modulus x+2 = 4
Thus x+2 = + 4
or
X + 2 = -4
Solving we have X = 6 or X = -2
From A. we know X = +2 or - 2
We can conclude x = -2
A is Sufficient.
From B. x = +6 or -6
thus x = 6
B. Sufficient.
Hence its D
variables
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anishprabhu
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kanha81
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I am going to say absolute value (abs) instead of modulus.
q: abs(x+2) = 4, x = ?
(i) x^2 <> 4
=> (x-2)(x+2) <> 0
=> x-2 <> 0 or x+2 <> 0
=> x <> 2 or x <> -2
=> we can conclude abs(x) <> 2
now, x>0: x+2 = 4
=> x = 2, but abs(x) <> 2. hence not possible
x<0: x+2 = -4
=> x = -6. our answer. So, [A] or [D]
(ii) x^2 = 36
=> x = 6 or x = -6
now, x>0: subst. x=6 in abs(x+2)=abs(8) <> 4. hence not possible
x<0: subst. x=-6 in abs(x+2)=abs(-6+2)=abs(-4)=4=rhs. our answer
in both the cases the answer MUST be the same. Hence x=-6
hence [D]
q: abs(x+2) = 4, x = ?
(i) x^2 <> 4
=> (x-2)(x+2) <> 0
=> x-2 <> 0 or x+2 <> 0
=> x <> 2 or x <> -2
=> we can conclude abs(x) <> 2
now, x>0: x+2 = 4
=> x = 2, but abs(x) <> 2. hence not possible
x<0: x+2 = -4
=> x = -6. our answer. So, [A] or [D]
(ii) x^2 = 36
=> x = 6 or x = -6
now, x>0: subst. x=6 in abs(x+2)=abs(8) <> 4. hence not possible
x<0: subst. x=-6 in abs(x+2)=abs(-6+2)=abs(-4)=4=rhs. our answer
in both the cases the answer MUST be the same. Hence x=-6
hence [D]
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