2001 2U HSC (note, I do not guarantee 100% that my answers/working are right, if someone finds a mistake please notify me
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Q5
(a) To work out the domain, observe that one can only take square roots of zero or positive numbers, i.e. numbers more than or equal to zero (we are only dealing with real numbers here, so complex numbers don't come into play). Thus,
domain:
25 - x^2 >= 0
x^2 - 25 <= 0
(x-5)(x+5) <= 0
Sketching the graph, it can be shown that:
d: {-5 <= x <= 5}
The range can be worked out in a similar way:
Note that y must be zero or a positive number, i.e. y >= 0
Now, (y/2)^2 = 25-x^2
x^2 = 25 - y^2/4
x = Sqrt(25 - y^2/4)
Thus,
25 - y^2/4 >= 0
y^2 - 100 <= 0
We have {y >= 0} U {-10 <= y <= 10}
Therefore,
r: {0 <= y <= 10}
The range can also be worked out by sketching a graph of the function y^2, y^2 >= 0.
(b) (i) log_10 (2^1000)
= 1000 log_10 (2)
= 1000 ln(2)/ln(10)
= 301.030 (3dp)
(ii) 10^x = 2^1000 (not 1024, thanks for the correction "Pop n' Fresh")
x ln(10) = 1000 ln (2)
x = 1000 ln(2) / ln(10) = 301.030 (3dp) [from section (i)]
Thus, 10^301.030 = 2^1000
So, 2^1000 has 301 zeros, and counting the 1, it has 302 digits.
Q6
(c) (iii) From previous parts, it was worked out that A = (-1,3) and B = (1/3, 49/27)
With the equation x^3 + x^2 - x + 2 = k, k represents the range of y values for which the graph x^3 + x^2 - x + 2 has 3 real solutions, that is the line y = k cuts the graph at 3 distinct points. Thus, reading off the graph we have:
(y-coordinate of B) < k < (y-coordinate of A)
Thus, the answer is: {49/27 < k < 3}
Hope this helps!
