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已知对任意

x \in \mathrm{R}

，均有不等式

a x^{2}+b x+c \geq 0

成立，其中

b<0

. 若存在 $t \in \...

Mar 8, 2024

已知对任意

x \in \mathrm{R}

，均有不等式

a x^{2}+b x+c \geq 0

成立，其中 b<0. 若存在

t \in \mathrm{R}

使得

(1-t) a+(1+2 t) b+3 c=0

成立，则

t

的最小值为

Solution by Steps

step 1

Given the inequality

a x^{2} + b x + c \geq 0

holds for all

x \in \mathbb{R}

and b < 0 , we need to find the minimum value of

t

such that

(1-t) a + (1+2 t) b + 3 c = 0

step 2

From the asksia-ll calculation list, we have the solution for

t

t = \frac{a + b + 3 c}{a - 2 b}

step 3

To find the minimum value of

t

, we need to consider the constraints on

a

b

, and

c

provided by the asksia-ll calculator

step 4

The constraints are a > 0 , b < 0 , and

c \geq -x (a x + b)

for all

x \in \mathbb{R}

step 5

Since b < 0 and a > 0 , the denominator

a - 2 b

is positive, and the expression for

t

is well-defined

step 6

The minimum value of

t

is obtained directly from the expression

t = \frac{a + b + 3 c}{a - 2 b}

, considering the given constraints

Answer

The minimum value of

t

t = \frac{a + b + 3 c}{a - 2 b}

given a > 0 , b < 0 , and

c \geq -x (a x + b)

for all

x \in \mathbb{R}

Key Concept

Solving a system of inequalities and equations to find a variable's value

Explanation

The minimum value of

t

is found by solving the equation

(1-t) a + (1+2 t) b + 3 c = 0

under the given constraints that the quadratic inequality holds for all real numbers and b < 0 . The solution for

t

is derived from the system of inequalities and the equation provided.