Continuity and Differentiability — Class 12 Maths Solution

exemplar sa SA NCERT Exemp. Ex.5.3 ,Q.10,Page 107
Question

$f(x) = |x| + |x - 1|$ at $x = 1$

Step-by-step Solution

We have, $f(x) = |x| + |x - 1|$ at $x = 1$
At $x = 1,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {1^ - }} [|x| + |x - 1|]$
$= \mathop {\lim }\limits_{h \to 0} [|1 - h| + |1 - h - 1|] = 1 + 0 = 1$
and ${\rm{RHL}} = \mathop {\lim }\limits_{x \to {1^ - }} [|x| + |x - 1|]$
$= \mathop {\lim }\limits_{h \to 0} [|1 + h| + |1 + h - 1|] = 1 + 0 = 1$

and $f(1) = |1| + |0| = 1$

therefore,${\rm{LHL}} = {\rm{RHL}} = f(1)$

Therefore we can say that $f(x)$ is continuous at $x = 1$.

Tip: Every modulus function is a continuous function at any real point.

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