Q1288: Check whether the function $f(x)=x^{2}|x|$ is differentiable...

Class CBSE Class 12 Mathematics Continuity and Differentiability Q #1288

KNOWLEDGE BASED

UNDERSTAND

2 Marks 2024 AISSCE(Board Exam) VSA

Check whether the function $f(x)=x^{2}|x|$ is differentiable at $x=0$ or not.

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AI Tutor Explanation

Detailed Solution

Step 1: Define the function

The given function is $f(x) = x^2|x|$. We need to check its differentiability at $x=0$.

Step 2: Rewrite the function

We can rewrite the function as follows: $f(x) = \begin{cases} x^3, & \text{if } x \geq 0 \\ -x^3, & \text{if } x < 0 \end{cases}$

Step 3: Find the left-hand derivative (LHD)

The left-hand derivative at $x=0$ is given by: $LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h}$ Since $h < 0$, we use $f(x) = -x^3$. Also, $f(0) = 0$. $LHD = \lim_{h \to 0^-} \frac{-h^3 - 0}{h} = \lim_{h \to 0^-} -h^2 = 0$

Step 4: Find the right-hand derivative (RHD)

The right-hand derivative at $x=0$ is given by: $RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h}$ Since $h > 0$, we use $f(x) = x^3$. Also, $f(0) = 0$. $RHD = \lim_{h \to 0^+} \frac{h^3 - 0}{h} = \lim_{h \to 0^+} h^2 = 0$

Step 5: Compare LHD and RHD

Since $LHD = 0$ and $RHD = 0$, we have $LHD = RHD$ at $x=0$. Therefore, the function is differentiable at $x=0$.

Final Answer: Differentiable at x=0

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the definition of differentiability and apply it to the given function.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concept of differentiability, limits, and how to apply them to a piecewise function.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's knowledge of differentiability and its application.