)=0 or log(x^2-x-1)=0 or both are 0
solving both we get
x+2=0 =>x=-2
and (x^2-x-1)=1 [since log 1=0]
=> x= -1,2
also x=0 satisfies the equation

hence we have total 4 values

Is this solution Helpfull? Yes (8) | No   

Aparna S M  (9 years ago)

ans is 3
when x=0,-1,2

Is this solution Helpfull? Yes (7) | No (10)  

Deepak  (9 years ago)

With due respect i would like to request all people answering wrong to check their ans on paper and then post it to site !!
it really generates a wasteful data and also waste time

Is this solution Helpfull? Yes (7) | No   

Pooja P  (9 years ago)

(x^2-x-1)^x=-1
only 1 will satisfy this equation
(1-1-1)^1=-1
ans=1

Is this solution Helpfull? Yes (4) | No (7)  

pawan   (9 years ago)

the ques is nt clr bcz its is nt clr that weather x+2 is in power or x is in power +2 seprate so both posssibilities gives diff reslts

Is this solution Helpfull? Yes (3) | No   

Abhinav Raj  (9 years ago)

if the equation is a^b=1 then it will be true only if either a=1 or b=0 as anything raised to the power 0 equals to 1. hence the equation x^2-x-1 will be equal to 1 only for the values 0, -1 and 2. and x+2 will be equal to 0 for x=-2 hence 4 will be the true answer.... sumit kumar is right about it...

Is this solution Helpfull? Yes (3) | No   

Narasapu Hari prasad  (9 years ago)

ans is only 1.
x=1 then (1^2-1-1)^1+2= (-1)^1+2=-1+2=1
if you take x=0,-1,2,-2 the condition not satisfied
when x=0=(0^2-0-1)^0+2=(-1)^0+2=1+2=3....... ........(-1)^0=+1 because anything power 0 is 1
when x=-1=(-1^2-(-1)-1)^-1+2=(1+1-1)^-1+2=1^-1+2=1+2=3.............1^-1=1/(1^1)=1
when x=2=(2^2-2-1)^2+2=(4-2-1)^2+2=1+2=3
when x=-2=(-2^2-(-2)-1)^-2+2=(4+2-1)^-2+2=5^-2+2=1/(5^2)+2=0.04+2=2.04
so only 1 is corect when x=1

Is this solution Helpfull? Yes (2) | No (8)  

Narasapu Hari prasad  (9 years ago)

Ans is 3.
When x=0,1,-1

Is this solution Helpfull? Yes (1) | No (8)  

Manish Parlani  (9 years ago)

when x= 0,-1,2,-2
ans is 4

Is this solution Helpfull? Yes (1) | No (1)  

Gurram Mahendra ranganath  (9 years ago)

(x^2-x-1)^(x+2)=1

The equation satisfies when x=0,-1,-2,2
therefore,number of integers are 4

Is this solution Helpfull? Yes (1) | No   

NIVEDA SIVAPRAKASAM  (9 years ago)

May be 2 alone.
4-2-1^4=1

Is this solution Helpfull? Yes | No (7)  

Palak Gupta  (9 years ago)

ans will be 4

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ujjwal panchakshri  (9 years ago)

ans. is 4.
for x=-2,-1,0,2

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ANIMESH GORAI  (9 years ago)

but if we take x=2,then that value will b 1/64,then how the answer is 4??

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abhinav singh  (9 years ago)

ans is 4.
x+2=0
x=-2,
and x^2-x-1=1,
then x=2,-1
and x=0

Is this solution Helpfull? Yes | No (1)  

Puja   (9 years ago)

ans=3
taking log on both side ans comes -2, 2, -1

Is this solution Helpfull? Yes | No (3)  

tirupathi reddy  (9 years ago)

If the base is one the equation will be correct.
Put x=2
Then 1^4=1

Is this solution Helpfull? Yes | No (2)  

Ajit Singh  (9 years ago)

(x^2-x-1)^x+2=1
(x^2-x-1)=1^(1/x+2)
x^2-x-1=1
x^2-x-2=0
By solving it
x=-1,2
-1 satisfy the given equation so x=-1

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ANIMESH GORAI  (9 years ago)

there is no such solution

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manish sukhadia  (9 years ago)

-1 is the answer.
when x = -1 this equation gives the result as 1.

Is this solution Helpfull? Yes | No (1)  

Sreetama Dam  (9 years ago)

2 intergers are there.
Explanation:
(x^2-x-1)^x+2 = 1
or,(x^2-x-1)^x = -1
(-1)^any odd positive number=-1
so x must not be any negative number or even number..and (x^2-x-1) should be -1.
x^2-x-1=-1
or, x^2-x=0
or,x(x-1)=0
so,x=0,x=1..
So,2 integers are there to solve the equation.
note:- (1)^negative number is not -1,it is +1.

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aman shrivastava  (9 years ago)

x=1 & x=-1 are the only solutions possible..therefore 2

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Soumyadip Hazra  (9 years ago)

How the answer can be 4 or 3?? Ans will be only 1.
If we put (x)^(-anything) it will be 1/(x) not -x. Then how can it be -1 or -2?????????
Ans is 1..

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