Key Concepts and Formulas

• Existence of Second Derivative: For $f''(a)$ to exist, the function $f$ must be twice differentiable at $a$. This implies that $f'(x)$ must exist in an open interval containing $a$, and $f'(a)$ must exist. Furthermore, $f''(a)$ is defined as the limit of the difference quotient for $f'(x)$ at $a$: $f''(a) = \lim_{h \to 0} \frac{f'(a+h) - f'(a)}{h}$ For a piecewise function, we need to check the differentiability from both sides at the point where the definition changes.
• Differentiation Rules: Standard differentiation rules for power functions, trigonometric functions (sine and cosine), and the product rule are required.
• Limit of Trigonometric Functions: We will use the fact that $\lim_{x \to 0} x^n \sin(1/x) = 0$ and $\lim_{x \to 0} x^n \cos(1/x) = 0$ for $n > 0$.

Step-by-Step Solution

Step 1: Define the function and analyze its continuity at x = 0. The function is given by: f\left( x \right) = \left\{ {\matrix{ {{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr {0,} & {x = 0} \cr {{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr } } \right. For $f''(0)$ to exist, $f(x)$ must be continuous at $x=0$. We check the limit as $x \to 0^-$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^5 \sin(1/x) + 5x^2) = 0 \cdot 0 + 5 \cdot 0 = 0$. We check the limit as $x \to 0^+$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^5 \cos(1/x) + \lambda x^2) = 0 \cdot 0 + \lambda \cdot 0 = 0$. Since $f(0) = 0$, the function is continuous at $x=0$.

Step 2: Calculate the first derivative, $f'(x)$, for $x < 0$ and $x > 0$. For $x < 0$: $f'(x) = \frac{d}{dx}(x^5 \sin(1/x) + 5x^2)$ Using the product rule for $x^5 \sin(1/x)$: $\frac{d}{dx}(x^5 \sin(1/x)) = 5x^4 \sin(1/x) + x^5 \cos(1/x) \cdot (-1/x^2) = 5x^4 \sin(1/x) - x^3 \cos(1/x)$. So, for $x < 0$, $f'(x) = 5x^4 \sin(1/x) - x^3 \cos(1/x) + 10x$.

For $x > 0$: $f'(x) = \frac{d}{dx}(x^5 \cos(1/x) + \lambda x^2)$ Using the product rule for $x^5 \cos(1/x)$: $\frac{d}{dx}(x^5 \cos(1/x)) = 5x^4 \cos(1/x) + x^5 (-\sin(1/x)) \cdot (-1/x^2) = 5x^4 \cos(1/x) + x^3 \sin(1/x)$. So, for $x > 0$, $f'(x) = 5x^4 \cos(1/x) + x^3 \sin(1/x) + 2\lambda x$.

Step 3: Calculate the first derivative at $x=0$, $f'(0)$. We use the definition of the derivative: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$. For $h < 0$: $\lim_{h \to 0^-} \frac{f(h)}{h} = \lim_{h \to 0^-} \frac{h^5 \sin(1/h) + 5h^2}{h} = \lim_{h \to 0^-} (h^4 \sin(1/h) + 5h) = 0 \cdot 0 + 0 = 0$. For $h > 0$: $\lim_{h \to 0^+} \frac{f(h)}{h} = \lim_{h \to 0^+} \frac{h^5 \cos(1/h) + \lambda h^2}{h} = \lim_{h \to 0^+} (h^4 \cos(1/h) + \lambda h) = 0 \cdot 0 + \lambda \cdot 0 = 0$. Since the left-hand limit and the right-hand limit are equal, $f'(0) = 0$.

Step 4: Calculate the second derivative, $f''(x)$, for $x < 0$ and $x > 0$. For $x < 0$: $f'(x) = 5x^4 \sin(1/x) - x^3 \cos(1/x) + 10x$. $f''(x) = \frac{d}{dx}(5x^4 \sin(1/x) - x^3 \cos(1/x) + 10x)$. Differentiating $5x^4 \sin(1/x)$: $20x^3 \sin(1/x) + 5x^4 \cos(1/x) \cdot (-1/x^2) = 20x^3 \sin(1/x) - 5x^2 \cos(1/x)$. Differentiating $-x^3 \cos(1/x)$: $-3x^2 \cos(1/x) - x^3 (-\sin(1/x)) \cdot (-1/x^2) = -3x^2 \cos(1/x) - x \sin(1/x)$. So, for $x < 0$, $f''(x) = (20x^3 \sin(1/x) - 5x^2 \cos(1/x)) - (3x^2 \cos(1/x) + x \sin(1/x)) + 10$ $f''(x) = 20x^3 \sin(1/x) - 6x^2 \cos(1/x) - x \sin(1/x) + 10$.

For $x > 0$: $f'(x) = 5x^4 \cos(1/x) + x^3 \sin(1/x) + 2\lambda x$. $f''(x) = \frac{d}{dx}(5x^4 \cos(1/x) + x^3 \sin(1/x) + 2\lambda x)$. Differentiating $5x^4 \cos(1/x)$: $20x^3 \cos(1/x) + 5x^4 (-\sin(1/x)) \cdot (-1/x^2) = 20x^3 \cos(1/x) + 5x^2 \sin(1/x)$. Differentiating $x^3 \sin(1/x)$: $3x^2 \sin(1/x) + x^3 (\cos(1/x)) \cdot (-1/x^2) = 3x^2 \sin(1/x) - x \cos(1/x)$. So, for $x > 0$, $f''(x) = (20x^3 \cos(1/x) + 5x^2 \sin(1/x)) + (3x^2 \sin(1/x) - x \cos(1/x)) + 2\lambda$ $f''(x) = 20x^3 \cos(1/x) + 8x^2 \sin(1/x) - x \cos(1/x) + 2\lambda$.

Step 5: Calculate the second derivative at $x=0$, $f''(0)$. We use the definition of the second derivative, which requires the first derivative to be differentiable at $x=0$. $f''(0) = \lim_{h \to 0} \frac{f'(0+h) - f'(0)}{h} = \lim_{h \to 0} \frac{f'(h)}{h}$.

For $h < 0$: $\lim_{h \to 0^-} \frac{f'(h)}{h} = \lim_{h \to 0^-} \frac{5h^4 \sin(1/h) - h^3 \cos(1/h) + 10h}{h}$ $= \lim_{h \to 0^-} (5h^3 \sin(1/h) - h^2 \cos(1/h) + 10)$ $= 5 \cdot 0 \cdot 0 - 0 \cdot 0 + 10 = 10$. So, $f''(0^-) = 10$.

For $h > 0$: $\lim_{h \to 0^+} \frac{f'(h)}{h} = \lim_{h \to 0^+} \frac{5h^4 \cos(1/h) + h^3 \sin(1/h) + 2\lambda h}{h}$ $= \lim_{h \to 0^+} (5h^3 \cos(1/h) + h^2 \sin(1/h) + 2\lambda)$ $= 5 \cdot 0 \cdot 0 + 0 \cdot 0 + 2\lambda = 2\lambda$. So, $f''(0^+) = 2\lambda$.

Step 6: Equate the left and right limits of the second derivative to find $\lambda$. For $f''(0)$ to exist, the left-hand limit and the right-hand limit of the difference quotient for $f'(x)$ must be equal. $f''(0^-) = f''(0^+)$ $10 = 2\lambda$ $\lambda = \frac{10}{2}$ $\lambda = 5$.

Common Mistakes & Tips

• Incorrectly applying limits: When differentiating functions like $x^n \sin(1/x)$ or $x^n \cos(1/x)$, remember that $\sin(1/x)$ and $\cos(1/x)$ do not have limits as $x \to 0$. However, when multiplied by $x^n$ with $n > 0$, their limits become 0.
• Forgetting to check continuity and first differentiability: For the second derivative to exist, the function must first be continuous and then differentiable at the point in question. Ensure $f(0)$ and $f'(0)$ are correctly evaluated.
• Algebraic errors in differentiation: Differentiating terms involving $1/x$ can be prone to sign errors or mistakes with the chain rule. Double-check each step of the differentiation process.

Summary

To ensure the existence of the second derivative $f''(0)$, we first verified the continuity of the function at $x=0$. Then, we calculated the first derivative $f'(x)$ for $x<0$ and $x>0$, and confirmed that $f'(0)=0$. Subsequently, we computed the limits of $\frac{f'(h)}{h}$ as $h \to 0^-$ and $h \to 0^+$. For $f''(0)$ to exist, these two limits must be equal. By equating the left-hand limit (which yielded 10) and the right-hand limit (which yielded $2\lambda$), we solved for $\lambda$ and found it to be 5.

The final answer is \boxed{5}.