CUET PG 2022 — Mathematics PYQ

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Question

CUET PG 2022 — Mathematics PYQ

CUET PG | Mathematics | 2022

Let $f (x) = ∣∣ x ∣ - 1∣$ then point (s) where $f (x)$ is not differentiable is (are):

Choose the correct answer:

Explanation

1. Identify Candidate Points

A function involving absolute values is typically non-differentiable where the argument of the absolute value is zero.

Inner absolute value: $|x|$

$x = 0$
Outer absolute value: $||x| - 1|$

$|x| - 1 = 0 \implies |x| = 1 \implies x = 1, -1$

So, the critical points to check are $x = -1, 0, 1$ .

2. Define the Piecewise Function

We can rewrite $f(x)$ by breaking it into intervals based on the critical points:

f(x) = \begin{cases} -( (-x) - 1 ) = x + 1 & \text{if } x < -1 \\ -x - 1 & \text{is not correct, let's re-evaluate:} \end{cases}

Correct Piecewise Form:

If $x \le -1$ , then $|x| = -x$ and $-x-1 \ge 0$ , so $f(x) = -x - 1$
If $-1 < x \le 0$ , then $|x| = -x$ and $-x-1 < 0$ , so $f(x) = 1 + x$
If $0 < x \le 1$ , then $|x| = x$ and $x-1 \le 0$ , so $f(x) = 1 - x$
If $x > 1$ , then $|x| = x$ and $x-1 > 0$ , so $f(x) = x - 1$

3. Check Derivatives (Slopes)

Now, find the derivative $f'(x)$ for each interval:

f'(x) = \begin{cases} -1 & \text{if } x < -1 \\ 1 & \text{if } -1 < x < 0 \\ -1 & \text{if } 0 < x < 1 \\ 1 & \text{if } x > 1 \end{cases}

At each critical point, compare the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD):

At $x = -1$ : LHD = $-1$ , RHD = $1$ . Since $LHD \neq RHD$ , it is not differentiable.
At $x = 0$ : LHD = $1$ , RHD = $-1$ . Since $LHD \neq RHD$ , it is not differentiable.
At $x = 1$ : LHD = $-1$ , RHD = $1$ . Since $LHD \neq RHD$ , it is not differentiable.

Final Answer:

The points where $f(x)$ is not differentiable are:

$x \in {- 1, 0, 1}$