Find the minimum value of the function fix) = log2 (x2 - 2x + 5).a)-4b)2c)4d)-2Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Method to Solve :

y= x2 – 2x + 5

Step 1 : Differentiate with respect to x
Step 2 : Equate to 0
Step 3 : Find the value of x
dy/dx=2x-2 =0 implies x=1
Hence f(1)= 12 – 2 + 5= 4
Thus minimum value of the argument of the log is 4.
So minimum value of the function is log 4 (base 2) =2

Answer

Given function: f(x) = log2(x2 - 2x + 5)

To find the minimum value of the function, we need to find the value of x for which f(x) is minimum.

Method: Completing the square

Step 1: Write the given function in the form of (x - a)2 + b.

f(x) = log2(x2 - 2x + 5)
f(x) = log2[(x - 1)2 + 4]

Step 2: The minimum value of (x - a)2 is 0, which occurs when x = a. Therefore, the minimum value of the function f(x) occurs when (x - 1)2 = 0.

(x - 1)2 = 0
x = 1

Step 3: Substituting x = 1 in the expression for f(x), we get:

f(1) = log2[(1 - 1)2 + 4]
f(1) = log2[4]
f(1) = 2

Therefore, the minimum value of the function f(x) is 2, which occurs when x = 1.

Answer: Option B (2)

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