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Who can help me with this high school math calculus test paper?
1.{ x | x> 1 or x

2.y=lgu,u=sinx

3.2+

4.(e^x-sinx)dx

5.(- 1,+infinity)

6.y=x^2(x>; =0)

7.2-

8.y=2x- 1

9. Be guided to a certain extent

1. Wrong

mistake

mistake

4. The increasing interval is C and the decreasing interval is Wu.

5- 12.CDBDDCDC

1. Original formula =ln 1+sin0+2=2.

2. The denominator of the topic is 2x 2+3x+ 1.

(x^3+ 1)/(2x^2+3x+ 1)=(x+ 1)(x^2-x+ 1)/(x+ 1)(2x+ 1)=(x^2-x+ 1)/(2x+ 1)

The original formula = (1-1+1)/(2+1) =1/3.

3. Original formula = 2x 3/3+3x 2/2+x+c

4. Original formula =0

5. Original formula = infinity

6. (x-3)/(x 2-9) =1/(x+3), the original formula = 1/6.

7. I can't see any trend.

8.[ 1-2/(x+ 1)]^(x)=[ 1-2/(x+ 1)]^[(x+ 1)/(-2)](-2x)/(x+ 1]

Original formula = e (-2)

1. Original formula = (4/5) x (5/2)

2. The original formula = ∫ lnxd (x2/2) = x2 * lnx/2-∫ (x/2) dx = x2 * lnx/2-x2/4+C.

3.∫ (4+2x) dx = x 2+4x+c, and the original formula = 1+4-0=5.

4. ∫ xcosxdx = ∫ xdsinx = xsinx-∫ sinxdx = xsinx+cosx+c, the original formula =- 1- 1=-2.

5. ∫ (COSX-EX+1/X) DX = SINX-EX+LNX+C, and the original formula = SIN2-E2+LN2-SIN1+E.

6. The original formula = ∫ [3-3/(1+x 2)] dx = 3x-3arctanx+c.

7. The original formula = ∫ e (x 2) d (x 2)/2 = e (x 2)/2+c.

8.∫ (x-2) dx = x 2/2-2x+c, the original formula =9/2-6-0=-3/2.

9.∫ xlnxdx = x 2 * lnx/2-x 2/4+c, the original formula = 2LN2-1-(0-1/4) = 2LN2-3/4.

1.y'=2x-(2/3)x^(- 1/3)

2.y'=ln3*3^x

3.y'= 1+ 1/x

4.y'=- 1/ radical sign (2x-x2)

5.y'= 12x^2+e^x-sinx

6.dy=(-2x)e^(-x^2)dx

7.2x+y'/y+cosy*y'=0,y'=2xy/(ycosy+ 1)

1.f' (x) =1-x (1/3), so that f' (x) = 0 and x = 1.

Therefore, f(x) increases monotonically at (-infinity, 1) and decreases monotonically at (1,+infinity), and the maximum value is =1-3/2 =-12 (x =/kloc-0)

2.f(0-)=2, f(0+)=f(0)=k, and for the sake of continuity, k=2.

3.L'(x)=2-0.002x, let L'(x)=0, x= 1000.

Therefore, L(x) monotonically increases at (0, 1000) and monotonically decreases at (1000,+infinity).

Therefore, the maximum value = 500+2000-1000 =1500 (when x =1000).

4.l(p)=pq-c(q)= 1000p- 100p^2-5( 1000- 100p)-200=- 100p^2+ 1500p-5200

L'(P)=-200P+ 1500=0,P=7.5

So L(P) increases monotonically at (0,7.5) and decreases monotonically at (7.5,+infinity).

So the maximum value = 425 (when P = 7.5 and Q=250).