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let f(x) = [ ln ( 7x-x 2 ) ] 3/2 , then 12 f is defined only on R + and is stricly increasing. f is defined on an interval of finite length and is strictly increasing. f is defined on an interval of finite length and is bounded. range of f contains 1.

let f(x) = [ ln ( 7x-x2) ]3/2 , then                      
                              12                    
  1. f is defined only on R+  and is stricly increasing.
  2. f is defined on an interval of finite length and is strictly increasing.
  3. f is defined on an interval of finite length and is bounded.
  4. range of f contains 1.

Grade:12

2 Answers

Vikas TU
14149 Points
7 years ago
1)  7x – x^2 should be strictly greater than zero for log property.
therfore, 7x – x^2 > 0 
x belongs to (0,7)
or
 0
Thus x is defined in this interval.
 
  1.  For strictly increasing and decreasing we check f’(x) 

    f’(x) = 3/2*[log(7x – x^2)]^1/2 * (7 – 2x)
    NOW log term in underroot will always be positive
    also 7 – 2x is > 0
    for  x lies in (0,7). 

    Hence strictly increasing.

    2nd option. is the appropriate one.
Riddhish Bhalodia
askIITians Faculty 434 Points
7 years ago
lets check the domain
as the term inside squareroot must be >= 0
=> term inside ln be >=1
7x-x^2 \geq 12 \quad \Rightarrow \quad (x-3)(x-4) \leq 0
thus
x \in [3,4]
now at x=3 and x=4 ,f(3)= f(4) = 0 hence it cannot be strictly increasing
and it’s bounded.
the maximum value that f(x) can get is at x = 3.5
evaluate that
so now it’s solved

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