Question: [Maximum mark: 5]
Find the least positive value of x for which cos \(\left ( \frac{x}{2} + \frac{\pi }{3} \right ) = \frac{1}{\sqrt{2}}.\)
▶️Answer/Explanation
Ans:
determines \(\frac{\pi }{4}\) ( or 450) as the first quadrant (reference) angle attempts to solve \(\frac{x }{2} + \frac{\pi }{3} = \frac{\pi }{4}\)
Note: Award M1 for attempting to solve \(\frac{x }{2} + \frac{\pi }{3} = \frac{\pi }{4}, \frac{7\pi }{4} (,…)\)
Question
Show that sin \(\frac{sinxtanx}{1-cosx}= 1+\frac{1}{cosx}\) x ≠ 2nπ, n ∈ R . [3]
Hence determine the range of values of k for which \(\frac{sinxtanx}{1-cosx}\)= K has no real solutions. [4]
▶️Answer/Explanation
Ans:
(a)
METHOD 1
METHOD 2
(b)
METHOD 1
consider \(1+\frac{1}{cosx}\)= k, leasing to cosx= \(\frac{1}{k-1}\)
consider graoh of y= \(\frac{1}{y-1}\)or range of solutions for y= cosx
(no solutions if y<-1or y>1)\(\Rightarrow 0<k<2\)
METHOD 2
consider graph of y= 1+ sec x
no real solutions if 0<k<2
METHOD 3
Question
Consider the equation ax2 = bx + c + 0 , where a ≠ 0 . Given that the roots of this equation
are x = sinθ and x = cosθ , show that b2 = a2 + 2ac . [Maximum mark: 4]
▶️Answer/Explanation
Ans:
METHOD 1
METHOD 2
Question
(a) Show that sin 2x + cos 2x – 1 = 2 sin x (cos x – sin x) . [2]
(b) Hence or otherwise, solve sin 2x + cos 2x – 1 + cos x – sin x = 0 for 0 < x < 2p . [6]
▶️Answer/Explanation
Ans:
(a)
METHOD 1 (LHS to RHS)
attempt to use double angle formula for sin 2x or cos 2x
LHS = 2 sin x cos x+ cos 2x – 1 OR
sin 2x+ 1 – 2 sin x OR
= 2 sin x cos x – 2 sin2 x
= sin 2x + cos 2 x- 1 = 2 sin (cos x- sinx ) =RHS
METHOD 2
(RHS to LHS)
RHS = 2sin x cos x – 2sin2x
attempt to use double angle formula for sin 2 x or cos 2x
= sin 2x + 1 – 2 sin 2 x-1
= sin 2x + cos 2x – 1 = LHS
(b)
attempt to factorise
(cosx- sinx)(2 sin x + 1)=0
recognition of cos x = sin x ⇒ \(\frac{sinx}{cosx}\) = tan x = 1 Or sin x =- \(\frac{1}{2}\)
one correct reference angle seen anywhere, accept degrees
\(\frac{\pi }{4}\) Or \(\frac{\pi }{6}\)
Note: This (M1)(A1) is independent of the previous M1A1.
x = \(\frac{7\pi }{6}\frac{11\pi }{6}\frac{\pi }{4}\frac{5\pi }{4}\)
Note: Award A1 for any two correct (radian) answers. Award A1A0 if additional values given with the four correct (radian) answers. Award A1A0 for four correct answers given in degrees.
Question
Consider the function f defined by f (x) = 6 + 6 cos x , for 0 ≤ x ≤ 4π .
The following diagram shows the graph of y = f (x) .
The graph of f touches the x-axis at points A and B, as shown. The shaded region is enclosed by the graph of y = f (x) and the x-axis, between the points A and B.
(a) Find the x-coordinates of A and B. [3]
(b) Show that the area of the shaded region is 12π . [5]
The right cone in the following diagram has a total surface area of 12π , equal to the shaded area in the previous diagram.
The cone has a base radius of 2, height h , and slant height l .
(c) Find the value of l . [3]
(d) Hence, find the volume of the cone. [4]
▶️Answer/Explanation
Ans:
(a) When $f\left(x\right)=0$,
$$\begin{eqnarray}
6\left(1+\cos x\right) = 0 \nonumber \\
\cos x = -1.
\end{eqnarray}$$
Thus, $x=\pi$ or $x=3\pi$.<br>
(b)
$$\begin{eqnarray}
\text{req’d area} &=& \int_{\pi}^{3\pi} f\left(x\right) \text{d}x \nonumber \\
&=& 6\int_{\pi}^{3\pi} 1+\cos x \text{d}x \nonumber \\
&=& 6\left[ 1+\sin x \right]_{\pi}^{3\pi} \nonumber \\
&=& 6\left(3\pi-\pi\right) \nonumber \\
&=& 12\pi.
\end{eqnarray}$$
(c)
$$\begin{eqnarray}
\text{total surface area} &=& \pi\left(2\right)^2+\pi\left(2\right)l \nonumber \\
12\pi &=& 4\pi+2\pi l \nonumber \\
2\pi l &=& 8\pi \nonumber \\
l &=& 4.
\end{eqnarray}$$
(d) By Pythagoras’ Theorem, we have $h=\sqrt{4^2-2^2}$, i.e., $h=\sqrt{12}$. Thus,
$$\begin{eqnarray}
\text{vol. of cone} &=& \frac{1}{3}\pi\left(2\right)^2\left(h\right) \nonumber \\
&=& \frac{4\pi\sqrt{12}}{3} \nonumber \\
&=& \frac{8\pi\sqrt{3}}{3}.
\end{eqnarray}$$
Question
Solve the equation 2 cos2 x + 5 sin x = 4 , 0 ≤ x ≤ 2π . [Maximum mark: 7]
▶️Answer/Explanation
Ans:
2cos2x+5sinx=4
2(1−sin2x)+5sinx=4
2sin2x−5sinx+2=0
(2sinx−1)(sinx−2)=0.
Thus, sinx=1/2 or sinx=2 (no solution)
For sinx=1/2, we have α=arcsin1/2=π/6. Since 0≤x≤2π, x=π/6 or x=5π/6.