2 advanced math problems to solve, 2 advanced math problems to solve in detail

Question 1: You can directly use Lobida's rule to directly derive the numerator and denominator of the previous test, and get f(x)=xf(x) (2*x), and then remove x, you can get f(x)=f(x) 2, and because f(0)=1, that is, f(x)=1 2;Since f(x) is continuous at x=0, i.e., a=1 2.

Question 2: f'(lnx)=-1 (x 2), then the integral test is -1 (x 3), and the answer obtained by directly finding the integral is: -3 8.

1) When x=0, it is 0 to 0, using the Lobida rule.

(xf(x)) (2*x).

It is known that f(0)=1

Since f(x) is continuous.

a=1 22) The answer is -1 2

Integral(1 2)(f'(inx) x)dx

Integral (in1 ln2) f'(inx)d(inx)e^（-in2）-e^0

Question 1 When x=0, it is 0 to 0, using Lobida's rule.

(xf(x)) (2*x).

It is known that f(0)=1

Since f(x) is continuous.

a=1 2 Question 2 The answer is -1 2

Integral(1 2)(f'(inx) x)dx

Integral (in1 ln2) f'(inx)d(inx)e^（-in2）-e^0

Lobida's rule, a=1 2

1/2.Someone has steps, I won't write about it.

Here's how, please refer to:

3. z = e^ucosv, u = x^2-y, v = 3xy

z/∂x = e^u (2x) cosv - e^u sinv (3y) =e^u(2xcosv-3ysinv)

e^(x^2-y)[2xcos(3xy)-3ysin(3xy)]

z/∂y = e^u (-1) cosv - e^u sinv (3x) =e^u(cosv+3xsinv)

e^(x^2-y)[cos(3xy)+3xsin(3xy)]

4. f = 2x^3-6x^2-18x-7, f'=6x 2-12x-18 = 6(x+1)(x-3), with a standing point x = 1, 3;

f''12x-12 = 12(x-1), let f''0, get x = 1

f''(1) =24 < 0, x = 1 is the maximum point, and the maximum f(-1) =3;

f''(3) =24 > 0, x = 3 is the minimum point, and the minimum f(3) =61;

Monotonic increase interval (-1), (3,+ monotonically decrease interval (-1,3).

Convex interval (-1), concave interval (1,+ inflection point (1,-29).

38 Let x=lnt, then t=e x, integral=sf(lnt)d(lnt)=s[ln(1+t) t]d(lnt)=s[ln(1+t) t]d(lnt)=s[ln(1+t) t 2]dt=-sln(1+t)d(1 t)=-ln(1+t) t+s(1 t)d[ln(1+t)=-ln(1+t) t+s(1+t)s(1+t)s(1+t)t=-ln(1+t)t=-ln(1+t). t） t） t+s（1 t）dt=-ln（1+t） t+s（1 t）dt=-ln（1 （1+t））dt=ln（1+t） t+ln | t/（1+t）|+c Note that logarithms are not reciprocal. Just replace t=e x with it. 66 After moving x 3 to d, it becomes 1 4dx 4, remember x 4 = too much, then the original integral = s(sect) 2 (sect) 4d t=s(cost) 2dt=sin(4t) 4+t 2+c, replace only t=arctan(x 4).

It's not an advanced math problem at all, it's obviously a high school problem.

In the first question, find the limit of the fraction, first find the derivative of the numerator and denominator, and then find the limit (e x-1).'/x'=e^x

x---0,e^x---1

So it's limit = 1

The same goes for the second question.

x-√a)'/√(x-a)'= (x-a) xx---a, (x-a) x---0 so its limit = 0

Hello Lopida Rule.

Type 0 0, can be applied.

lim(x→0）(e^x-1)/x

The numerator and denominator are derivatives respectively.

lim(x→0）e^x/1

1 The second question, I have to think about it.

The first one is equal to 1 and the second one is equal to 0. You see, right?

The first one is very simple, just turn arctanx and ln1+x 2 into a power series.

The second puts f(1 n) at x=0 Taylor, i.e. f(1 n) = f(0) + f'(0)/n +f"(a)/2n^2

Then put f(-1 n) at x=0 Taylor, i.e. f(-1 n)=f(0)-f'(0)/n +f"(b)/2n^2

f(1 n) = f(-1 n), and the addition of the two formulas gives f(1 n) = 1+[f"(a)+f"(b)]/4n^2f"(a)+f"(b) <=m (with continuous partial derivation) because 1 n 2 converges to the sum of the above equation and completes the verification.

You just arctanx, and then multiply the whole formula by an x...

The first question is very simple, each of them is just that, you can get:

f(x)=x^2/2 - x^4/12 + x^6/30 - x^8/56 + x^10/90 - x^12/132;

The second question, how do I feel wrong?,Did I miss F?'(0) = 0 condition, because.

f(1n) is obtained by f(0).

f(1/n)=f(0)+f'(0)/n+f"(0)/2/n^2+..

Bring in to get the desired series.

.=sum(f'(0)(1/1+1/2+..1/n...f"(0)/2(1/1+1/4+..1/n^2)+.

Obviously this is a divergent progression, unless f'(0), you missed this condition.

If f'If (0)=0 is true, it can be two levels.

abs (to be demanded. =sum(abs(1/2*(f"(r1)/1+f"(r2)/4+..f"(rn)/n^2+..

sum(m/2*(1/1+1/4+..1/n^2+..

m/2*pi^2/6;

where m=max(abs(f.)"(rn)))0<=r1,r2,..rn,<=1)

Hence the series converges.