Question

Function, \(f(x) = -|x-1|+5, \forall x \in R\) attains maximum value at x =

• A. 1
• B. 5
• C. 2
• D. 9

Hint:

For functions of the form \(k|x-a|+c\), the vertex (and the extremum) is always at \(x=a\). If \(k\) is negative, it's a maximum. If \(k\) is positive, it's a minimum. The value of the extremum is \(c\). Here, \(k=-1\), so it's a maximum at \(x=1\).

Answer 1

The Correct Option is A

Step 1: Concept Overview:
This function includes an absolute value. We can usually find the maximum or minimum by looking closely at the absolute value part.

Step 2: Core Principle:
1. \(|x-a|\) is always non-negative: \(|x-a| \geq 0\). 2. \(|x-a|\)'s smallest value is 0, when \(x=a\). 3. We use this to find the largest value of \(f(x)\).

Step 3: Detailed Explanation:
We have \(f(x) = -|x-1| + 5\).
The part \(|x-1|\) is always zero or more. \[ |x-1| \geq 0 \] \(|x-1|\) is smallest (0) when \(x-1 = 0\), so \(x=1\).
Now, \(-|x-1|\). When multiplying by -1, the inequality flips: \[ -|x-1| \leq 0 \] So, \(-|x-1|\)'s biggest value is 0. This happens when \(|x-1|\) is smallest, at \(x=1\).
\(f(x)\) is just \(-|x-1|\) moved up by 5. So, \(f(x)\)'s biggest value is at the same x-value.
\(f(x)\)'s maximum is \((\text{max value of } -|x-1|) + 5 = 0 + 5 = 5\).
This happens when \(x=1\).

Step 4: Conclusion:
\(f(x)\) is highest when \(x=1\).