If the function f(x) = begin{cases} frac{cos | vedclass

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(B) For the function to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = -1$. Evaluating the limit: $\lim_{x 	o 0} \frac{\cos ax -

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Arithmetic progression

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(B) For the function to be continuous at $x = 0$,we must have $\lim_{x 	o 0} f(x) = f(0) = -1$. Evaluating the limit: $\lim_{x 	o 0} \frac{\cos ax - \cos bx}{\cos cx - \cos bx}$. Using the expansion $\cos 	heta \approx 1 - \frac{	heta^2}{2}$ for small $	heta$: $\lim_{x 	o 0} \frac{(1 - \frac{a^2x^2}{2}) - (1 - \frac{b^2x^2}{2})}{(1 - \frac{c^2x^2}{2}) - (1 - \frac{b^2x^2}{2})} = \lim_{x 	o 0} \frac{\frac{b^2x^2}{2} - \frac{a^2x^2}{2}}{\frac{b^2x^2}{2} - \frac{c^2x^2}{2}} = \frac{b^2 - a^2}{b^2 - c^2}$. Setting this equal to $-1$: $\frac{b^2 - a^2}{b^2 - c^2} = -1$. $b^2 - a^2 = -(b^2 - c^2) \implies b^2 - a^2 = c^2 - b^2$. $2b^2 = a^2 + c^2$. This condition implies that $a^2, b^2, c^2$ are in Arithmetic progression.

If the function $f(x) = \begin{cases} \frac{\cos ax - \cos bx}{\cos cx - \cos bx} & , x \neq 0 \\ -1 & , x = 0 \end{cases}$ is continuous at $x = 0$,then $a^2, b^2, c^2$ are in

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