Understanding the Function \(f(x)\)

The given function is \(f(x)=\displaystyle\sum_{n=0}^{2k}\left(a_n|x|^n+b_n\ \sin^2x\right)\). We can rewrite this function by separating the terms involving \(|x|^n\) and \(\sin^2x\):

\[f(x) = \sum_{n=0}^{2k} a_n |x|^n + \sum_{n=0}^{2k} b_n \sin^2 x\]

Let \(B = \sum_{n=0}^{2k} b_n\), which is a real constant since \(b_i\) are real constants. Then the function becomes:

\[f(x) = \sum_{n=0}^{2k} a_n |x|^n + B \sin^2 x\]

We need to determine when \(f(x)\) is differentiable at \(x=0\).

Analyzing Differentiability at \(x=0\)

A function \(g(x)\) is differentiable at \(x=0\) if the limit \(\lim_{h \to 0} \frac{g(h) - g(0)}{h}\) exists.

Differentiability of \(B \sin^2 x\) Term

Let \(h_B(x) = B \sin^2 x\). To check differentiability at \(x=0\), we evaluate the limit:

\[\lim_{h \to 0} \frac{h_B(h) - h_B(0)}{h} = \lim_{h \to 0} \frac{B \sin^2 h - B \sin^2 0}{h} = \lim_{h \to 0} \frac{B \sin^2 h}{h}\]

We can rewrite the limit:

\[\lim_{h \to 0} B \frac{\sin h}{h} \sin h = B \left(\lim_{h \to 0} \frac{\sin h}{h}\right) \left(\lim_{h \to 0} \sin h\right) = B (1)(0) = 0\]

Since the limit exists and is finite (equal to 0), the term \(B \sin^2 x\) is always differentiable at \(x=0\), regardless of the value of \(B\). Its derivative at \(x=0\) is 0.

Differentiability of \(a_n |x|^n\) Terms

Let \(g_n(x) = a_n |x|^n\). We need to check the differentiability of each term in the sum \(\sum_{n=0}^{2k} a_n |x|^n\) at \(x=0\).

For \(n=0\), the term is \(g_0(x) = a_0 |x|^0 = a_0\) (for \(x \ne 0\)). At \(x=0\), \(|0|^0\) is typically taken as 1 in this context, so \(g_0(0) = a_0\). This is a constant function, which is differentiable everywhere, including at \(x=0\). The derivative is 0.

For \(n \ge 1\), \(g_n(0) = a_n |0|^n = 0\). The differentiability is determined by the limit:

\[\lim_{h \to 0} \frac{g_n(h) - g_n(0)}{h} = \lim_{h \to 0} \frac{a_n |h|^n - 0}{h} = \lim_{h \to 0} a_n \frac{|h|^n}{h}\]

Case 1: \(n\) is even (\(n=2m\) for \(m \ge 1\))

The term is \(g_{2m}(x) = a_{2m} |x|^{2m} = a_{2m} x^{2m}\). This is a polynomial term.

\[\lim_{h \to 0} a_{2m} \frac{|h|^{2m}}{h} = \lim_{h \to 0} a_{2m} \frac{h^{2m}}{h} = \lim_{h \to 0} a_{2m} h^{2m-1}\]

Since \(n=2m \ge 2\), \(2m-1 \ge 1\). Thus, the limit is \(a_{2m} \cdot 0 = 0\). The term \(a_{2m}|x|^{2m}\) is differentiable at \(x=0\) for all \(m \ge 1\) (i.e., \(n \ge 2\) even), regardless of \(a_{2m}\). Its derivative at \(x=0\) is 0.

Case 2: \(n\) is odd (\(n=2m+1\) for \(m \ge 0\))

The term is \(g_{2m+1}(x) = a_{2m+1} |x|^{2m+1}\).

\[\lim_{h \to 0} a_{2m+1} \frac{|h|^{2m+1}}{h} = \lim_{h \to 0} a_{2m+1} \frac{|h| h^{2m}}{h} = \lim_{h \to 0} a_{2m+1} h^{2m} \frac{|h|}{h}\]

Focus on the Critical Term \(a_1|x|\)

For \(m=0\), \(n=1\). The term is \(g_1(x) = a_1 |x|\). The limit is:

\[\lim_{h \to 0} a_1 h^{2(0)} \frac{|h|}{h} = \lim_{h \to 0} a_1 \frac{|h|}{h}\]

This limit exists if and only if \(a_1 = 0\). If \(a_1 \ne 0\), the left limit (\(h \to 0^-\), \(\frac{|h|}{h}=-1\)) is \(-a_1\) and the right limit (\(h \to 0^+\), \(\frac{|h|}{h}=1\)) is \(a_1\). For the limit to exist, we need \(-a_1 = a_1\), which implies \(2a_1 = 0\), so \(a_1=0\). Thus, \(a_1|x|\) is differentiable at \(x=0\) if and only if \(a_1=0\).

For \(m \ge 1\), \(n = 2m+1 \ge 3\). The limit is \(\lim_{h \to 0} a_{2m+1} h^{2m} \frac{|h|}{h}\).

• As \(h \to 0^+\): \(\lim_{h \to 0^+} a_{2m+1} h^{2m} \frac{h}{h} = \lim_{h \to 0^+} a_{2m+1} h^{2m}\). Since \(m \ge 1\), \(2m \ge 2\), so this limit is 0.
• As \(h \to 0^-\): \(\lim_{h \to 0^-} a_{2m+1} h^{2m} \frac{-h}{h} = \lim_{h \to 0^-} -a_{2m+1} h^{2m}\). Since \(m \ge 1\), \(2m \ge 2\), so this limit is 0.

Since the left and right limits are equal (both 0), the limit exists for \(n \ge 3\) odd, and is equal to 0, regardless of the value of \(a_{2m+1}\). Thus, \(a_{2m+1}|x|^{2m+1}\) is differentiable at \(x=0\) for all \(m \ge 1\) (i.e., \(n \ge 3\) odd).

Combining Terms for \(f(x)\) Differentiability

\(f(x)\) is the sum of \(a_n|x|^n\) terms and \(B \sin^2 x\). The sum of functions differentiable at a point is differentiable at that point.

We found that:

• \(a_0\) is differentiable at \(x=0\).
• \(a_{2m}|x|^{2m}\) is differentiable at \(x=0\) for \(m \ge 1\).
• \(a_{2m+1}|x|^{2m+1}\) is differentiable at \(x=0\) for \(m \ge 1\).
• \(a_1|x|\) is differentiable at \(x=0\) if and only if \(a_1=0\).
• \(B \sin^2 x\) is differentiable at \(x=0\).

Therefore, \(f(x)\) is differentiable at \(x=0\) if and only if the sum of the terms that are not guaranteed to be differentiable is differentiable. The only such term is \(a_1|x|\). Thus, \(f(x)\) is differentiable at \(x=0\) if and only if \(a_1 = 0\).

Evaluating the Given Options

We need to find which option states a condition under which \(f(x)\) is differentiable at \(x=0\).

Option 1 Analysis

The condition is "all a_2m+1 = 0, 0 ≤ m ≤ k - 1". The indices \(2m+1\) for \(0 \le m \le k-1\) are \(1, 3, 5, \dots, 2k-1\). So the condition is \(a_1=0, a_3=0, a_5=0, \dots, a_{2k-1}=0\).

If this condition holds, then specifically \(a_1=0\). As we determined that \(f(x)\) is differentiable at \(x=0\) if and only if \(a_1=0\), having \(a_1=0\) is sufficient for differentiability. The additional conditions \(a_3=0, \dots, a_{2k-1}=0\) are not necessary for differentiability at \(x=0\), but if they hold, then \(a_1=0\) also holds, and thus \(f(x)\) is differentiable. So, this is a sufficient condition.

Option 2 Analysis

The condition is "all a_2m = 0, 0 ≤ m ≤ k". The indices \(2m\) for \(0 \le m \le k\) are \(0, 2, 4, \dots, 2k\). So the condition is \(a_0=0, a_2=0, \dots, a_{2k}=0\). If these coefficients are zero, \(f(x)\) becomes \(\sum_{m=0}^{k-1} a_{2m+1} |x|^{2m+1} + B \sin^2 x\). This is \(a_1|x| + a_3|x|^3 + \dots + a_{2k-1}|x|^{2k-1} + B \sin^2 x\). This function is differentiable at \(x=0\) if and only if \(a_1=0\). The condition \(a_{2m}=0\) does not guarantee \(a_1=0\). Thus, Option 2 is not a sufficient condition.

Option 3 Analysis

The statement is "not differentiable at x = 0 for all a_i's and b_i's". This claims that \(f(x)\) is never differentiable at \(x=0\). This is false, because if we take \(a_1=0\) (and any values for other coefficients), \(f(x)\) becomes differentiable at \(x=0\).

Option 4 Analysis

The statement is "differentiable at x = 0 for all a_i's and b_i's". This claims that \(f(x)\) is always differentiable at \(x=0\). This is false, because if we take \(a_1=1\) (and other \(a_i=0\) for \(i \ne 1\), and all \(b_i=0\)), the function is \(f(x) = |x|\) (assuming \(2k \ge 1\)), which is not differentiable at \(x=0\).

Conclusion

Based on our analysis, \(f(x)\) is differentiable at \(x=0\) if and only if \(a_1=0\). Option 1 provides the condition \(a_1=0, a_3=0, \dots, a_{2k-1}=0\). This set of conditions includes \(a_1=0\), and as such, it is a sufficient condition for \(f(x)\) to be differentiable at \(x=0\). The other options state conditions that do not guarantee differentiability or make false claims about the function's differentiability.

Thus, the correct statement is that \(f(x)\) is differentiable at x = 0 for all a_2m+1 = 0, 0 ≤ m ≤ k - 1.

Revision Table: Key Concepts for Differentiability

Additional Information: Properties of Absolute Value and Differentiability

The absolute value function \(|x|\) is a classic example of a continuous function that is not differentiable at \(x=0\). This non-differentiability arises from the sharp corner or cusp at \(x=0\), where the slope changes abruptly from -1 (for \(x<0\)) to 1 (for \(x>0\)).

However, functions involving higher powers of \(|x|\) can be differentiable at \(x=0\).

• Consider \(g(x) = |x|^n\).
• If \(n\) is even, \(g(x) = x^n\), which is a polynomial and differentiable everywhere.
• If \(n\) is odd and \(n \ge 3\), \(g(x) = |x|^n = x^{n-1}|x|\). Since \(n-1 \ge 2\), the factor \(x^{n-1}\) goes to 0 faster than \(|x|\) causes the difference quotient to become unbounded or have differing left/right limits. The limit \(\lim_{h \to 0} \frac{|h|^n}{h} = \lim_{h \to 0} |h| h^{n-2}\) is 0 for \(n \ge 3\). This means \(|x|^n\) is differentiable at \(x=0\) with derivative 0 for odd \(n \ge 3\).

The term \(\sin^2 x\) is differentiable everywhere because \(\sin x\) is differentiable everywhere, and the square of a differentiable function is differentiable (using the chain rule).

When combining functions, the sum of differentiable functions is differentiable. Non-differentiability typically arises from terms that are individually not differentiable, like \(|x|\), unless those terms cancel out or their coefficients are zero.