You are here:

Advertisement

For info only: The answers required by the setter for the last set of questions were: A=6, B=5, F=17, D=19, E=12. With your excellent explanation before me I was able to spot your error in the last line of the third answer. With 4 out of 5 it was then simple enough to deduce what B had to be (though not why!)

--------------------------------------------------------------------------

1. The coefficient of the x^3 term in the binomial expansion of (8-3x)^(1/3) is –A/768. Find A.

2. The area under the curve with equation (Sin x)^2, between x=pi/4 and x=pi/2 is 0.64B7 to 4 decimal places. Find B.

3. The equation x^2+kx+8=k has no real solutions for x. C=the largest integer value possible for k.

4. The gradient of y=(x^2)(Ln 3x) at x=5 is 32.0D to 2 decimal places. Find D.

5. A cube with edge length Xcm expands uniformly. The rate of increase of total surface area is 0.0E when X=10. Given that the volume is increasing at a rate of 0.05 cubic centimeters per second, find E.

--------------------------------------------------------------------------

This is the last in this circuit of caches but I have remembered another set of four, also based around mathematics of increasing difficulty. Two solved by me and another two coming your way in due course.....

This set is very easy! Basic calculus and algebra.

When you expand any power, you get binomial coefficients:

(a+b)^n = a^n + n a^(n-1) b^1 + ... + (n choose r) a^(n-r) b^r + ... + n a b^(n-1) + b^n

That is why multiplying out (a+b)^10 gives you row 10 (or 11, depending on how you count) of Pascal's triangle. Oddly enough, that also works with non-integer powers in an analogous way.

The x^3 term will basically just take a = 8, b = -3x, and r = 3. That gives you:

(1/3 choose 3) (8)^(1/3-3) (-3x)^3

So the question is really, what is this coefficient?

(1/3 choose 3) 8^(-8/3) (-3)^3 ?

Well, it's easy to work out most of it:

8^(-8/3) (-3)^3 = - 2^(-8) 3^3 = - 27 / 256

The question "what is 1/3 choose 3" is answered in the link above. It is:

(1/3)(1/3-1)(1/3-2) / 3! = 5/81

Multiplying those two quantities gives the coefficient as - 5/786, and so A=5.

This one can be tackled by Wolfram, giving 0.642699. So B=2.

This requires you to know the quadratic formula, specifically the discriminant (stuff in the square root), which is "b^2 - 4ac" or in this case k^2 - 32.

You want no real solutions, so you need this quantity to be negative:

k^2 - 32 < 0

k^2 < 32

What's the biggest (integer) value of k? Well, I think you know it must be 5.

Here you just compute a derivative (not so much a gradient). You use the product rule to obtain:

y' = 2xln(3x) + x

and at x=5 you have 10ln(15)+5 = 32.080502, so D = 8.

Wolfram can do this one as well.

The surface area is given by 6x^2, while the volume is x^3. These two quantities are related, so their rates of change are also related (hence the standard term "related rates").

You can easily eliminate the variable X to obtain obtain:

SA = 6 V^(2/3)

and this gives you the following:

SA' = 6 (2/3) V^(-1/3) V' = 4 V^(-1/3) V'

At X=10 you have V' = 0.05 and V = 1000, so you get:

SA' = 4 (1000)^(-1/3) (0.05) = 4/10 * 0.05 = 0.02. So E = 2.

A = 5

B = 2

C = 5

D = 8

E = 2

- View Follow-Ups
- Add to this Answer
- Ask a Question

Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 10 | Politeness = 10 |

Comment | This guy has been so, so helpful to me. |

Advanced Math

Answers by Expert:

I can answer all questions up to, and including, graduate level mathematics. I am more likely to prefer questions beyond the level of calculus. I can answer any questions, from basic elementary number theory like how to prove the first three digits of powers of 2 repeat (they do, with period 100, starting at 8), all the way to advanced mathematics like proving Egorov's theorem or finding phase transitions in random networks.

I am a PhD educated mathematician working in research at a major university.**Organizations**

AMS**Publications**

Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.**Education/Credentials**

BA mathematics & physics, PhD mathematics from a top 20 US school.**Awards and Honors**

Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.**Past/Present Clients**

In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.