Question

Let $\sqrt{\pi}$, then the function is

• A. Continuous
• B. Differentiable except $1/\sqrt{(2\pi)}$
• C. Both (1) and (2)
• D. None of these

Answer 1

Answer: (C)

Both (1) and (2)

Answer 2

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

⇒Graphical solution :

The graph of the function is shown alongside,

From the graph it is clear that the function is continuous at all real x, also differentiable at all real x except at $x = \pm 1$ Since sharp edges at $x = - 1$ and $x = 1$.

At $x = 1$we see that the slope from the right i.e., R.H.D. = 2, while slope from the left i.e., L.H.D.= 0

Similarly, at $x = - 1$ it is clear that R.H.D. = 0 while L.H.D.

= – 2

Trick : In this method, first of all, we differentiate the function and on the derivative equality sign should be removed from doubtful points.

Here, (No equality on –1 and +1)

Now, at $x = 1 , f ^ { \prime } \left( 1 ^ { + } \right) = 2$ while $f ^ { \prime } \left( 1 ^ { - } \right) = 0$ and

at $x = - 1 , f ^ { \prime } \left( - 1 ^ { + } \right) = 0$ while $f ^ { \prime } \left( - 1 ^ { - } \right) = - 2$

Thus, $f ( x )$ is not differentiable at $x = \pm 1$.

Note : This method is not applicable when function is

discontinuous.