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Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1
Question 11
Use a SEPARATE writing booklet. (a) Solve \( \left( x + \frac{2}{y} \right)^2 - 6 \left( x + \frac{2}{y} \right) + 9 = 0 \). (b) The probability that it rains on ... show full transcript
Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1
Step 1
Solve \( \left( x + \frac{2}{y} \right)^2 - 6 \left( x + \frac{2}{y} \right) + 9 = 0 \)
Answer
To solve the equation, substitute ( z = x + \frac{2}{y} ). The equation becomes:
[ z^2 - 6z + 9 = 0 ]
Factoring, we get:
[ (z - 3)^2 = 0 ]
Thus, ( z = 3 ), which gives us:
[ x + \frac{2}{y} = 3 ]
This is rearranged to find ( x ):
[ x = 3 - \frac{2}{y} ]
Substituting back into the original expression, we find ( x = 1 ) and ( x = 2 ) after considering the values for y.
Step 2
Write an expression for the probability that it rains on fewer than 3 days in November.
Answer
The probability of rain on any day is ( p = 0.1 ). For fewer than 3 days, we can use the binomial formula:
[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) ]
Where: [ P(X = k) = {30 \choose k} p^k (1-p)^{30-k} ]
Calculating: [ P(X = 0) = {30 \choose 0} (0.1)^0 (0.9)^{30} ] [ P(X = 1) = {30 \choose 1} (0.1)^1 (0.9)^{29} ] [ P(X = 2) = {30 \choose 2} (0.1)^2 (0.9)^{28} ]
Thus: [ P(X < 3) = (0.9)^{30} + 30(0.1)(0.9)^{29} + 435(0.1)^2(0.9)^{28} ]
Step 3
Sketch the graph \( y = 6 \tan^{-1}(x) \), clearly indicating the range.
Answer
The function ( y = 6 \tan^{-1}(x) ) ranges from:
- As ( x \to -\infty ), ( y \to -6 \frac{\pi}{2} = -3\pi )
- As ( x \to +\infty ), ( y \to 6 \frac{\pi}{2} = 3\pi )
The graph should be sketched as an S-shaped curve approaching these horizontal asymptotes.
Step 4
Evaluate \( \int_{2}^{5} \frac{x}{\sqrt{x-1}} \, dx \) using the substitution \( x = u^2 + 1 \).
Answer
Let ( x = u^2 + 1 ) then ( dx = 2u , du ).
Change the limits:
- When ( x = 2 ), ( u = 1 )
- When ( x = 5 ), ( u = 2 )
The integral becomes:
[ \int_{1}^{2} \frac{u^2 + 1}{\sqrt{u^2}} (2u) , du = \int_{1}^{2} (u^2 + 1)(2) , du]
Evaluating gives: [ \int_{1}^{2} (2u^2 + 2) , du = \left[ \frac{2u^3}{3} + 2u \right]_{1}^{2} = \left( \frac{16}{3} + 4 \right) - \left( \frac{2}{3} + 2 \right) = \frac{14}{3} ]
Step 5
Step 6