Assertion (A): f(x)=|x-a|+|x-b| is continuous | vedclass

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(B) $1$. The function $f(x) = |x-a| + |x-b|$ is a sum of two absolute value functions. Since $|x-c|$ is continuous for all $x \in R$, the sum of two c

Vedclass · Accepted Answer

$A$ is true, $R$ is true but $R$ is not the correct explanation for $A$

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(B) $1$. The function $f(x) = |x-a| + |x-b|$ is a sum of two absolute value functions. Since $|x-c|$ is continuous for all $x \in R$, the sum of two continuous functions is also continuous on $R$. Thus, Assertion $(A)$ is true.$2$. The function $g(x) = \frac{|x-\alpha|}{x-\alpha}$ is defined as $1$ for $x > \alpha$ and $-1$ for $x $3$. While both statements are true, the continuity of $f(x)$ is a property of absolute value functions, and the continuity of $g(x)$ is a separate property of rational-like expressions involving absolute values. $R$ does not explain why $f(x)$ is continuous. Therefore, $R$ is not the correct explanation for $A$.

Assertion $(A)$: $f(x)=|x-a|+|x-b|$ is continuous on $R$. Reason $(R)$: $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in R-\{\alpha\}$. The correct option among the following is:

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