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3 Answers

Best answer
27 votes
27 votes
$f(x)$ will be minimum at $x = 2,$
Here is another approach.
Since log and exponent are monotonically increasing functions, the problem of minimizing $f(x)$ can be reduced to just minimizing the quadratic expression $x^{2} - 4x + 5,$
This quadratic expression can be written as $(x^{2} - 4x + 4) + 1$ which is equal to $(x - 2)^{2} + 1.$
Now since $(x - 2)^{2}$ can not be less than $0$, so $(x - 2)^{2} + 1$ can not be less than $1.$
Also $(x - 2)^{2} + 1$ will be at its minimum value $(= 1)$, when $x = 2.$

So, value of $f(x)$ will be minimum at $x = 2.$
4 votes
4 votes
Just differentiate whole function and put value =0, x comes out to be 2.
1nd derivative +ve here, so minima
edited by
1 votes
1 votes

In this question it is not necessary to either use differentiation or anything much different.

Just see that how this expression simplifies =>

After cancelling out log & Exponent we get something like =>

2.5ln2 + 2.5 ( x2-4x+5 )

Now 2.5ln2 will be there same in all x values.

Just put in all values one by one, it is not hard to see that , At x = 2 we get minimum value !

Answer c

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