If the image of the quadratic function y = - x square + MX-1 has two different intersections with the line segment AB whose two ends are a (0,3) (3,0), the value range of M is obtained

If the image of the quadratic function y = - x square + MX-1 has two different intersections with the line segment AB whose two ends are a (0,3) (3,0), the value range of M is obtained

It is easy to obtain the analytic expression of the straight line AB: y = - x + 3, which is substituted into the analytic expression of the quadratic function. The univariate quadratic equation about X can be obtained: - x + 3 = - x + MX - 1, that is, X - (M + 1) x + 4 = 0. There are two different equations, and they are all between 0 and 3. This is equivalent to: △ > 0 9 - 3 (M + 1) + 4 > 0. The range of M can be obtained by solving the inequality system
Proving tan2 / X-1 / TaNx / 2 = - 2 / TaNx
Tan (x / 2) - 1 / Tan (x / 2) = sin (x / 2) / cos (x / 2) - cos (x / 2) / sin (x / 2) = Sin & # 178; (x / 2) / sin (x / 2) cos (x / 2) - cos & # 178; (x / 2) / sin (x / 2) cos (x / 2) = [Sin & # 178; (x / 2) - cos & # 178; (x / 2)] / [sin (x / 2) cos (x / 2)] = - cosx / [(SiNx) / 2] = - 2 / TaNx
Would you like to add brackets? There is no bracket
A. The coordinates of B are (1,0) (2,0) respectively. If there is only one intersection point between the image of quadratic function y = x & sup2; + (A-3) x + 3 and line segment AB, then the value of a is determined
The value range of a
Because there is an intersection point with AB, the solution of B ^ 2-4ac > 0 is a > 3 + 2 times root sign 3 or a
When TaNx = 3, tan2 / x =?
Is it Tan (x / 2)?
Because TaNx = 2tan (x / 2) / [1 - (Tan (x / 2)) ^ 2] = 3
So 3 (Tan (x / 2)) ^ 2 + 2tan (x / 2) - 3 = 0
Then Tan (x / 2) = [- 2 ± √ (4 + 36)] / 6 = (- 1 ± √ 10) / 3
There are infinitely many answers... There are infinitely many x satisfying TaNx = 3
Why not TaNx / 2 = (1 + 1 radical 10) / 3
It is known that the two zeros of the quadratic function f (x) = the square of AX + BX + C are one and three, and the intersection point with the Y axis is (0,3) to find the analytic expression of F (x)
The two zeros of F (x) = the square of AX + BX + C are one and three
So let f (x) = a (x-1) (x-3)
3 = a (0-1) (0-3), a = 1
f(x)=(x-1)(x-3)=x^2-4x+3
Since the zero point is 1,3, f can be expressed as
f(x)=a(x-1)(x-3)
Substituting (0,3) to get
3=a*(-1)*(-3)=3a
So a = 1
Then f (x) = (x-1) (x-3) = x ^ 2-4x + 3
From (0, 3) we can know C = 3. From the problem we can know f (1) = O, f (3) = 0. Substituting it, we can get a, B. Answer: F (x) = the square of X - 4x + 3
Obviously, 1 and 3 satisfy the equation AX ^ 2 + BX + C = 0
∴a+b+c=0 (1), 9a+3b+c=0 (2)
The point of intersection with y axis is (0,3)
∴c=3 (3)
The equations are composed of (1), (2), (3),
The solution is a = 1, B = - 4, C = 3
So f (x) = x ^ 2-4x + 3
a+b+c=o
9A + 3B + C = 0, a = 1, B = - 4, C = 3
C=3
If x = 3, then SiNx
Tan (2 / 2 x) = 3,
sin(x/2)=3cos(x/2)
9cos²(x/2)+cos²(x/2)=1
Cos (x / 2) = ± (radical 10) / 10
Sin (x / 2) = ± 3 (root 10) / 10
sinx=2sin(x/2)cos(x/2)=2*3/10=3/5
three out of five
Given that f (x) is a quadratic function and f (0) = 0, f (x + 1) = f (x) + X + 1, find f (x)
Tan2 / x = 2, find TaNx =?
Tan2 / x = 2 should be TaNx / 2 = 2,
tanx=2(tanx/2)/[1-(tanx/2)^2]=4/(1-4)=-4/3.
If f [f (x)] = x ^ 4-6x ^ 2 + 6, then the quadratic function f (x) = ()
If f [f (x)] = x ^ 4-6x ^ 2 + 6, then the quadratic function f (x) = ()
Who can write more detailed steps
Write down any formula
If f [f (x)] = x ^ 4-6x ^ 2 + 6 = (x ^ 2-3) ^ 2-3, Let f (x) = a (x + b) ^ 2 + C, then f [f (x)] = a (a (x + b) ^ 2 + C + B) ^ 2 + C = a ^ 3 [(x + b) ^ 2 + (c + b) / a] ^ 2 + C comparison coefficient a = 1 (x + b) ^ 2 + (c + b) / a = x ^ 2-3, C = - 3, the solution is (x + b) ^ 2 + B = x ^ 2, B = 0  f (x) = x ^ 2-3
limx→0 e^sinx(x-sinx)／(x-tanx)
0 / 0 type limit
limx→0 e^sinx(x-sinx)／(x-tanx)
=limx→0 [e^sinx cosx (x-sinx)+e^sinx (1-cosx) ]/1-1/(x^2+1))
=limx→0 e^sinx [(xcosx-cosxsinx)+(1-cosx)]*(x^2+1)/x^2
=limx→0 e^sinx [(xcosx-cosxsinx)+(1-cosx)]+e^sinx [(xcosx-cosxsinx)+(1-cosx)]/x^2
=0+limx→0 e^sinx [(xcosx-cosxsinx)+(1-cosx)]/x^2
=limx→0 {e^sinx cosx [(xcosx-cosxsinx)+(1-cosx)] +e^sinx [cosx-xsinx-cosxcosx+sinxsinx+sinx] } /(2x)
=limx→0 e^sinx {cosx [cosx (xsinx)+2(1-cosx) ] -xsinx +sinxsinx+sinx] } /(2x)
=limx→0 (e^sinx cosx {cosx [cosx (xsinx)+2(1-cosx) ] -xsinx +sinxsinx+sinx] }+ e^sinx {-sinx[cosx (xsinx)+2(1-cosx) ]+cosx [-sinx(xsinx)+cosx (sinx+xcosx)+2sinx] -sinx-xcosx+2sinxcosx+cosx} )/2
=1/2