harshvinayak wrote:
If \frac{4}{x} <\frac{1}{3} , what is the possible range of values of x?
The question is from an MGMAT guide and the solution provided multiplies the inequality by x and flips the sign to arrive at the solution x < 0 or x > 12 . However the way I naturally would solve the given equation is:
\frac{4}{x} - \frac{1}{3} < 0
\frac{(12 - x)}{(3x)} < 0
Two cases:
a) 12 - x < 0 &3x > 0 :: x > 12 & x > 0 therefore x > 12
b) 12 - x > 0 & 3x < 0 :: x < 12 & x < 0 therefore x < 12
Hence my solution: x < 12 or x > 12
Can someone kindly explain where I went wrong here?
Another way to solve this type of inequality:
From
\frac{4}{x}<\frac{1}{3} it follows that
x\neq0, therefore we can multiply both sides by
3x^2, which is positive.
We obtain
12x<x^2 or
x^2-12x>0.Now, imagine the graph of the quadratic function,
y=12x^2-x, which is an upward parabola. See the attached drawing.
This parabola intercepts the X-axis at
x=0 and
x=12, the two "arms" are above the X-axis, meaning the values of
y are positive when
x<12 or
x>0, and the values of
y are negative (graph under the X-axis) when
0<x<12. Hence the solution
x<0 or
x>12.Note: Just remember the shape of the parabola, then you can easily deduce the sign of the quadratic function
y=x^2+bx+c. If the quadratic equation
x^2+bx+c=0 has two roots
x_1<x_2 then
y<0 (negative)
between the two roots and
y>0 (positive)
outside the roots.
Or succinctly,
y>0 if
x_1<x<x_2 and
y>0 if
x<x_1 or
x>x_2.
Attachments

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PhD in Applied Mathematics
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