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(B) Since $f(x)$ is differentiable on $R$,it must be continuous at $x = 0$.For continuity at $x = 0$:$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightar

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$a = 1, b = 0$

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(B) Since $f(x)$ is differentiable on $R$,it must be continuous at $x = 0$.For continuity at $x = 0$:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^+} f(x) = f(0)$$\lim_{x ightarrow 0^-} \frac{\sin x^2}{x} = \lim_{x ightarrow 0^+} (x^2 + ax + b)$Using the limit $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$,we have $\lim_{x ightarrow 0^-} (x \cdot \frac{\sin x^2}{x^2}) = 0 \cdot 1 = 0$.Thus,$0 = b$,so $b = 0$.For differentiability at $x = 0$,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$:$LHD = \lim_{h ightarrow 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h ightarrow 0^+} \frac{\frac{\sin(-h)^2}{-h} - 0}{-h} = \lim_{h ightarrow 0^+} \frac{\sin h^2}{h^2} = 1$.$RHD = \lim_{h ightarrow 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0^+} \frac{h^2 + ah + b - b}{h} = \lim_{h ightarrow 0^+} (h + a) = a$.Equating $LHD$ and $RHD$,we get $a = 1$.Therefore,$a = 1$ and $b = 0$.

Define a function $f: R \rightarrow R$ by $f(x) = \begin{cases} \frac{\sin x^2}{x}, & \text{for } x < 0 \\ x^2 + ax + b, & \text{for } x \geq 0 \end{cases}$. Suppose $f(x)$ is differentiable on $R$. Then,

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