15) I'm not sure if this is the intended working out, but anyway...
Consider that at the point of intersection, the
equation of the tangent to both graphs is y = x. In other words, the gradient is 1!
We just need to solve for the value of k in which
both derivatives are equal to 1!
The inverse is
=\frac{1}{k}\ln{x} \\<br />\text{The derivatives are }<br />f'(x)=ke^{kx} \text{ and } \frac{d}{dx}f^{-1}(x)=\frac{1}{kx} \\<br />\text{Now solve (simultaneously) for when both derivatives equal 1 }\\<br />ke^{kx}=1 \text{ and } \frac{1}{kx}=1)
This should give you k=1/e, which is E.
16) If the area under the graph is 8, consider what the transformations will do to change the area.
A dilation of 2 from the x-axis will double the area i.e. A = 2*8 = 16
A transformation of 1 unit down will reduce the area by a rectangle with height 1 and length 3 (the 3 comes from the 3 units between x = 1 and x = 4), which is 3 sq units
So now after removing the rectangle, 16 - 3 = 13, we get B as the answer.
17) I should add a caveat for when you use the method you described -
beware logarithms.
When you have typed in option C using the method you described, you will get an equation (rather than the expected 'true'). Now, you need to solve this equation for x. You will get x>=1. This tells us option C is only true when x is greater than or equal to 1. Nevertheless, option C is still true (for numbers greater than or equal to 1, anyway)! Hope this makes sense
In general, I would solve all equations involving logs for the purpose of the 'true/false' method.
19) The probability we want is
}=\frac{\Pr{(D>8 \cap D>7)}}{\Pr{(D>7)}}=\frac{\Pr{(D=9)}}{\Pr{(D=8)}+\Pr{(D=9)}})
Now we use the formula given in the question
}}{\Pr{(D=8)}+\Pr{(D=9)}}=\frac{\log_{10}{(1+\frac{1}{9})}}{\log_{10}{(1+\frac{1}{8})}+\log_{10}{(1+\frac{1}{9})}} \\<br />=\frac{\log_{10}{\frac{10}{9}}}{\log_{10}{\frac{5}{4}}})
which after a bit of manipulation (hint: subtracting logs with the same logs means you divide what's in the brackets) should give option B.
20)
Please see here for a discussion about this same question (discussion lasts for about 4-5 comments)
Home
Help
Search
Login
