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Given that a, B and C belong to R, the function f (x) = ax ^ 3 + BX ^ 2 + CX satisfies f (1) = 0 Let the derivative of F (x) be f '(x), and f' (0) * f '(1) > 0 Finding the range of C / A
f(1)=a+b+c=0,
b=-a-c
f'(x)=3ax²+2bx+c
f'(0)·f'(1)>0
That is, C (3a + 2B + C) > 0
c(3a-2a-2c+c)>0
c(a-c)>0
Divide by C and 178; to get
a/c -1>0
That is, a / C > 1
So C / a ∈ (0,1)
Let f (x) = ax-3, G (x) = BX ^ - 1 + CX ^ - 2 (a, B ∈ R) and G (- 1 / 2) - G (1) = f (0)
If B = 1, set a = {x | f (x) > G (x), and G (x)
When B = 1, G (- 1 / 2) - G (1) = f (0)
We get: G (x) = 1 / x + C / (x ^ 2)
[-2+4c]-[1+c]=-3
-3+3c=-3
∴c=0
That is g (x) = 1 / X
The set a can be expressed as {x | ax-3 ＞ 1 / X and 1 / X
Let f (x) = X3 + bx2 + CX, G (x) = f (x) - F ′ (x). If G (x) is an odd function, find the value of B and C
From F (x) = X3 + bx2 + CX, f ′ (x) = 3x2 + 2bx + C, then G (x) = f (x) - F ′ (x) = X3 + (B-3) x2 + (c-2b) x-C, ∵ g (x) is an odd function, ∵ G (0) = - C = 0, C = 0. ∵ g (x) = X3 + (B-3) x2-2bx
A triangle and a parallelogram have the same base and area. The parallelogram is 16 cm high and the triangle is () cm high
32. It can be calculated by equal area
The product of any two odd numbers must be an odd number a, an even number B, a prime number D
A odd number
If one side of an isosceles triangle is 4cm long and the other side is 9cm long, then the circumference of the isosceles triangle is
9+4+9
As shown in the figure, PA is tangent to ⊙ o at point a, and the extension line of Po intersects with ⊙ o at point C. if the radius of ⊙ o is 3, PA = 4. The length of chord AC is______ ．
Connect OA, make ad ⊥ CP through a, ∵ PA is tangent line of circle O, ≁ PA ⊥ OA, in RT △ AOP, OA = 3, PA = 4, according to Pythagorean theorem: OP = 5, ∵ s △ AOP = 12ap · Ao = 12op · ad, ∵ ad = AP · AOOP = 4 × 35 = 125, according to Pythagorean theorem: PD = pa2 − ad2 = 165, ∵ CD = pc-Pd = 8-165 = 325, according to Pythagorean theorem: AC = ad2 + DC2 = 4735
As shown in the picture, the two banks of the river form parallel lines. A and B are the two workshops located on both banks of the river. To build a bridge CD on the river, make the bridge CD perpendicular to the river bank, and make the path acdb between a and B the shortest, please determine the location of the bridge
Make a vertical line from a to the river bank and cross PQ and Mn to f and g respectively. Take AE = FG on Ag and connect them EB.EB Cross Mn to D. make a vertical DC at d to the opposite bank, then DC is the location of the bridge
: ∵ AE ⊥ PQ, CD ⊥ PQ (known)
‖ AE ‖ CD (two lines perpendicular to the same line are parallel to each other)
∵ PQ = Mn (known)
‖ CD = FG (the vertical lines in parallel lines are equal)
And ∵ AE = FG (known)
‖ AE = CD (equivalent substitution)
∵ AE = CD, and AE ‖ CD (proved)
ACDE is a parallelogram ()
ACDE is a parallelogram (proved)
‖ AC = de (the opposite sides of parallelogram are parallel and equal)
CD is quantitative
The shortest is AC + dB and the shortest is AC + CD + dB
∵ segment EB connects point E and point B (as shown in the figure)
The EB segment is the shortest line from point e to point B (the shortest segment between two points)
And ∵ AC = de (proved)
‖ AC + DB = de + dB (equivalent substitution)
The shortest is AC + dB
The shortest path is AC + CD + dB
As shown in the figure, angle AOB = 120 degrees, OC bisecting angle AOB, OD bisecting angle AOC, calculate the degree of angle BOD
∠COD=1/2*∠AOC=1/4*∠AOB=30°
∴∠BOD=∠BOC+∠COD=120°/2+30°=90°
A sufficient and necessary condition for Ax 2 + 2x + 1 = 0 (a ≠ 0) to have at least one negative root is obtained
It is proved that the equation with at least one negative root has one positive root and one negative root or two negative roots △ = 4 − 4A ＞ 0x1x2 = 1a ＜ 0 {a ＜ 0, which is equivalent to △ = 4 − 4A ≥ 0 − 2A ＜ 01A ＞ 0; In conclusion, the necessary condition for the original equation to have at least one negative root is a ＜ 0 or 0 ＜ a ≤ 1 sufficiency: from the reversibility of the above reasoning, we know that when a ＜ 0, the equation has two different signs; when 0 ＜ a ≤ 1, the equation has two negative roots. Therefore, a ＜ 0 or 0 ＜ a ≤ 1 is the sufficient condition for the equation AX2 + 2x + 1 = 0 to have at least one negative root
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