A Level Mathematics 9709/11 May/June 2025 Past Paper with Answers

Question 1

Solve the equation $8\sin\theta = 2 + \dfrac{3}{\sin\theta}$ for $-180^\circ < \theta < 180^\circ$. How many solutions does this equation have?

1. $4$
2. $2$
3. $3$
4. $1$

Question 2

A curve has $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3(2x-3)^3 - 8x^{1/2}$. The curve passes through the point $A(4, 7)$. What is the gradient of the normal to the curve at $A$?

1. $-\dfrac{1}{359}$
2. $359$
3. $-359$
4. $\dfrac{1}{359}$

Question 3

A curve has $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4(2x-5)^3 - 9x^{1/2}$ and passes through the point $(4, -\frac{11}{2})$. What is the constant of integration $c$ in the equation of the curve?

1. $2$
2. $-2$
3. $\dfrac{7}{2}$
4. $0$

Question 4 · Answer A

Step 1: Let the first term be $a$ and common ratio $r$. Then $ar^2 = 12$ so $a = 12/r^2$. Step 2: Sum of first 3 terms: $a + ar + ar^2 = 28$, so $12/r^2 + 12/r + 12 = 28$. Step 3: Multiply by $r^2$: $12 + 12r + 12r^2 = 28r^2$, giving $16r^2 - 12r - 12 = 0$, i.e. $4r^2 - 3r - 3 = 0$. Step 4: Using the quadratic formula: $r = \dfrac{3 \pm \sqrt{9 + 48}}{8} = \dfrac{3 \pm \sqrt{57}}{8}$. Since $r < 0$: $r = \dfrac{3 - \sqrt{57}}{8} \approx -0.5686$. Step 5: $a = 12/(-0.5686)^2 \approx 37.11$. Step 6: Tenth term $= 37.11 \times (-0.5686)^9 \approx -0.901$ (3 s.f.).

Question 5 · Answer A

Step 1: The sum to infinity formula is $S_\infty = \dfrac{a}{1 - r}$, valid when $|r| < 1$. Step 2: Here $a = 48$, $r = -\dfrac{2}{3}$, so $|r| = \dfrac{2}{3} < 1$. Step 3: $S_\infty = \dfrac{48}{1 - (-2/3)} = \dfrac{48}{5/3} = \dfrac{48 \times 3}{5} = \dfrac{144}{5}$.

Question 6 · Answer A

Step 1: Determine which function is greater on $[1, 16]$. At $x = 4$: line gives $-12$, curve gives $5(8) - 80 = -40$. So line is above curve. Step 2: Integrand: $(x - 16) - (5x^{3/2} - 20x) = -5x^{3/2} + 21x - 16$. Step 3: Integrate: $\int_{1}^{16}(-5x^{3/2} + 21x - 16)\,\mathrm{d}x = \left[-2x^{5/2} + \dfrac{21}{2}x^2 - 16x\right]_{1}^{16}$. Step 4: Evaluate at $x = 16$: $-2(1024) + \dfrac{21}{2}(256) - 256 = -2048 + 2688 - 256 = 384$. Step 5: Evaluate at $x = 1$: $-2 + \dfrac{21}{2} - 16 = -2 + 10.5 - 16 = -7.5$. Step 6: Area $= 384 - (-7.5) = 391.5 = \dfrac{783}{2}$.

Question 7 · Answer A

Step 1: Use the binomial expansion $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}a^{n-r}b^r$. Step 2: First term: $\binom{4}{0} \cdot 3^{4} \cdot (-qx)^{0} = 81$. Step 3: Second term: $\binom{4}{1} \cdot 3^{3} \cdot (-qx)^{1} = 4(27)(-qx) = -108qx$. Step 4: Third term: $\binom{4}{2} \cdot 3^{2} \cdot (-qx)^{2} = 6(9)(q^{2}x^{2}) = 54q^{2}x^{2}$. Step 5: First 3 terms: $81 - 108qx + 54q^{2}x^{2}$.

Question 8 · Answer A

Step 1: First term: $\binom{5}{0}(1)^5 = 1$. Step 2: Second term: $\binom{5}{1}(1)^4\left(-\dfrac{x}{3}\right) = 5\left(-\dfrac{x}{3}\right) = -\dfrac{5x}{3}$. Step 3: Third term: $\binom{5}{2}(1)^3\left(-\dfrac{x}{3}\right)^2 = 10 \times \dfrac{x^2}{9} = \dfrac{10x^2}{9}$. Step 4: First 3 terms: $1 - \dfrac{5x}{3} + \dfrac{10x^2}{9}$.

Question 9 · Answer A

Step 1: The $x^2$ coefficient in the product comes from three pairs: (constant from first) $\times$ ($x^2$ term from second) + ($x$ term from first) $\times$ ($x$ term from second) + ($x^2$ term from first) $\times$ (constant from second). Step 2: $= 32 \times \dfrac{3}{2} + (-80p)(-2) + 80p^2 \times 1 = 48 + 160p + 80p^2$. Step 3: Set equal to 93: $80p^2 + 160p + 48 = 93$, so $80p^2 + 160p - 45 = 0$. Step 4: Divide by 5: $16p^2 + 32p - 9 = 0$, giving $(4p - 1)(4p + 9) = 0$. Step 5: $p = \dfrac{1}{4}$ or $p = -\dfrac{9}{4}$.

Question 10 · Answer A

Step 1: Substitute $y = 11x + 3$ into $2x^2 - 2xy + 2 = 0$: $2x^2 - 2x(11x + 3) + 2 = 0$. Step 2: Expand: $2x^2 - 22x^2 - 6x + 2 = 0$, giving $-20x^2 - 6x + 2 = 0$. Step 3: Multiply by $-1$: $20x^2 + 6x - 2 = 0$, divide by 2: $10x^2 + 3x - 1 = 0$. Step 4: Factorise: $(5x - 1)(2x + 1) = 0$, so $x = \dfrac{1}{5}$ or $x = -\dfrac{1}{2}$.

Question 11 · Answer A

Step 1: Substituting $y = 4x + 3$ into $2x^2 - kxy + 2 = 0$ gives $(2 - 4k)x^2 - 3kx + 2 = 0$. Step 2: For no intersection, the discriminant must be negative: $(-3k)^2 - 4(2 - 4k)(2) < 0$. Step 3: Expand: $9k^2 - 8(2 - 4k) < 0$, so $9k^2 - 16 + 32k < 0$, giving $9k^2 + 32k - 16 < 0$. Step 4: Factorise: $(k + 4)(9k - 4) < 0$. Step 5: The quadratic is negative between its roots: $-4 < k < \dfrac{4}{9}$.

Question 12 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6x - \dfrac{8}{x^3}$. Step 2: At $x = 2$: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12 - \dfrac{8}{8} = 12 - 1 = 11$. Step 3: Use the chain rule: $\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t}$. Step 4: $-7 = 11 \times \dfrac{\mathrm{d}x}{\mathrm{d}t}$, so $\dfrac{\mathrm{d}x}{\mathrm{d}t} = -\dfrac{7}{11}$.

Question 13 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 8x - \dfrac{18}{x^3}$. Step 2: Set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: $8x = \dfrac{18}{x^3}$, so $8x^4 = 18$, giving $x^4 = \dfrac{9}{4}$. Step 3: $x^2 = \dfrac{3}{2}$ (taking the positive root of $x^4$), so $x = \pm\sqrt{\dfrac{3}{2}}$. Step 4: Substitute into $y$: $y = 4\left(\dfrac{3}{2}\right) + 9\left(\dfrac{2}{3}\right) - 8 = 6 + 6 - 8 = 4$. Step 5: Both stationary points have $y = 4$. The second derivative $8 + 54/x^4 > 0$ for all $x$, so both are minima.

Question 14 · Answer A

Step 1: Rewrite the circle in standard form: $(x-3)^2 + (y+5)^2 = 9 + 25 + 27 = 61$. Centre $(3, -5)$, radius $\sqrt{61}$. Step 2: Substitute $x = -2$: $4 + y^2 + 12 + 10y - 27 = 0$, so $y^2 + 10y - 11 = 0$, giving $(y - 1)(y + 11) = 0$. So $P(-2, 1)$ and $Q(-2, -11)$. Step 3: Gradient of $CP = \dfrac{-5-1}{3-(-2)} = -\dfrac{6}{5}$. Tangent at $P$ is perpendicular: gradient $= \dfrac{5}{6}$. Step 4: Tangent at $P$: $y - 1 = \dfrac{5}{6}(x + 2)$. Step 5: Gradient of $CQ = \dfrac{-5-(-11)}{3-(-2)} = \dfrac{6}{5}$. Tangent at $Q$: $y + 11 = -\dfrac{5}{6}(x + 2)$. Step 6: Find intersection of tangents. From tangent at $P$: $y = \dfrac{5}{6}(x+2) + 1$. From tangent at $Q$: $y = -\dfrac{5}{6}(x+2) - 11$. Set equal: $\dfrac{5}{6}(x+2) + 1 = -\dfrac{5}{6}(x+2) - 11$, so $\dfrac{10}{6}(x+2) = -12$, giving $x + 2 = -\dfrac{36}{5}$, so $x = -\dfrac{46}{5}$. Step 7: $y = \dfrac{5}{6}(-\dfrac{36}{5}) + 1 = -6 + 1 = -5$. Tangents meet at $(-\dfrac{46}{5}, -5)$. Step 8: Base $PQ = |1 - (-11)| = 12$. Height = distance from $(-\dfrac{46}{5}, -5)$ to line $x = -2$: $|-\dfrac{46}{5} + 2| = \dfrac{36}{5}$. Step 9: Area $= \dfrac{1}{2} \times 12 \times \dfrac{36}{5} = \dfrac{216}{5}$.

Question 15 · Answer A

Step 1: Sector area $= \dfrac{1}{2}r^2\alpha = 8\alpha$, so $r^2 = 16$, giving $r = 4$. Step 2: Triangle area $= \dfrac{1}{2}r^2\sin\alpha = \dfrac{1}{2}(16)\sin\alpha = 8\sin\alpha = 4$, so $\sin\alpha = \dfrac{1}{2}$, giving $\alpha = \dfrac{\pi}{6}$. Step 3: Sector area $= \dfrac{1}{2}(16)\left(\dfrac{\pi}{6}\right) = \dfrac{8\pi}{6} = \dfrac{4\pi}{3}$. Step 4: Segment area $=$ sector area $-$ triangle area $= \dfrac{4\pi}{3} - 4$.

Question 16 · Answer A

Step 1: By the cosine rule: $BC^2 = r^2 + r^2 - 2r^2\cos\alpha$, so $\dfrac{r^2}{2} = 2r^2 - 2r^2\cos\alpha$. Step 2: Divide by $r^2$: $\dfrac{1}{2} = 2 - 2\cos\alpha$, so $2\cos\alpha = \dfrac{3}{2}$, giving $\cos\alpha = \dfrac{3}{4}$, so $\alpha = \cos^{-1}(0.75) \approx 0.7227$ rad. Step 3: Triangle area: $\dfrac{1}{2}r^2\sin\alpha = 4$, so $r^2 = \dfrac{8}{\sin(0.7227)} \approx \dfrac{8}{0.6614} \approx 12.10$. Step 4: Segment area $= \dfrac{1}{2}r^2\alpha - 4 = \dfrac{1}{2}(12.10)(0.7227) - 4 \approx 4.371 - 4 = 0.371$.

Question 17 · Answer A

Step 1: Starting from $f(x) = \sqrt{x}$, apply a stretch of factor 3 parallel to the $y$-axis: $y = 3\sqrt{x}$. Step 2: Then apply a translation $\binom{-2}{-5}$: replace $x$ with $x + 2$ and subtract 5: $y = 3\sqrt{x+2} - 5 = g(x)$. Step 3: This matches the required function exactly.

Question 18 · Answer A

Step 1: The domain of $g^{-1}$ equals the range of $g$. Step 2: Since $g(x) = 3\sqrt{x+2} - 5$ with $x \ge -2$: when $x = -2$, $g(-2) = 3(0) - 5 = -5$. As $x \to \infty$, $g(x) \to \infty$. Step 3: So the range of $g$ is $g(x) \ge -5$, meaning the domain of $g^{-1}$ is $x \ge -5$.

Question 19 · Answer A

Step 1: Let $y = 3\sqrt{x+2} - 5$. Add 5: $y + 5 = 3\sqrt{x+2}$. Step 2: Divide by 3: $\dfrac{y+5}{3} = \sqrt{x+2}$. Step 3: Square both sides: $\left(\dfrac{y+5}{3}\right)^2 = x + 2$. Step 4: Subtract 2: $x = \left(\dfrac{y+5}{3}\right)^2 - 2$. Step 5: Swap $x$ and $y$: $g^{-1}(x) = \left(\dfrac{x+5}{3}\right)^2 - 2$.

Question 20 · Answer A

Step 1: The range of $g^{-1}$ equals the domain of $g$. Step 2: The domain of $g$ is $x \ge -2$. Step 3: Therefore the range of $g^{-1}$ is $g^{-1}(x) \ge -2$.

Question 21 · Answer A

Step 1: First find $h(4) = 4 - 2 = 2$. Step 2: Then find $g^{-1}(2) = \left(\dfrac{2+5}{3}\right)^2 - 2 = \left(\dfrac{7}{3}\right)^2 - 2 = \dfrac{49}{9} - 2 = \dfrac{49 - 18}{9} = \dfrac{31}{9}$.

Question 22 · Answer A

Step 1: For $hg^{-1}(x)$ to be defined, the output of $g^{-1}$ must lie within the domain of $h$. Step 2: The range of $g^{-1}$ is $y \ge -2$, which includes values in $[-2, 0)$. Step 3: The domain of $h$ is $x \ge 0$, so values from $g^{-1}$ in $[-2, 0)$ are not valid inputs for $h$. Step 4: Therefore $hg^{-1}$ cannot be formed.