NIMCET 2023 — Mathematics PYQ

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Question

NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

A real valued function f is defined as f(x)= ${- 1 x - 1 am p; - 2 \leq x \leq 0 am p; 0 \leq x \leq 2$ . Which of the following statement is FALSE?

Choose the correct answer:

Explanation

1. Analysis of $f(|x|)$ at $x = 0$

f(|x|) = |x| - 1

f(|x|) = \begin{cases} -x - 1, & x < 0 \\ x - 1, & x \geq 0 \end{cases}

Continuity:

\lim_{x \to 0^-} f(|x|) = 0 - 1 = -1

\lim_{x \to 0^+} f(|x|) = 0 - 1 = -1

f(|0|) = -1

$\implies$ Continuous at $x = 0$ .

Differentiability:

LHD = \frac{d}{dx}(-x - 1) = -1

RHD = \frac{d}{dx}(x - 1) = 1

$LHD \neq RHD \implies$ Not differentiable at $x = 0$ .

2. Analysis of $|f(x)|$ at $x = 0$

|f(x)| = \begin{cases} |-1| = 1, & -2 \leq x \leq 0 \\ |x - 1|, & 0 \leq x \leq 2 \end{cases}

Continuity:

\lim_{x \to 0^-} |f(x)| = 1

\lim_{x \to 0^+} |f(x)| = |0 - 1| = 1

$\implies$ Continuous at $x = 0$ .

Differentiability:

LHD = \frac{d}{dx}(1) = 0

RHD = \frac{d}{dx}(1 - x) = -1 \quad (\text{since } x \to 0^+, x-1 < 0)

$LHD \neq RHD \implies$ Not differentiable at $x = 0$ .

Conclusion:

Statement (B) is FALSE.

Correct Option: (B)