Application of DerivativesmediumFree

Local Maximum of a Function Redefined at a Single Point | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Given
f(x)={∣x−1∣+5if x≠18if x=1f(x) = \begin{cases} |x-1|+5 & \text{if } x \ne 1 \\ 8 & \text{if } x = 1 \end{cases}
 then at x=1x = 1, f(x)f(x) has
Aa local maximumcorrect
Ba local minimum
Cneither maximum nor minimum
Da unique tangent with finite slope
Solution
Step 1: Write down the value at the point itself.
f(1)=8.f(1) = 8 .
 Step 2: Write down the values nearby. For any x≠1x \ne 1,
f(x)=∣x−1∣+5.f(x) = |x-1| + 5 .
 Step 3: Bound those nearby values. Taking xx within a distance 11 of the point, so 0<∣x−1∣<10 < |x-1| < 1,
5<f(x)<6.5 < f(x) < 6 .
 Step 4: Compare with f(1)f(1).
f(x)<6<8=f(1)for every x≠1 near 1.f(x) < 6 < 8 = f(1) \quad \text{for every } x \ne 1 \text{ near } 1 .
 Step 5: Apply the definition. Since f(1)f(1) is greater than the value at every nearby point, x=1x = 1 is a point of local maximum. So (1) is true and (2), (3) are false. Step 6: Deal with (4). ff is not even continuous at x=1x = 1 (the nearby values approach 55, not 88), so it has no tangent there at all. False. Answer: (1).
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