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Question

The range of the function fx=ln3x2-4x+5


A

ln113,

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B

ln10,

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C

ln116,

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D

ln1112,

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Solution

The correct option is A

ln113,


Explanation for the correct option

Step 1: Solve for the upper bound of the given function

Given equation is fx=ln3x2-4x+5

Range of a function is the set of values that are possible outputs of the function.

We observe that the input for the ln function is a quadratic polynomial with its coefficient of x2 being positive.
We know that a quadratic polynomial with positive coefficient of x2 has as the upper limit of range.
So, when 3x2-4x+5=, ln=.

Thus, the upper bound of the range of the given function is .

Step 2: Solve for the required range

We know that a polynomial of degree n has n-1 critical points (Because critical points are found by differentiating and differentiation decreases the degree by 1).
Since the quadratic polynomial has a degree of 2, it has 1 critical point.
We have already established that the upper bound of the quadratic is , thus its critical point must be its global minima.
We know that a is a critical point if f'a=0
So, the critical point of the quadratic is,
ddx3x2-4x+5=06x-4=0x=23

So the least possible value of the quadratic is when x=23 which is,
3232-423+5=43-83+5=113

So, the least value of the given function is ln113

Therefore, the range of the given function is ln113,.

Hence, option(A) is correct.


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