Q110 / 180 · COMEDK 2019 · Mathematics

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

  • A
    $f(x)$ is discontinuous at $x \simeq a$
  • B
    $f(x)$ is not differentiable at $x$
  • C
    $f(x)$ is differentiable at all $x \geq a$
  • Correct answer
    D
    $f(x)$ is contimuous at all $x < a$

Solution

We have, At $x=a$ $$ \begin{aligned} &\mathrm{LHL}=\lim _{x \rightarrow a}(2 a-x)=2 a-a=a \\ &\mathrm{RHL}=\lim _{x \rightarrow a}(3 x-2 a)=3 a-2 a=a \\ &f(a)=3 a-2 a=a \\ &\because \mathrm{LHL}=\mathrm{RHL}=f(a) . \end{aligned} $$ So, $f(x)$ is continuous at $x=a$. Now, when $x < a, f(x)=2 a-x$, which is continuous for all $x < a$. Again, when $x>a, f(x)=3 x-2 a$, which is continuous for all $x>a$. So, $f(x)$ is continuous for all $x$. Now, at $x=a$ $$ \begin{aligned} \mathrm{LHD} &=\lim _{x \rightarrow a} \frac{(2 a-x)-a}{x-a} \\ &=\lim _{x \rightarrow a} \frac{-(x-a)}{(x-a)}=-1 \\ \text { and } \mathrm{RHD} &=\lim _{x \rightarrow a} \frac{(3 x-2 a)-a}{x-a} \\ &=\lim _{x \rightarrow a} \frac{3(x-a)}{x-a}=3 \\ \therefore \quad \mathrm{LHD} & \neq \mathrm{RHD} \end{aligned} $$ So, $f(x)$ is not differentiable at $x=a$.

Hence, the correct answer is option: 'D'.

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