Standard +0.3 This is a straightforward modulus inequality requiring squaring both sides (valid since both sides are non-negative) to eliminate the modulus signs, then solving a linear inequality. The presence of surds makes it slightly less routine than integer coefficients, but the method is standard and requires only 2-3 steps with no conceptual difficulty.
Attempt to square and expand brackets with 3 terms resulting from each. M1
Obtain \(x^2 - 2\sqrt{3}\,x + 3\) A1
Obtain \(x^2 + 4\sqrt{3}\,x + 12\) A1
Rearrange to make \(x\) the subject. M1
\(x > \dfrac{-\sqrt{3}}{2}\) aef. A1
Alternative 1: an approach based on a piecewise function
Consider at least two intervals:
\(-(x-\sqrt{3}) - -(x+2\sqrt{3}) < 0\)
\(-(x-\sqrt{3}) - (x+2\sqrt{3}) < 0\)
\((x-\sqrt{3}) - (x+2\sqrt{3}) > 0\) M1
Specify the intervals \((-\infty, -2\sqrt{3}),(-2\sqrt{3}, \sqrt{3}),(\sqrt{3}, \infty)\) A1
Discard the first and last intervals, may be without comment M1
Solve \(-(x-\sqrt{3}) - (x+2\sqrt{3}) < 0\) or equiv M1
\(x > \dfrac{-\sqrt{3}}{2}\) with no incorrect working. Extra intervals M1M0M1 only A1
Alternative 2: an approach based on graphs only
Answer
Marks
Guidance
\(y =
x - \sqrt{3}
\) drawn with intersections with axes shown M1A1
\(y =
x + 2\sqrt{3}
\) drawn with intersections with axes shown M1A1
\(x > \dfrac{-\sqrt{3}}{2}\) A1
Total: 5 marks
Attempt to square and expand brackets with 3 terms resulting from each. **M1**
Obtain $x^2 - 2\sqrt{3}\,x + 3$ **A1**
Obtain $x^2 + 4\sqrt{3}\,x + 12$ **A1**
Rearrange to make $x$ the subject. **M1**
$x > \dfrac{-\sqrt{3}}{2}$ aef. **A1**
**Alternative 1: an approach based on a piecewise function**
Consider at least two intervals:
$-(x-\sqrt{3}) - -(x+2\sqrt{3}) < 0$
$-(x-\sqrt{3}) - (x+2\sqrt{3}) < 0$
$(x-\sqrt{3}) - (x+2\sqrt{3}) > 0$ **M1**
Specify the intervals $(-\infty, -2\sqrt{3}),(-2\sqrt{3}, \sqrt{3}),(\sqrt{3}, \infty)$ **A1**
Discard the first and last intervals, may be without comment **M1**
Solve $-(x-\sqrt{3}) - (x+2\sqrt{3}) < 0$ or equiv **M1**
$x > \dfrac{-\sqrt{3}}{2}$ with no incorrect working. Extra intervals M1M0M1 only **A1**
**Alternative 2: an approach based on graphs only**
$y = |x - \sqrt{3}|$ drawn with intersections with axes shown **M1A1**
$y = |x + 2\sqrt{3}|$ drawn with intersections with axes shown **M1A1**
$x > \dfrac{-\sqrt{3}}{2}$ **A1**
**Total: 5 marks**