Which of the following statements are correct? (A) If , then is continuous everywhere. (B) If , then is continuous everywhere but not differentiable at . (C) Let then is continuous everywhere. (D) Let , then is continuous everywhere but not diffe

Question

Which of the following statements are correct?

(A) If $f : R \to R$ , then $f (x) =$ $| x |$ is continuous everywhere.

(B) If $f : R \to R$ , then $f (x) =$ $| x |$ is continuous everywhere but not differentiable at $x = 0$ .

(C) Let $f : R - {0} \to R,$ then $f (x) = \frac{1}{x}$ is continuous everywhere.

(D) Let $f : R \to R$ , then $f (x) =$ $| x - 1 |$ $+$ $| x - 2 |$ is continuous everywhere but not differentiable at exactly two points.

(E) If $f : R \to R$ , then $f (x) = \cot x$ is continuous everywhere.

Choose the correct answer from the options given below:

Smart Achievers · Accepted Answer

(3)

Ans. (A), (B), (C), (D) only

Explanation:

(A) - A modulus function is continuous everywhere.

(B) - For $f (x) =$ $| x |$ to be differentiable at $x = 0$ ,

$L.H.S. = \lim_{h \to 0} \frac{f (0 - h) - f (0)}{- h}$

$= \lim_{h \to 0} \frac{| 0 - h | - | 0 |}{- h}$

$f (x) =$

$R.H.S. = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h}$

$= \lim_{h \to 0} \frac{| 0 + h | - | 0 |}{h}$

$= \lim_{h \to 0} \frac{h}{h} = 1$

$L.H.S. \neq R.H.S.$

So, $f (x) =$ $| x |$ is not differentiable at $x = 0$

(C) - $f (x) = \frac{1}{x}$ is continuous everywhere except at $x = 0$

(D) - Yes, and those points are $x = 1$ and $x = 2$

(E) - No, $\cot x$ is not continuous everywhere.

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