Let f, g and h be the real valued functions defined on R as f(x) = arrayll x|x|, & x 0 array.

Given: the solution concludes that the correct option is A. Find: Which statement about continuity at x = 0 is correct. From the extracted working, the left-hand limit and right-hand limit being checked are: LHL = k 0, k0 g(h(-k)) and RHL = k 0, k0 g(h(k)) Using the shown fact, f(x) = -1 x < 0 so the left-side evaluation gives g(-1) = 1 Also, using the shown fact, f(x) = 1 x 0 so the right-side evaluation also gives g(-1) = 1 and hence the limit value obtained from both sides is the same. Therefore, as per the provided the solution, the correct option is A.

Let f, g and h be the real valued functions defined on R as f(x)… — Solution | JEE Mathematics

MCQMediumJEE 2023Continuity

Let $f$ , $g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as $f(x) = \left\lbrace \begin{array}{ll} \frac{x}{|x|}, & x \neq 0\end{array} \right.$

A
$f$ is continuous at $x = 0$
B
$g$ is continuous at $x = 0$
C
$h$ is continuous at $x = 0$
D
$f, g, h$ are continuous at $x = 0$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: the solution concludes that the correct option is A.

Find: Which statement about continuity at $x = 0$ is correct.

From the extracted working, the left-hand limit and right-hand limit being checked are:

\text{LHL} = \lim_{k \to 0,\; k>0} g(h(-k))

and

\text{RHL} = \lim_{k \to 0,\; k>0} g(h(k))

Using the shown fact,

f(x) = -1 \quad \forall x < 0

so the left-side evaluation gives

g(-1) = 1

Also, using the shown fact,

f(x) = 1 \quad \forall x > 0

so the right-side evaluation also gives

g(-1) = 1

and hence the limit value obtained from both sides is the same.

Therefore, as per the provided the solution, the correct option is A.

Common mistakes

Using the incomplete question statement as if all definitions of $g$ and $h$ were available. This is wrong because the given question is truncated. Use the solution as the authority for the final answer extraction.
Assuming $f(x)=\frac{x}{|x|}$ is continuous at $x=0$ directly. This is wrong because $f$ is not even defined at $x=0$ in the displayed definition. Do not infer continuity without the full original question context.
Confusing one-sided values of the sign-type function near $0$ . For $x<0$ , $\frac{x}{|x|}=-1$ and for $x>0$ , $\frac{x}{|x|}=1$ . Keep left-hand and right-hand behavior separate.

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Answer

Step-by-step solution

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