f(x)= begin{cases} 2a-x & text{in } -a

Question

(B) Given,$f(x) = \begin{cases} 2a-x & 	ext{in } -a < x < a \ 3x-2a & 	ext{in } a \leq x \end{cases}$ Check continuity at $x=a$: $LHL = \lim_{x 	o

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$f(x)$ is not differentiable at $x=a$

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(B) Given,$f(x) = \begin{cases} 2a-x & 	ext{in } -a Check continuity at $x=a$: $LHL = \lim_{x 	o a^-} f(x) = \lim_{x 	o a} (2a-x) = 2a-a = a$ $RHL = \lim_{x 	o a^+} f(x) = \lim_{x 	o a} (3x-2a) = 3a-2a = a$ $f(a) = 3(a)-2a = a$ Since $LHL = RHL = f(a)$,the function is continuous at $x=a$. Check differentiability at $x=a$: $LHD = \lim_{h 	o 0} \frac{f(a-h)-f(a)}{-h} = \lim_{h 	o 0} \frac{2a-(a-h)-a}{-h} = \lim_{h 	o 0} \frac{h}{-h} = -1$ $RHD = \lim_{h 	o 0} \frac{f(a+h)-f(a)}{h} = \lim_{h 	o 0} \frac{3(a+h)-2a-a}{h} = \lim_{h 	o 0} \frac{3h}{h} = 3$ Since $LHD 
eq RHD$,the function is not differentiable at $x=a$. Thus,option $B$ is correct.

$f(x)= \begin{cases} 2a-x & \text{in } -a < x < a \\ 3x-2a & \text{in } a \leq x \end{cases}$
Then,which of the following is true?

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$f(x)= \begin{cases} 2a-x & \text{in } -a < x < a \\ 3x-2a & \text{in } a \leq x \end{cases}$ Then,which of the following is true?