Here are my solutions: d), c), c), b), a), b), a), b), c), a)
For question B, note that setting x=y=z=1 on identity (iii) shows it is wrong. For question E, note symmetry tells us (a) and (c) are the same. I disagree on H
For question B, note that setting x=y=z=1 on identity (iii) shows it is wrong. For question E, note symmetry tells us (a) and (c) are the same. I disagree on H
Spoiler
Thanks! Really helpful! So the rest is correct right?
If (iii) is true then both (ii) and (iv) are true which is a contradiction so the answer is not (b) or (d). If (i) is true then (ii) is false (since if (i) is true, for (ii) LHS < RHS) and so the answer is not (a), leaving (c).
If you know de moivre's theorem, you can directly find the correct expression by taking the imaginary part of (cos a + i sin a)^5 (and then replacing cos^2 a by (1 - sin^2 a)).
However, more pragmatic is to rule out incorrect solutions:
when a = pi/2, sin a = sin 5a = 1. But (b) evaluates to -1, ruling out (b).
when a = pi/6, sin a = sin 5a = 1/2. For (c) and (d), each term before the sin^5 term can be written as 2n/32 for some integer n. But the sin^5 term has an odd coefficient. So you'd end up with (2n+1)/32 (for some integer n), which can't possibly equal 1/2. (You could also just churn through the arithmetic, but I think this is easier).
This leaves (a) as the only option.
As a sanity check: when a=pi/2, (a) evaluates to 5-20+16 = 1 as we'd expect. And when a = pi/6, (a) evaluates to 5/2 - 20/8 +16/32 = 1/2 as we'd expect.
And as I finish, I notice there's an easier way of ruling out (c) and (d):
since sin 5a is an odd function of a, any equivalent expression must also be an odd function of a. This immediately rules out (c) and (d) as they we can write them as the sum of odd and even functions of a. (By grouping the odd and even powers of sin a).
I said for 3 roots at x>0, both sps must have x>0. And since the sps have x coords (-a ± √ (a²+3))/3 This is not possible for both x coords to be greater than 0.
So then it is between 1 or 2 roots for x>0. When you sketch these graphs, you can see that since the y intercept is at -2, and there is a turning point as x<0, there must only be 1 root.
I said for 3 roots at x>0, both sps must have x>0. And since the sps have x coords (-a ± √ (a²+3))/3 This is not possible for both x coords to be greater than 0.
So then it is between 1 or 2 roots for x>0. When you sketch these graphs, you can see that since the y intercept is at -2, and there is a turning point as x<0, there must only be 1 root.
Suppose the roots are alpha, beta, gamma so (x - alpha)(x - beta)(x - gamma) = 0
As the constant coefficient is negative (-2), there must be two or zero negative roots, so either one or three positive roots. -alpha*beta*gamma = -2
Also the double product (x coefficient) of roots must be negative alpha*beta + beta*gamma + gamma*alpha = -1
The can't all be positive. So there must be one positive root.
Nothing particlarly wrong with what you've done, but you don't need to assume the quadratic formula.
I said for 3 roots at x>0, both sps must have x>0. And since the sps have x coords (-a ± √ (a²+3))/3 This is not possible for both x coords to be greater than 0.
So then it is between 1 or 2 roots for x>0. When you sketch these graphs, you can see that since the y intercept is at -2, and there is a turning point as x<0, there must only be 1 root.
Yes this looks fine. (And I was obviously too hasty in agreeing with the OP on this question. although as I said, I really didn't spend long on each question)..
If you know de moivre's theorem, you can directly find the correct expression by taking the imaginary part of (cos a + i sin a)^5 (and then replacing cos^2 a by (1 - sin^2 a)).
However, more pragmatic is to rule out incorrect solutions:
when a = pi/2, sin a = sin 5a = 1. But (b) evaluates to -1, ruling out (b).
when a = pi/6, sin a = sin 5a = 1/2. For (c) and (d), each term before the sin^5 term can be written as 2n/32 for some integer n. But the sin^5 term has an odd coefficient. So you'd end up with (2n+1)/32 (for some integer n), which can't possibly equal 1/2. (You could also just churn through the arithmetic, but I think this is easier).
This leaves (a) as the only option.
As a sanity check: when a=pi/2, (a) evaluates to 5-20+16 = 1 as we'd expect. And when a = pi/6, (a) evaluates to 5/2 - 20/8 +16/32 = 1/2 as we'd expect.
And as I finish, I notice there's an easier way of ruling out (c) and (d):
since sin 5a is an odd function of a, any equivalent expression must also be an odd function of a. This immediately rules out (c) and (d) as they we can write them as the sum of odd and even functions of a. (By grouping the odd and even powers of sin a).
Referring to the last part of your argument on discarding option C and D. Is it necessarily the case that an odd function summed with an even function yields an even function? Take for example, g(x)= x^4+x^2 , g(x) is an even function. f(x)=x^3+x, f(x) is an odd function. Summing these two functions yields an odd function. Therefore, your argument is not true in all cases and is perhaps not the best way of eliminating options
Referring to the last part of your argument on discarding option C and D. Is it necessarily the case that an odd function summed with an even function yields an even function? Take for example, g(x)= x^4+x^2 , g(x) is an even function. f(x)=x^3+x, f(x) is an odd function. Summing these two functions yields an odd function. Therefore, your argument is not true in all cases and is perhaps not the best way of eliminating options
f+g is not an odd function. (f+g)(1) = 4. (f+g)(-1) = 0.
To answer your more general question: if f is odd, g is even, and both f and g are not identically zero, then f+g cannot be either odd or even.