Any number can be written in the form 5q+r where 0r<5

 

n²+n+1 = n(n+1)+1 = (5q + r)(5q + r+1)+1

 

From this we can see that if n²+n+1 is to be divisible by 5, then so must r(r+1)+1 = r²+r+1.

 

It is easy to check for r =0,1,2,3,4 that the expression is not divisible by 5. Hence it is not divisible by 5 for any n.

 

Method II:

 

First eliminate the cases where n = 5k or 5k+1 using the above reasoning or by expanding.

 

So, assume that n is not of the form 5k or 5k+1

Now, consider n⁴-1 which is divisible by 5 for n prime to 5 by Fermat's theorem.

n⁴-1 = (n-1)(n³+n²+n+1). Since n is not of the form 5k+1, n-1 is not divisible by 5. This means that n³+n²+n+1 is divisible by 5.

 

Now, if n²+n+1 is divisible by 5, then n³+n²+n+1 is divisible by 5 if n³ is divisible by 5 which means n is divisible by 5 which contradicts our assumption that n is not divisible by 5.

 

Hence  n²+n+1 cannot be divisible by 5 for any n.

 

So, now you can also answer the question: what is the probability that for a random number n, n³+n²+n+1 is divisible by 5?

 

The result is obviously true for n = 5k.

Suppose n 5k

 

Let's assume that n²+n+1 is divisible by 5. Then (n²+n+1)  (n²-n+1) = n⁴+n²+1 is also divisible by 5.

 

n⁴+n²+1 = (n⁴-1)+n²+2

 

 (n⁴-1) is divisible by 5 by Fermat's theorem. Hence n²+2 is divisble by 5.

 

Hence n²+2 = 5k or n² = 5k - 2, which is absurd as any square is of the form 5k, or 5k1

 

Consider f(x) = ax³+bx²+cx

we can see that x = 0 is a zero of f(x).

 

also it is given that f(1) = a+b+c = 0. Hence x =1 is another zero.

 

f'(x) = 3ax²+2bx+c by Rolle's theorem will therefore have a zero in the interval [0,1] which is a subset of the intervals in options (b) and (c).

 

Hence, (a), (b) and (c) are all correct.

tan^-1(e^x) = cot^-1(e^-x) = /2 - tan^-1(e^-x)

Hence tan^-1(e^x)/e^x dx =   (/2 - tan^-1(e^-x))/e^x dx

= /2e^-x dx - tan^-1(e^-x)/e^x dx

For the 2nd integral, let e^-x = t. Then dt = -dx/e^x.

So the 2nd part becomestan^-1(t)dt which I suppose is a standard integral.

 

PS: On second thoughts, we could have just stopped at cot^-1(e^-x)

Another approach:

A+B=-C

hence dA+dB = 0.

Also c = acosB+bcosA

dc = (cosBda+cosAdb) - (asinBdB+bsinAdA) = 0

 

But asinB = bsinA. Hence, the 2nd expression works out to asinB(dB+dA) = 0.

 

Hence CosBda+CosAdb = 0 or da/cosA+db/cosB = 0.

aks, if you want modulus function call it |x| and not [x] bcos [x] is the floor function (greatest integer less than x). The key '|' is just below 'backspace'

The method in such cases is to resort to an alternative definition of a limit by Heine (usually, Cauchy's definition is used). (Ref: I.A. Maron)

 

Heine said that a number l is the limit of a function f(x) when xa, if for all sequences {x_i}a,

the sequence {f(x_i)} l

This definition is more often used to prove that a limit does not exist.

 

So here consider two sequences {x_n} = n and  {y_n} = 2n + /2 both tend to inf. In the first case {f(x_n)} is zero. In the second case the sequence definitely does not tend to zero. Hence the limit does not exist.

 

 

 

L'Hospital's rule is quite alright. But the fastest way to getting at this one is:

let f(x) = 4cos²x/2+cosx. Then f() = 2

 

So, the problem is lim (f(x) - f())/(x-) which is nothing but f'(x) at x=.

                          x

 

You get the answer in no time as zero.

 

(x-) in the denominator when x gives the 1st clue that some mischief like this is intended.

I think the method would be something like this:

1/1+sinx+sin²x dx

We have 1+sinx+sin²x = (1-sinx) (1-²sinx)

Hence 1/1+sinx+sin²x = 1/(1-sinx) (1-²sinx) = A/(1-sinx) + B/(1-²sinx)

Comparing LHS and RHS we get
A+B = 1 and
A²+B = 0

Eqn 2 gives on multiplying by ². A+B = 0 or A+1 = A giving
A = 1/1-  and B = -/1-

Now the integral can be evaluated using the integral dx/a+bsinx which I've unfortunately forgotten how to do.I leave it for the students to finish.
 

PS: I should have said: "The rest is left as an exercise for  the student".