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Relation between Continuity and Differentiability

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नीचे दिए गए प्रश्न के लिए अनुच्छेद

Let $f (x) = {\begin{matrix} x | x |; & - \infty \leq x \leq 1 2 x^{3} - 1; & x > 1 \end{matrix}$

Choose the correct answer.

माना $f (x) = {\begin{matrix} x | x |; & - \infty \leq x \leq 1 2 x^{3} - 1; & x > 1 \end{matrix}$

सही उत्तर का चयन कीजिए।

f(x) is discontinuous at
f(x) असतत है

A

x = 1

x = 1 पर

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B

x = –1

x = –1 पर

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C

x = 0

x = 0 पर

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D

No value of x

x के किसी भी मान पर नहीं

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Solution

The correct option is D No value of x

x के किसी भी मान पर नहीं
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Similar questions

f (x) = {\begin{matrix} x | x |; & - \infty \leq x \leq 1 2 x^{3} - 1; & x > 1 \end{matrix}

Choose the correct answer.

माना

f (x) = {\begin{matrix} x | x |; & - \infty \leq x \leq 1 2 x^{3} - 1; & x > 1 \end{matrix}

सही उत्तर का चयन कीजिए।

f(x) is non-differentiable at
f(x) अन-अवकलनीय है

f (x)

[- 2, 2]

, and is given by

f (x) = {\begin{matrix} - 1, - 2 \leq x \leq 0 x - 1, 0 \leq x \leq 2 \end{matrix} a n d g (x) = f (| x |) + | f (x) |

g (x)

g (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 (x + 1), & - \infty < x \leq - 1 \sqrt{1 - x^{2}}, & - 1 < x < 1 | x + 1 |, & 1 \leq x < \infty \end{matrix}

F (x) = \frac{1}{π} (\frac{π}{2} + {tan}^{- 1} x)

- \infty < x < \infty

is a distribution function of a continuous variable

X

P (0 \leq x \leq 1)

f (x) = x^{4} - 8 x^{3} + 22 x^{2} - 24 x

g (x) = {\begin{matrix} \begin{matrix} min (f (t)), & x \leq t \leq x + 1, & - 1 \leq x \leq 1 \end{matrix} \begin{matrix} x - 10 & x > 1 \end{matrix} \end{matrix}

Then, in the interval

[- 1, \infty)

g (x)

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