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The solution of inequality (x+1)(x2+1)1/2>x2-1 is (a) {-1} U [2,infinite) (b) [-1,infinite) (c){-4}u[2,3] (d)(-5,infinite)
The solution of inequality (x+1)(x2+1)1/2>x2-1 is
(a) {-1} U [2,infinite)
(b) [-1,infinite)
(c){-4}u[2,3]
(d)(-5,infinite)
Hi, I am assuming your question is (x+1) sqrt(x2+1) > x2-1. This can be re-written as (x+1) sqrt(x2+1) > (x+1)(x-1) Case 1 : (x+1)>0, or x > -1 In this case, we can cancel (x+1) in both sides. We get sqrt(x2+1)>(x-1) This is valid for all x>(-1). So Case 1 gives a solution x> -1 Case 2: (x+1)<0 or x < -1 We can still cancel (x+1) from both sides, but since (x+1) is negative, we have to change the > sign to <. SO now we have sqrt(x2+1) < (x-1). This is not possible because x-1< -2 (because x< -1 in this case). And sqrt() will always be positive, so it can never be less than -1. Hence Case 2 gives no results. Therefore the answer is x > -1 or x = (-1,inf). This is not mentioned in any of the options so all options are INCORRECT. You can verify this by putting x=-1 and x=-3. Putting x=-1, we get LHS=0 and RHS=0, SoLHS>RHS is not true. So x=-1 is not a solution. So options A and B are wrong. Similarly, we find that x=-3 soes NOT satisfy the equation, so options C and D are also wrong. Correct range is (-1,inf)
Hi,
I am assuming your question is (x+1) sqrt(x2+1) > x2-1.
This can be re-written as (x+1) sqrt(x2+1) > (x+1)(x-1)
Case 1 : (x+1)>0, or x > -1
In this case, we can cancel (x+1) in both sides.
We get sqrt(x2+1)>(x-1)
This is valid for all x>(-1). So Case 1 gives a solution x> -1
Case 2: (x+1)<0 or x < -1
We can still cancel (x+1) from both sides, but since (x+1) is negative, we have to change the > sign to <.
SO now we have sqrt(x2+1) < (x-1).
This is not possible because x-1< -2 (because x< -1 in this case).
And sqrt() will always be positive, so it can never be less than -1.
Hence Case 2 gives no results.
Therefore the answer is x > -1 or x = (-1,inf). This is not mentioned in any of the options so all options are INCORRECT.
You can verify this by putting x=-1 and x=-3. Putting x=-1, we get LHS=0 and RHS=0, SoLHS>RHS is not true. So x=-1 is not a solution. So options A and B are wrong. Similarly, we find that x=-3 soes NOT satisfy the equation, so options C and D are also wrong.
Correct range is (-1,inf)
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