Question:
Let $f(x)=10-|x-3|,\; x\in\mathbb{R}$. The maximum of $f(x)$ occurs at:
Let $f(x)=10-|x-3|,\; x\in\mathbb{R}$. The maximum of $f(x)$ occurs at:
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Functions of the form $f(x) = K - |x-a|$ always reach their maximum value $K$ at $x=a$.
Updated On: Apr 28, 2026
- $x=0$
- $x=3$
- $x=-3$
- $x=10$
- $x=1$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The absolute value $|x-3|$ is always $\ge 0$.
Step 2: Analysis
For $f(x) = 10 - |x-3|$ to be maximum, the term $|x-3|$ must be at its minimum value.
Step 3: Conclusion
The minimum value of $|x-3|$ is 0, which occurs when $x-3=0 \implies x=3$. Final Answer: (B)
The absolute value $|x-3|$ is always $\ge 0$.
Step 2: Analysis
For $f(x) = 10 - |x-3|$ to be maximum, the term $|x-3|$ must be at its minimum value.
Step 3: Conclusion
The minimum value of $|x-3|$ is 0, which occurs when $x-3=0 \implies x=3$. Final Answer: (B)
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