Question

Let \([t]\) denote the greatest integer less than or equal to t.
Let \(f(x)=x-[x]\), \(g(x)=1-x+[x]\), and \(h(x) = \min\{f(x), g(x)\}\), \(x \in [-2, 2]\).
Then h is :

• A. not continuous at exactly four points in [-2, 2]
• B. not continuous at exactly three points in [-2, 2]
• C. continuous in [-2, 2] but not differentiable at more than four points in (-2, 2)
• D. continuous in [-2, 2] but not differentiable at exactly three points in (-2, 2)

Hint:

The graph of \(h(x) = \min(\{x\}, 1-\{x\})\) is a periodic triangular wave. Visualizing this graph makes it easy to see that it's continuous everywhere but has sharp corners (non-differentiable points) at all integers and half-integers.

Answer 1

The Correct Option is C

Step 1: Understanding the Functions
We are given \(f(x) = x-[x] = \{x\}\), the fractional part function. And \(g(x) = 1 - x + [x] = 1 - (x-[x]) = 1 - \{x\}\). The function \(h(x)\) is the minimum of these two functions: \(h(x) = \min(\{x\}, 1-\{x\})\). The domain is \([-2, 2]\).
Step 2: Analyze Continuity of h(x)
The functions \(f(x)\) and \(g(x)\) are known to be discontinuous at all integer values of \(x\). Let's check the continuity of \(h(x)\) at an integer \(n\). Left Hand Limit (LHL): \(\lim_{x \to n^-} h(x) = \min\left(\lim_{x \to n^-}\{x\}, \lim_{x \to n^-}(1-\{x\})\right) = \min(1, 1-1) = \min(1,0) = 0\). Right Hand Limit (RHL): \(\lim_{x \to n^+} h(x) = \min\left(\lim_{x \to n^+}\{x\}, \lim_{x \to n^+}(1-\{x\})\right) = \min(0, 1-0) = \min(0,1) = 0\). Function value at n: \(h(n) = \min(\{n\}, 1-\{n\}) = \min(0, 1-0) = 0\). Since LHL = RHL = \(h(n)\) = 0, the function \(h(x)\) is continuous at all integer points. Since \(f(x)\) and \(g(x)\) are continuous at non-integer points, \(h(x)\) is also continuous at non-integer points. Thus, \(h(x)\) is continuous everywhere in \([-2, 2]\).
Step 3: Analyze Differentiability of h(x)
The function \(h(x)\) will be non-differentiable at points where the "minimum" switches from one function to the other, i.e., where \(f(x) = g(x)\). \[ \{x\} = 1 - \{x\} \implies 2\{x\} = 1 \implies \{x\} = \frac{1}{2} \] This occurs when \(x = n + \frac{1}{2}\) for any integer \(n\). In the interval \((-2, 2)\), these points are: \(x = -2 + 0.5 = -1.5\) \(x = -1 + 0.5 = -0.5\) \(x = 0 + 0.5 = 0.5\) \(x = 1 + 0.5 = 1.5\) So there are 4 points of non-differentiability of this type. Now, let's check differentiability at the integer points in \((-2, 2)\), which are \(x = -1, 0, 1\). Right Hand Derivative at n: \(\lim_{x \to n^+} \frac{h(x)-h(n)}{x-n}\). For \(x\) just greater than \(n\), \(x-n=\{x\}\) is small, so \(h(x)=\min(\{x\}, 1-\{x\}) = \{x\} = x-n\). \[ RHD = \lim_{x \to n^+} \frac{x-n-0}{x-n} = 1 \] Left Hand Derivative at n: \(\lim_{x \to n^-} \frac{h(x)-h(n)}{x-n}\). For \(x\) just less than \(n\), \(x-(n-1)=\{x\}\) is close to 1, so \(h(x)=\min(\{x\}, 1-\{x\}) = 1-\{x\} = 1-(x-(n-1)) = n-x\). \[ LHD = \lim_{x \to n^-} \frac{n-x-0}{x-n} = \lim_{x \to n^-} \frac{-(x-n)}{x-n} = -1 \] Since LHD \(\neq\) RHD, \(h(x)\) is not differentiable at integers \(x = -1, 0, 1\). Total points of non-differentiability in \((-2, 2)\) are \(\{-1.5, -1, -0.5, 0, 0.5, 1, 1.5\}\), which is a total of 7 points.
Step 4: Final Answer
The function is continuous in \([-2, 2]\) but not differentiable at 7 points in \((-2, 2)\). Since 7 is more than 4, option (C) is the correct description.