1. **Problem statement:** I) Find real values of $a$ and $b$ such that the function $$f(x) = \begin{cases} a x + \frac{b^{x-2}}{x+1} - 1, & x \leq 1 \\ b \sqrt{2x - 1} - a e^{x-1}, & x > 1 \end{cases}$$ is differentiable on $\mathbb{R}$. II) Study differentiability of $$f(x) = \begin{cases} x^2 \cos\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$$ --- 2. **Key formulas and rules:** - Differentiability at a point requires continuity and matching left and right derivatives. - For piecewise functions, check continuity and derivative limits at the boundary point. --- ### Part I 3. **Continuity at $x=1$:** Set left and right limits equal: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$ Left side: $$f(1) = a(1) + \frac{b^{1-2}}{1+1} - 1 = a + \frac{b^{-1}}{2} - 1 = a + \frac{1}{2b} - 1$$ Right side: $$f(1) = b \sqrt{2(1) - 1} - a e^{1-1} = b \sqrt{1} - a e^0 = b - a$$ Equate: $$a + \frac{1}{2b} - 1 = b - a$$ Rearranged: $$2a + \frac{1}{2b} - 1 = b$$ --- 4. **Differentiability at $x=1$:** Compute left derivative: $$f'_-(1) = \frac{d}{dx} \left(a x + \frac{b^{x-2}}{x+1} - 1\right) \Big|_{x=1}$$ Derivatives: - $\frac{d}{dx} (a x) = a$ - For $g(x) = \frac{b^{x-2}}{x+1}$, use quotient rule: $$g'(x) = \frac{(b^{x-2} \ln b)(x+1) - b^{x-2} \cdot 1}{(x+1)^2} = \frac{b^{x-2} ((x+1) \ln b - 1)}{(x+1)^2}$$ Evaluate at $x=1$: $$g'(1) = \frac{b^{-1} (2 \ln b - 1)}{4} = \frac{(1/b)(2 \ln b - 1)}{4} = \frac{2 \ln b - 1}{4b}$$ So, $$f'_-(1) = a + \frac{2 \ln b - 1}{4b}$$ --- 5. **Right derivative:** $$f'_+(1) = \frac{d}{dx} \left(b \sqrt{2x - 1} - a e^{x-1}\right) \Big|_{x=1}$$ Derivatives: - $\frac{d}{dx} b \sqrt{2x - 1} = b \cdot \frac{1}{2 \sqrt{2x - 1}} \cdot 2 = \frac{b}{\sqrt{2x - 1}}$ - $\frac{d}{dx} (-a e^{x-1}) = -a e^{x-1}$ Evaluate at $x=1$: $$f'_+(1) = \frac{b}{\sqrt{1}} - a e^0 = b - a$$ --- 6. **Set derivatives equal:** $$a + \frac{2 \ln b - 1}{4b} = b - a$$ Rearranged: $$2a + \frac{2 \ln b - 1}{4b} = b$$ --- 7. **Solve system:** From continuity: $$2a + \frac{1}{2b} - 1 = b$$ From differentiability: $$2a + \frac{2 \ln b - 1}{4b} = b$$ Subtract second from first: $$\left(2a + \frac{1}{2b} - 1\right) - \left(2a + \frac{2 \ln b - 1}{4b}\right) = b - b$$ $$\frac{1}{2b} - 1 - \frac{2 \ln b - 1}{4b} = 0$$ Multiply both sides by $4b$: $$2 - 4b - (2 \ln b - 1) = 0$$ $$2 - 4b - 2 \ln b + 1 = 0$$ $$3 - 4b - 2 \ln b = 0$$ Rearranged: $$4b + 2 \ln b = 3$$ Divide by 2: $$2b + \ln b = \frac{3}{2}$$ This transcendental equation can be solved numerically. --- 8. **Numerical approximation:** Try $b=1$: $$2(1) + \ln 1 = 2 + 0 = 2 > 1.5$$ Try $b=0.5$: $$2(0.5) + \ln 0.5 = 1 - 0.693 = 0.307 < 1.5$$ Try $b=0.7$: $$1.4 + \ln 0.7 \approx 1.4 - 0.357 = 1.043 < 1.5$$ Try $b=0.8$: $$1.6 + \ln 0.8 \approx 1.6 - 0.223 = 1.377 < 1.5$$ Try $b=0.85$: $$1.7 + \ln 0.85 \approx 1.7 - 0.163 = 1.537 > 1.5$$ By interpolation, $b \approx 0.83$. --- 9. **Find $a$:** Use continuity equation: $$2a + \frac{1}{2b} - 1 = b$$ Plug $b=0.83$: $$2a + \frac{1}{2 \times 0.83} - 1 = 0.83$$ $$2a + 0.602 - 1 = 0.83$$ $$2a - 0.398 = 0.83$$ $$2a = 1.228$$ $$a = 0.614$$ --- ### Part II 10. **Function:** $$f(x) = \begin{cases} x^2 \cos\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$$ Check differentiability at $x=0$. 11. **Continuity at 0:** $$\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right)$$ Since $|\cos(1/x)| \leq 1$, $$|x^2 \cos(1/x)| \leq x^2 \to 0$$ So, $f$ is continuous at 0. 12. **Derivative at 0:** $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \cos(1/h) - 0}{h} = \lim_{h \to 0} h \cos(1/h)$$ Since $|\cos(1/h)| \leq 1$, $$|h \cos(1/h)| \leq |h| \to 0$$ So, $$f'(0) = 0$$ 13. **Differentiability for $x \neq 0$:** $f$ is differentiable as product of differentiable functions. 14. **Conclusion:** $f$ is differentiable everywhere on $\mathbb{R}$ with $$f'(0) = 0$$ --- **Final answers:** $$a \approx 0.614, \quad b \approx 0.83$$ and $f$ in part II is differentiable on $\mathbb{R}$ with $f'(0) = 0$.