gps5441 wrote:

x=1 and x=9 both are valid

|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)

2x+2=2*1+2=4.

Hence valid

ArunpriyanJ wrote:

Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid value. In this question why hasn't something similar been done

Both x=1 and x=9 are valid for (1). Please elaborate what you mean?

Bunel, I too have the same doubt in my mind. I will try to explain it.

By taking condition,

X>0, i got the value of X as 9

And by taking the condition X<0, I have got the solution as X=1, Which is not a valid solution with the given condition.

So i took X=9 only and got the answer as A.

Pls explain how X=1 with the condition X<0 is valid.

Thanks in advance

Statement 1

X = 1 or 9

(Both the values satisfy the equation)If x = 1, then \(\sqrt{x}\) = 1 => Not a prime number.

If x = 9, then \(\sqrt{x}\) = 3 => Prime Number

2 different answers, so this is insufficient.

Statement 2

x = 0 or 9

(Both the values satisfy the equation)If x = 0, then \(\sqrt{x}\) = 0 => Not a prime number.

If x = 9, then \(\sqrt{x}\) = 3 => Prime Number

2 different answers, so this is insufficient.

Statement 1 + Statement 2

x = 9

(The only common value)=> \(\sqrt{x}\) = 3 => Prime Number

Unique Answer and hence, Sufficient.

C is the answer.

Hope this helps!

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