If f(x) = operatorname{sgn} left( 3cos x - fr | vedclass

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Question

(C) The signum function $\operatorname{sgn}(u)$ is discontinuous at $u = 0$. For $f(x) = \operatorname{sgn}\left(3 \cos x - \frac{a}{3}\right)$ to be

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) The signum function $\operatorname{sgn}(u)$ is discontinuous at $u = 0$. For $f(x) = \operatorname{sgn}\left(3 \cos x - \frac{a}{3}ight)$ to be continuous for all $x$,the argument $3 \cos x - \frac{a}{3}$ must never be equal to $0$. This implies $3 \cos x - \frac{a}{3} 
eq 0$ for all $x$,which means $\cos x 
eq \frac{a}{9}$. Since the range of $\cos x$ is $[-1, 1]$,the equation $\cos x = \frac{a}{9}$ must have no solution. This occurs if $\frac{a}{9} > 1$ or $\frac{a}{9} For $\frac{a}{9} > 1$,we have $a > 9$. For $\frac{a}{9} We are looking for the least positive integral value of $a$. Since $a > 9$,the smallest integer $a$ is $10$.

If $f(x) = \operatorname{sgn} \left( 3\cos x - \frac{a}{3} \right)$ is continuous for all $x$,then the least positive integral value of $a$ is - (where $\operatorname{sgn}(x)$ denotes the signum function of $x$)

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