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प्रश्न
Let \[f\left( x \right) = \left( x + \left| x \right| \right) \left| x \right|\]
विकल्प
f is continuous
f is differentiable for some x
f' is continuous
f'' is continuous
उत्तर
(a) f is continuous
(c) f' is continuous
\[\text{ We have }, \]
\[f\left( x \right) = \left( x + \left| x \right| \right) \left| x \right|\]
\[ = x\left| x \right| + \left( \left| x \right| \right)^2 \]
\[ = x\left| x \right| + x^2 \]
`f(x) = {(2x^2 ,xge0),(0, x<0):}`
\[\text{To check continuity of} f\left( x \right) \text { at x } = 0\]
\[\left( \text {LHL at x } = 0 \right) = \lim_{x \to 0^-} f\left( x \right)\]
\[ = \lim_{x \to 0^-} 0\]
\[ = 0\]
\[\left( \text { RHL at x } = 0 \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[ = \lim_{x \to 0^+} 2 x^2 \]
\[ = 0\]
\[\text { And } f\left( 0 \right) = 0\]
\[\text { Here, LHL = RHL } = f\left( 0 \right)\]
Therefore,f (x) is continous at x = 0
Hence,f(x) is continous everywhere.
\[\text{To check the differentiability of} f\left( x \right) \text { at} x = 0\]
\[\left(\text { LHD at x } = 0 \right) = \lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ = \lim_{x \to 0^-} \frac{0 - 0}{x} = 0\]
\[\left( \text { RHD at x } = 0 \right) = \lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0}\]
\[ = \lim_{x \to 0^-} \frac{2 x^2 - 0}{x}\]
\[ = \lim_{x \to 0^-} \frac{2 x^2 - 0}{x}\]
\[ = \lim_{x \to 0^-} 2x = 0\]
LHD = RHD
Therefore,f(x) is derivative at x = 0
Hence,f(x) is differeentiable everywhere.
f' (x) = `{(4x,xge 0),(0 ,x<0):}`
\[\text{To check continuity of }f'\left( x \right) \text{ at }x = 0\]
\[\left( \text { LHL at x } = 0 \right) = \lim_{x \to 0^-} f'\left( x \right)\]
\[ = \lim_{x \to 0^-} 0\]
\[ = 0\]
\[\left( \text { RHL at x } = 0 \right) = \lim_{x \to 0^+} f'\left( x \right)\]
\[ = \lim_{x \to 0^+} 4x\]
\[ = 0\]
\[\text { And } f'\left( 0 \right) = 0\]
\[\text { Here, LHL = RHL } = f'\left( 0 \right)\]
Therefore,f" (x) is not continous at x = 0
Hence ,f" (x) is not continous everywhere.
f" (x) =` {(4,xge0),(0, x<0):}`
\[\text { To check continuity of } f''\left( x \right) \text { at x }= 0\]
\[\left( \text { LHL at x } = 0 \right) = \lim_{x \to 0^-} f''\left( x \right)\]
\[ = \lim_{x \to 0^-} 0\]
\[ = 0\]
\[\left( \text { RHL at x } = 0 \right) = \lim_{x \to 0^+} f''\left( x \right)\]
\[ = \lim_{x \to 0^+} 4\]
\[ = 4\]
\[\text { Therefore, LHL} \neq \text { RHL } \]
\[\text { Therefore }, f''\left( x \right) \text { is not continuous at x = 0 }\]
\[\text { Hence, } f''\left( x \right) \text { is not continuous everywhere }.\]
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