2020 H2 Maths Paper 1, Question 9

Let $l_{1}$ be the line through the origin with gradient $2$, making an angle $\tan^{- 1}(2)$ with the positive $x$-axis, and $l_{2}$ the line through the origin with gradient $- \frac{1}{2}$, making an angle $\tan^{- 1}\left( - \frac{1}{2} \right)$ with the positive $x$-axis. We check the product of the two gradients.

Since the product of the gradients is $- 1$, $l_{1}$ and $l_{2}$ are perpendicular, so the angle between them is $\frac{1}{2}\pi$. This angle is exactly the difference between the two inclinations, measured from $l_{2}$ up to $l_{1}$:

Setting $C_{1}$ and $C_{2}$ equal,

Cross-multiplying,

Since $k > 0$, this is a genuine quadratic in $x$, so real intersection requires its discriminant to be non-negative.

Factorising the quadratic in $k$,

so

Combined with the given condition $k > 0$,

With $k = 2$: $C_{1}:y = \frac{1}{x^{2} + 1}$ has $y$-intercept $(0,1)$ (its maximum, since $C_{1}$ decreases as $|x|$ increases), no $x$-intercept, and horizontal asymptote $y = 0$.

$C_{2}:y = \frac{2}{3x + 4}$ has $y$-intercept $\left( 0,\frac{1}{2} \right)$, no $x$-intercept, a vertical asymptote $x = - \frac{4}{3}$, and horizontal asymptote $y = 0$.

With $k = 2$, intersections satisfy $2x^{2} - 3x - 2 = 0$.

From the sketch in part (iii), $C_{1}$ lies above $C_{2}$ on $\left\lbrack - \frac{1}{2},2 \right\rbrack$, so

Using $\int\frac{1}{x^{2} + 1}\, dx = \tan^{- 1}x$ and $\int\frac{2}{3x + 4}\, dx = \frac{2}{3}\ln|3x + 4|$,

Using the result of part (i), $\tan^{- 1}(2) - \tan^{- 1}\left( - \frac{1}{2} \right) = \frac{\pi}{2}$, and $\ln 10 - \ln\frac{5}{2} = \ln 4 = 2\ln 2$,