Question 9
Problem:
(a) Using formulae, show that:
- (i) $ \cos^2 A = \frac{\cos 2A + 1}{2} $
- (ii) $ \sin^2 A = \frac{1 - \cos 2A}{2} $
(b) Show that
$ (2 \sin x - \cos x)(\sin x - 3 \cos x) = \frac{1}{2} (\cos 2x - 7 \sin 2x + 5) $
(c) Solve for $0^\circ \lt x \lt 180^\circ$ the equation $ \frac{dy}{dx} = 0 $.
Solution:
(a)(i)
$ \cos(A+B) = \cos A \cos B - \sin A \sin B $
Let $A = B$:
$ \cos 2A = \cos^2 A - \sin^2 A $
$ \cos 2A = \cos^2 A - (1 - \cos^2 A) $
$ \cos 2A = 2\cos^2 A - 1 $
$ \cos^2 A = \frac{\cos 2A + 1}{2} $
(a)(ii)
$ \cos 2A = 1 - 2\sin^2 A $
$ 2\sin^2 A = 1 - \cos 2A $
$ \sin^2 A = \frac{1 - \cos 2A}{2} $
(b)
$ \text{LHS} = (2 \sin x - \cos x)(\sin x - 3 \cos x) $
$ = 2\sin^2 x - 6\sin x \cos x - \sin x \cos x + 3\cos^2 x $
$ = 2\sin^2 x - 7\sin x \cos x + 3\cos^2 x $
$ \text{RHS} = \frac{1}{2}(\cos 2x - 7\sin 2x + 5) $
$ = \frac{1}{2}(\cos^2 x - \sin^2 x) - \frac{7}{2}(2\sin x \cos x) + \frac{5}{2} $
$ = \frac{1}{2}\cos^2 x - \frac{1}{2}\sin^2 x - 7\sin x \cos x + \frac{5}{2}(\sin^2 x + \cos^2 x) $
$ = 2\sin^2 x - 7\sin x \cos x + 3\cos^2 x $
Thus LHS = RHS.
(c)
$ y = \frac{1}{2}(\cos 2x - 7\sin 2x + 5) $
$ \frac{dy}{dx} = \frac{1}{2}(-2\sin 2x - 14\cos 2x) $
$ \frac{dy}{dx} = -\sin 2x - 7\cos 2x $
Set $ \frac{dy}{dx} = 0 $:
$ -\sin 2x - 7\cos 2x = 0 $
$ \sin 2x = -7\cos 2x $
$ \tan 2x = -7 $
$ 2x \approx -81.87^\circ + 180^\circ n $
For $0^\circ \lt x \lt 180^\circ$:
$ 2x \approx 98.13^\circ,\; 278.13^\circ $
$ x \approx 49.06^\circ,\; 139.06^\circ $
Final Answer:
$ x \approx 49^\circ,\; 139^\circ $
Question 10
Points $O$, $A$, and $B$ are fixed such that:
$$ \overrightarrow{OA} = (b+1) \mathbf{i} + b \mathbf{j} $$ $$ \overrightarrow{AB} = 3 \mathbf{i} $$ $$ \text{The unit vector parallel to } \overrightarrow{OB} \text{ is } \frac{\sqrt{17}}{34} \left[ (3a + 2) \mathbf{i} + b \mathbf{j} \right]. $$ We need to find the vector $ \overrightarrow{OB} $ and the relationship between $a$ and $b$.
Step 1: Expression for $ \overrightarrow{OB} $
First, note that: $$ \overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB}. $$ Substitute: $$ \overrightarrow{OB} = \left( (b+1) \mathbf{i} + b \mathbf{j} \right) + 3 \mathbf{i}. $$ Simplify: $$ \overrightarrow{OB} = (b+4) \mathbf{i} + b \mathbf{j}. $$ Thus: $$ \overrightarrow{OB} = (b+4) \mathbf{i} + b \mathbf{j}. $$Step 2: Unit Vector Parallel to $ \overrightarrow{OB} $
$$ \hat{u} = \frac{\overrightarrow{OB}}{|\overrightarrow{OB}|} $$ Magnitude: $$ |\overrightarrow{OB}| = \sqrt{(b+4)^2 + b^2} $$ $$ |\overrightarrow{OB}| = \sqrt{2b^2 + 8b + 16} $$ Unit vector: $$ \hat{u} = \frac{(b+4) \mathbf{i} + b \mathbf{j}}{\sqrt{2b^2 + 8b + 16}} $$Step 3: Given Unit Vector
$$ \frac{\sqrt{17}}{34} \left[ (3a + 2) \mathbf{i} + b \mathbf{j} \right] $$ Equate: $$ \frac{(b+4) \mathbf{i} + b \mathbf{j}}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34} \left[ (3a + 2) \mathbf{i} + b \mathbf{j} \right] $$Step 4: $ \mathbf{i} $ components
$$ \frac{b+4}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34}(3a + 2) $$Step 5: $ \mathbf{j} $ components
$$ \frac{b}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34}b $$ Since $b \neq 0$: $$ \frac{1}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34} $$ Square: $$ \frac{1}{2b^2 + 8b + 16} = \frac{17}{1156} $$ $$ 1156 = 17(2b^2 + 8b + 16) $$ $$ 34b^2 + 136b - 884 = 0 $$ $$ b^2 + 4b - 26 = 0 $$ $$ b = \frac{-4 \pm \sqrt{120}}{2} $$ $$ b = -2 \pm \sqrt{30} $$ Thus: $$ b_1 = -2 + \sqrt{30}, \quad b_2 = -2 - \sqrt{30} $$ Since $b>0$, $ b = -2 + \sqrt{30} $Step 6: Find $a$
Substitute into: $$ \frac{b+4}{\sqrt{2b^2 + 8b + 16}} = \frac{\sqrt{17}}{34}(3a + 2) $$ For $b = -2 + \sqrt{30}$: $$ b+4 = 2 + \sqrt{30} $$ $$ b^2 = 34 - 4\sqrt{30} $$ $$ 2b^2 + 8b + 16 = 68 $$ $$ \sqrt{68} = 2\sqrt{17} $$ So: $$ \frac{2 + \sqrt{30}}{2\sqrt{17}} = \frac{\sqrt{17}}{34}(3a + 2) $$ Multiply: $$ 34 + 17\sqrt{30} = 17(3a + 2) $$ $$ 17\sqrt{30} = 51a $$ $$ a = \frac{\sqrt{30}}{3} $$ Final Answer: $$ a = \frac{\sqrt{30}}{3} $$Question 11
Solution
(a) Writing $ f(x) = 10 + 6x - x^2 $ in the form $ A(x + B)^2 + C $
We start with: $$ f(x) = -x^2 + 6x + 10 $$ Factor out $-1$: $$ f(x) = -(x^2 - 6x) + 10 $$ Complete the square: $$ x^2 - 6x = (x - 3)^2 - 9 $$ Substitute: $$ f(x) = -((x - 3)^2 - 9) + 10 $$ Simplify: $$ f(x) = -(x - 3)^2 + 19 $$ Thus, $ A = -1 $, $ B = -3 $, and $ C = 19 $.(b) (i) Value of $ x $ for maximum $ f(x) $
Since: $$ f(x) = -(x - 3)^2 + 19 $$ Maximum occurs at: $$ x = 3 $$(b) (ii) Greatest value of $ f(x) $
$$ f(3) = 19 $$(c) Intersection points of $ f(x) $ and $ S(x) $
$$ f(x) = 10 + 6x - x^2, \quad S(x) = x^2 - x + 13 $$ Set equal: $$ 10 + 6x - x^2 = x^2 - x + 13 $$ Simplify: $$ -2x^2 + 7x - 3 = 0 $$ $$ 2x^2 - 7x + 3 = 0 $$ Factorize: $$ (2x - 1)(x - 3) = 0 $$ Solutions: $$ x = \frac{1}{2}, \quad x = 3 $$(d) Area between $ f(x) $ and $ S(x) $
$$ \text{Area} = \int_{\frac{1}{2}}^3 [f(x) - S(x)] \, dx $$ $$ f(x) - S(x) = -2x^2 + 7x - 3 $$ So: $$ \text{Area} = \int_{\frac{1}{2}}^3 (-2x^2 + 7x - 3) \, dx $$ Integrate: $$ \left[-\frac{2x^3}{3} + \frac{7x^2}{2} - 3x \right]_{\frac{1}{2}}^3 $$ At $x = 3$: $$ -18 + \frac{63}{2} - 9 = 4.5 $$ At $x = \frac{1}{2}$: $$ -\frac{1}{12} + \frac{7}{8} - \frac{3}{2} = -\frac{17}{24} $$ Final area:Exact area = $\boxed{\frac{125}{24}}$
Post a Comment