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(a) The vectors \( \begin{pmatrix} \frac{a^2}{2} \\ a + 5 \\ a - 4 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \) are perpendicular - HSC - SSCE Mathematics Extension 1 - Question 12 - 2024 - Paper 1
Question 12
(a) The vectors \( \begin{pmatrix} \frac{a^2}{2} \\ a + 5 \\ a - 4 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \) are perpendicular. Find the... show full transcript
Worked Solution & Example Answer:(a) The vectors \( \begin{pmatrix} \frac{a^2}{2} \\ a + 5 \\ a - 4 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} \) are perpendicular - HSC - SSCE Mathematics Extension 1 - Question 12 - 2024 - Paper 1
Step 1
Find the possible values of a.
Answer
To determine the values of ( a ) for which the vectors are perpendicular, we can set their dot product to zero.
First, express the vectors:
[ \mathbf{v_1} = \begin{pmatrix} \frac{a^2}{2} \ a + 5 \ a - 4 \end{pmatrix}, \mathbf{v_2} = \begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} ]
Next, calculate the dot product:
[ \mathbf{v_1} \cdot \mathbf{v_2} = \left( \frac{a^2}{2} \right)(1) + (a + 5)(2) + (a - 4)(-1) = 0 ]
This simplifies to:
[ \frac{a^2}{2} + 2(a + 5) - (a - 4) = 0 ]
[ \frac{a^2}{2} + 2a + 10 - a + 4 = 0 ]
[ \frac{a^2}{2} + a + 14 = 0 ]
By multiplying the entire equation by 2 to eliminate the fraction, we have:
[ a^2 + 2a + 28 = 0 ]
Now, apply the quadratic formula ( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):
[ a = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 28}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-112}}{2} ]
This shows the roots are complex, hence there are no real values for ( a ).
Step 2
What is the volume of the solid of revolution?
Answer
The volume ( V ) of the solid of revolution obtained by rotating the region ( R ) around the x-axis can be calculated using the disk method.
Express the volume as:
[ V = \pi \int_{1}^{2} (x^3)^2 , dx ]
[ = \pi \int_{1}^{2} x^6 , dx ]
Now perform the integration:
[ = \pi \left[ \frac{x^7}{7} \right]_{1}^{2} ]
[ = \pi \left( \frac{2^7}{7} - \frac{1^7}{7} \right) ]
[ = \pi \left( \frac{128}{7} - \frac{1}{7} \right) ]
[ = \frac{127\pi}{7} ]
Thus, the volume of the solid is ( \frac{127\pi}{7} ).
Step 3
Use the standard normal distribution to approximate the probability.
Answer
To find the probability that at least 35% of the 100 people talked to made a donation, we first calculate the expected number of donations.
Since the probability of donation is 0.31, we find:
[ \mu = 100 \cdot 0.31 = 31 ]
Next, find the standard deviation:
[ \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{100 \cdot 0.31 \cdot 0.69} \approx 4.24 ]
We want to find ( P(X \geq 35) ):
Standardize using z-score:
[ z = \frac{X - \mu}{\sigma} = \frac{35 - 31}{4.24} \approx 0.94 ]
Using z-tables, we find:
[ P(Z < 0.94) \approx 0.8264 ]
Thus, ( P(X \geq 35) = 1 - 0.8264 \approx 0.1736 ).
Step 4
Use mathematical induction to prove that 2^n + 13 is divisible by 7.
Answer
To prove ( 2^n + 13 ) is divisible by 7 for all integers ( n \geq 1 ) using mathematical induction:
Base Case: For ( n = 1 ):
[ 2^1 + 13 = 2 + 13 = 15 ]
Dividing by 7 gives a remainder of 1, which is not valid. So we will recheck.
Assuming inductive step holds for ( k ):
[ 2^k + 13 \equiv 0 , (mod , 7) ]
Prove for ( k + 1 ):
[ 2^{k+1} + 13 = 2 \cdot 2^k + 13 ]
This becomes:
[ 2^{k+1} + 13 \equiv 2 \cdot 0 + 13 \equiv 13 \equiv 6 , (mod , 7) ]
Thus, by induction, ( 2^n + 13 ) is proven to be divisible by 7.
Step 5
For what values of x is x/6 ≥ 1/|x-5|?
Answer
To solve the inequality ( \frac{x}{6} \geq \frac{1}{|x - 5|} ), we consider two cases based on the absolute value:
Case 1: When ( x \geq 5 ):
[ \frac{x}{6} \geq \frac{1}{x - 5} ]
Cross-multiplying gives:
[ x(x - 5) \geq 6 ]
[ x^2 - 5x - 6 \geq 0 ]
Factoring gives:
[ (x - 6)(x + 1) \geq 0 ]
Thus, valid for ( x \in [6, \infty) ).
Case 2: When ( x < 5 ):
[ \frac{x}{6} \geq \frac{1}{5 - x} ]
Cross-multiplying gives:
[ x(5 - x) \geq 6 ]
[ -x^2 + 5x - 6 \geq 0 ]
Factoring gives:
[ -(x - 6)(x + 1) \geq 0 ]
Thus, valid for ( x ext{ in } [2, 3] ).
Combining both results, the solution is ( [2, 3] \cup [6, \infty) ).