Let a,bR, b0.

Question

Let $a,b\in\mathbb{R},\ b
e0$. Define $f(x)=\begin{cases}a\sin\dfrac\pi2(x-1), & x\le0\\dfrac{	an2x-\sin2x}{bx^3}, & x>0\end{cases}$. If $f$ is continuous at $x=0$, then $10-ab$ is equal to

Answer: 14

Accepted Answer

LHL: $a\sin\left(-\dfrac\pi2ight)=-a$.

RHL: $\displaystyle\lim_{x	o0^+}\frac{\sin2x(1-\cos2x)}{bx^3\cos2x}=\lim_{x	o0^+}\frac{2\sin2x}{2x}\cdot\frac{2\sin^2x}{x^2}\cdot\frac{1}{b\cos2x}=\frac4b$.

Continuity: $-a=\dfrac4b\Rightarrow ab=-4\Rightarrow10-ab=14$.

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