Math help please

Question

If x1 and x2 with x1<x2 meets the equation line of log3 a=2x^2+x , log9 b=5x-x^2 and 9a=b then the value of x2-x1

 

a.1/4

b.1/2

c.3/4

d.1

e.7/4

 

The maximum value of a function f(x)=ax^2+4x+a is 3,the value of a that meets the function is

A.-4
B.-2
C.-1
D.1
E.4

 

 

 

 

Answer 1

log₃a = 2x² + x = x(2x+1)     So, a = 3^x(2x+1)

 

log₉b = 5x - x² = x(5-x)     So, b = 9^x(5-x)

 

But, b = 9a      So, 9a = 9^x(5-x)   Thus, a = 9^x(5-x)/9 = 9^x(5-x)-1

 

We then have 3^x(2x+1) = 9^x(5-x)-1

 

                         3^x(2x+1) = (3²)^x(5-x)-1

 

                         3^x(2x+1) = 3^2x(5-x)-2

 

Equating the exponents, we get:  2x²+x = 10x-2x²-2

 

So, 4x² - 9x + 2 = 0

 

(4x - 1)(x - 2) = 0     x = 2, 1/4

 

If x₁ = 1/4 and x₂ = 2, then x₂ - x₁ = 7/4

-------------------------------------------------------------------------

f(x) = ax² + 4x + a

 

The graph of f(x) is a parabola.  So, in order for there to be a maximum, the value of a must be negative and the maximum occurs when x = -4/(2a) = -2/a.

 

Since the maximum value of the function is 3, we have:                       

     f(-2/a) = 3

 

So, a(-2/a)² + 4(-2/a) + a = 3

 

Simplify to get 4/a -8/a + a = 3

 

Multiply through by a:  4 - 8 + a² = 3a

 

a² - 3a - 4 = 0

 

(a - 4)(a + 1) = 0

 

Therefore, a = 4 or a = -1

 

But, the quadratic has a maximum, so a, the coefficient of x² must be negative.  

 

Consequently, a = -1