x^2 < x^4
Let x = -1
(-1)^2 < (-1)^4
1 < 1 , FALSE
Let x= -2
(-2)^2 < (-2)^4
4 < 16 , TRUE
Let x = -3
(-3)^2 < (-3)^4
9 < 81 , TRUE
Therefore, the inequality is FALSE for all values of x.x^3 < x^4
Let x= -1
(-1)^3 < (-1)^4
-1 < 1, TRUE
Let x= -2
(-2)^3 < (-2)^4
-8 < 4 , TRUE
Let x= -3
(-3)^3 < (-3)^4
-27 < 9, TRUE
Therefore, the inequality is TRUE for all values of x.x + (1/x) < 0
Let x= -1
(-1) + (1/(-1)) < 0
-1 - 1 < 0
-2 < 0 , TRUE
Let x= -2
(-2) + (1/(-2)) < 0
-2 - (1/2) < 0
-(5/2) < 0, TRUE
Let x= -3
(-3)+ (1/(-3)) < 0
-3 - (1/3) < 0
-(10/3), TRUE
Therefore the inequality is TRUE for all values of x.x = √ x^2
Let x= -1
(-1) = √(-1)^2
-1 = -1, TRUE
Let x = -2
(-2) = √(-2)^2
-2 = -2, TRUE
Let x= -3
(-3) = √(-3)^2
-3 = -3, TRUE
Therefore the inequality is TRUE for all values of x.
To summarize, The inequalities of #2, #3, and #4 are TRUE except #1 which is FALSE.
Your 4 th option explaination is wrong
Hi @satyamsoni, What makes it wrong? :D
As far as I know we can cancel the radical sign since the x is raised by 2. so if you are going to simplify the given it will look like this,
x = √x^2
x = (x^2)^1/2
x = x
so it means, what ever the value is your x, it will satisfy the given inequality. :D
You are correct we can cancel square with square root, but it can be only apply when x is positive integer, in the case with x is negative it cannot be applied.
In short sqrt(x^2) = |x|