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CBSE Class 12 Math 2023 All Sets Solved Paper

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Question : 11 of 20
Marks: +1, -0
The function f(x) = x | x | is:
Solution:  
f(x)={x2x≥0−x2x<0f(x)=\begin{cases} x^2 & x \geq 0 \\ -x^2 & x < 0 \end{cases}
At x=0x=0
lim⁡x→0−−x2=0 and lim⁡x→0+x2=0\lim\limits_{x \rightarrow 0^{-}} -x^2 = 0 \text{ and } \lim\limits_{x \rightarrow 0^{+}} x^2 = 0
As lim⁡x→0−f(x)=lim⁡x→0+f(x)=f(0)\lim\limits_{x \rightarrow 0^{-}} f(x) = \lim\limits_{x \rightarrow 0^{+}} f(x) = f(0)
∴f(x)\therefore \text{f}(\text{x}) is continuous at x=0\text{x}=0
f′(x)={2xx>0−2xx<0f'(x)=\begin{cases} 2x & x>0 \\ -2x & x<0 \end{cases}
At x=0x=0
ddxf(x)∣0−=ddxf(x)∣0+=0\left.\frac{d}{dx} f(x)\right|_{0^{-}} = \left.\frac{d}{dx} f(x)\right|_{0^{+}} = 0
∴f(x)\therefore \text{f}(\text{x}) is differentiable at x=0\text{x}=0
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