Question

The global maximum value of $f\left( x \right) = {\text{lo}}{{\text{g}}_{10}}\left( {4{x^2} - 12{x^2} + 11x - 3} \right),$ $x \in \left( [{2,3}] \right)$ is
a)$\dfrac{{ - 3}}{2}{\log _{10}}3$ b.) $1 + {\log _{10}}3$ c.) ${\text{lo}}{{\text{g}}_{10}}3$ d.) $\dfrac{3}{2}{\log _{10}}3$

Answer 1

Hint:- In the above question, we are provided with the function to find the maximum value of that function.To get the maximum value, at first we need to differentiate the given function on both the sides with respect to 'x'.
Now we see the nature of the function which we get upon differentiation, and hence by putting the value of 'x' in the function, we get the global maximum value.

Complete step-by-step answer:
We are given to find the global maximum value of:-
$f(x) = {\log _{10}}\left( {4{x^3} - 12{x^2} + 11x - 3} \right),x, \in \left[ {2,3} \right]$
So, the value where ‘x’ will be maximum, log value will also be maximum at that place as it is given that $x \in \left[ {2,3} \right]$
Let:- $g\left( x \right) = 4{x^3} - 12{x^2} + 11x - 3$. Differentiating the above equation on both the side with respect to ‘x’;
$\Rightarrow g'\left( x \right) = 4 \times 3{\left( x \right)^2} - 12 \times 2\left( x \right) + 11 \times 1\left( \right)$
$\Rightarrow g'\left( x \right) = 12{x^2} - 24x + 11$
$\Rightarrow g'\left( x \right) = 12\left( {{x^2} - 2x} \right) + 11$
$g'\left( x \right) = 12{\left( {{x} - 1} \right)^2} - 1+11$
Now, we are given the question that $x \in \left( {2,3} \right);$
So, $12{\left( {{x} - 1} \right)^2}$ is +ve value;
So, $g'\left( x \right)\; > \;0$
$\Rightarrow g\left( x \right)$ is an increasing function.
If, $g\left( x \right) = 4{x^3} - 12{x^2} + 11x - 3$ is the increasing function, then, it’s ,maximum value will be at ‘3’, so the maximum
Value of ‘x’ is also 3.
$\therefore f\left( 3 \right) = \;{\text{answer}}$
So, $f\left( x \right) = {\log _{10}}\left( {4{x^3} - 12{x^2} + 11x - 3} \right)$
$\Rightarrow f\left( 3 \right) = {\log _{10}}\left( {4{{\left( 3 \right)}^3} - 12{{\left( 3 \right)}^2} + 11\left( 3 \right) - 3} \right)$
$\Rightarrow f\left( 3 \right) = {\log _{10}}\left[ {\left( {4 \times 27} \right) - \left( {12 \times 9} \right) + 33 - 3} \right]$
$\Rightarrow f\left( 3 \right) = {\log _{10}}\left[ {108 - 108 + 33 - 3} \right]$
$\Rightarrow f\left( 3 \right) = {\log _{10}}\left( {30} \right)$
$\Rightarrow f\left( 3 \right) = {\log _{10}}\left( {10 \times 3} \right)$
$\Rightarrow f\left( 3 \right) = {\log _{10}}10 + {\log _{10}}3$
$\Rightarrow f\left( 3 \right) = 1 + {\log _{10}}3$
$\therefore$ option B is the correct answer.

Note:- If the squared-value of any of the equations is given then the equation is always positive, which means it will always be greater than zero.
The maximum or minimum over the entire function is called an "Absolute" or "Global" maximum or minimum. There is only one global maximum (and one global minimum) but there can be more than one local maximum or minimum.