Moderate -0.3 This is a standard substitution problem (let u = 1/y² or u = y²) that reduces to a straightforward quadratic equation. While it requires recognizing the quartic structure and solving in steps, it's a routine C1 technique with no conceptual difficulty beyond basic algebraic manipulation, making it slightly easier than average.
Use a correct substitution or pair of substitutions to obtain a quadratic or factorise into 2 brackets each containing \(\frac{1}{y^2}\). No marks if straight to quadratic formula to get \(y=-\frac{2}{3}\), \(y=4\) unless correct substitution applied later
\(3k^2-10k-8=0\), \((3k+2)(k-4)=0\)
M1dep
Correct method to solve a quadratic
\(k=-\frac{2}{3}\) or \(k=4\)
A1
\(k=4\) from correct method. If other root stated it must be correct. No marks if quadratic found from incorrect substitution. SC If M0 spotted solutions www B1; justifies 2 solutions exactly B3
\(y^2=-\frac{3}{2}\) or \(y^2=\frac{1}{4}\)
M1
Attempt to reciprocal and square root to obtain \(y\) (either term)
\(y=\pm\frac{1}{2}\)
A1
No other roots given. Must be from \(k=4\) from correct method
Alternative method: \(3-10y^2-8y^4=0\), \(k=y^2\), \(8k^2+10k-3=0\), \((4k-1)(2k+3)=0\) M1\*, M1dep; \(k=\frac{1}{4}\) or \(k=-\frac{3}{2}\) A1 (\(k=\frac{1}{4}\) from correct method); \(y=\pm\frac{1}{2}\) M1 A1
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = \frac{1}{y^2}$ | M1* | Use a correct substitution or pair of substitutions to obtain a quadratic or factorise into 2 brackets each containing $\frac{1}{y^2}$. **No marks** if straight to quadratic formula to get $y=-\frac{2}{3}$, $y=4$ unless correct substitution applied later |
| $3k^2-10k-8=0$, $(3k+2)(k-4)=0$ | M1dep | Correct method to solve a quadratic |
| $k=-\frac{2}{3}$ or $k=4$ | A1 | $k=4$ from correct method. If other root stated it must be correct. **No marks** if quadratic found from incorrect substitution. SC If M0 spotted solutions **www B1**; justifies 2 solutions exactly **B3** |
| $y^2=-\frac{3}{2}$ or $y^2=\frac{1}{4}$ | M1 | Attempt to reciprocal and square root to obtain $y$ (either term) |
| $y=\pm\frac{1}{2}$ | A1 | No other roots given. Must be from $k=4$ from correct method |
**Alternative method:** $3-10y^2-8y^4=0$, $k=y^2$, $8k^2+10k-3=0$, $(4k-1)(2k+3)=0$ **M1\*, M1dep**; $k=\frac{1}{4}$ or $k=-\frac{3}{2}$ **A1** ($k=\frac{1}{4}$ from correct method); $y=\pm\frac{1}{2}$ **M1 A1**
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