Calculus & Functions

Cheatsheet Content

### Curve Analysis: $y = |\sin x| + 1$ - **Function Definition:** $y = |\sin x| + 1$ for $0 \le x \le 2\pi$. - The graph consists of two main parts: - For $0 \le x \le \pi$, $\sin x \ge 0$, so $y = \sin x + 1$. - For $\pi ### Tangent Line Intersections #### Part (b): Range of gradient for a line through $P$ - **Line properties:** A straight line passes through $P(a, b)$ with a positive gradient, $m > 0$. - **Condition:** The line intersects the curve at exactly three distinct points. - **Visualizing the curve:** - $y = \sin x + 1$ for $0 \le x \le \pi$ (peak at $(\frac{\pi}{2}, 2)$) - $y = -\sin x + 1$ for $\pi \frac{3\sqrt{3}}{4\pi}$. - **Consider the local maximum at $(\frac{\pi}{2}, 2)$:** - Line passing through $P(\frac{5\pi}{3}, 1 + \frac{\sqrt{3}}{2})$ and $(\frac{\pi}{2}, 2)$. - Gradient $m_2 = \frac{2 - (1 + \frac{\sqrt{3}}{2})}{\frac{\pi}{2} - \frac{5\pi}{3}} = \frac{1 - \frac{\sqrt{3}}{2}}{-\frac{7\pi}{6}} = \frac{6(\sqrt{3} - 2)}{14\pi} = \frac{3(\sqrt{3} - 2)}{7\pi}$. This is negative. - **The other critical line:** The line through P that is tangent to the curve in the interval $(0, \pi)$. - Let the tangent point be $Q(x_0, \sin x_0 + 1)$. The gradient at $Q$ is $\cos x_0$. - The equation of the tangent at $Q$ is $y - (\sin x_0 + 1) = (\cos x_0)(x - x_0)$. - This line must pass through $P(\frac{5\pi}{3}, 1 + \frac{\sqrt{3}}{2})$. - $(1 + \frac{\sqrt{3}}{2}) - (\sin x_0 + 1) = (\cos x_0)(\frac{5\pi}{3} - x_0)$ - $\frac{\sqrt{3}}{2} - \sin x_0 = (\cos x_0)(\frac{5\pi}{3} - x_0)$. - This equation is hard to solve analytically. - **Graphical interpretation:** A positive gradient line through P. - If the gradient is 0, it intersects at P and two points on the left part of the curve. This is not 3. - As the gradient increases from 0, it will eventually hit the minimum at $(\pi, 1)$. This gives 3 points. - If the gradient continues to increase, it will eventually become tangent to the left hump at some point $Q(x_Q, y_Q)$ where $x_Q \in (0, \pi)$. This line will have 2 intersection points (P and Q). - If the gradient increases further, it will cut through the left hump twice, giving 4 points (P and two on the left). - Therefore, the range of gradients for 3 distinct points is when the line passes through $(\pi, 1)$ or is tangent to the left hump. - **Upper bound:** The line that is tangent to the curve at a point $Q(x_Q, \sin x_Q + 1)$ for some $x_Q \in (0, \pi)$, and also passes through $P$. - The gradient of this tangent is $\cos x_Q$. - This line will intersect the curve at P, Q, and one other point on the right side of Q on the $0 0$. - As $m$ increases from 0, the line $y-b=m(x-a)$ will eventually pass through $(\pi, 1)$. This gradient is $m_1 = \frac{3\sqrt{3}}{4\pi}$. - For $0 m_1$, the line intersects the first hump at two points, and also at $P$. So 3 points. No, this would be 4 points if it cuts the first hump twice. - The line that gives exactly 3 points is tangent to the curve at a point $Q$ on the first section $y = \sin x + 1$ ($0 m_1$, the line will intersect the first hump twice, giving 4 points. - If $m 0$), the line will also intersect the first hump twice. Unless it's tangent. - The range of gradient for 3 distinct points is for $m$ such that $0 0$. - The line passes through P. - The line must be tangent to the part of the curve $y = \sin x + 1$ for $0 m_1$, which contradicts the previous logic. - The line through $P$ with positive gradient. - If $m$ is very small positive, the line will intersect the curve at P, and two points on $y = \sin x + 1$ (since it goes above $y=1$). So 3 points. - As $m$ increases, it will eventually become tangent to $y = \sin x + 1$. This tangent point $Q$ and $P$ are 2 points. - So the range is $0