Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 28

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Which of the following function(s) defined below has/have single point continuity?

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Important Questions on Continuity and Differentiability

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 29

The function

f (x) = \{\begin{matrix} | x - 3 |, & x \geq 1 \\ (\frac{x^{2}}{4}) - (\frac{3 x}{2}) + (\frac{13}{4}), & x < 1 \end{matrix}

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 30

If

f (x) = \frac{1}{2} x - 1,

then on the interval

[0, π]

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 31

Let

f (x)

and

g (x)

be defined by

f (x) = [x]

and

g (x) = \{\begin{array}{lc} 0, & x \in I \\ x^{2}, & x \in R - I \end{array} (where [.]

denotes the greatest integer function), then

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 32

Let

f (x) = [x] + \sqrt{x - [x]}

, where

[.]

denotes the greatest integer function. Then

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 33

The points at which the function,

f (x) = |x - 0.5| + |x - 1| + \tan x

does not have a derivative in the interval

(0, 2)

are

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 34

f (x) = {(\sin^{- 1} x)}^{2} \cdot \cos (\frac{1}{x})

if

x \neq 0, f (0) = 0, f (x)

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 35

If

f (x) = a_{0} + \sum_{k = 1}^{n} a_{k} {|x|}^{k}

, where

a_{i}^{'} s

are real constants, then

f (x)

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 26 - Continuity and Differentiability>Exercise-2>Q 36

Let

f : R \to R

be a function such that

f (0) = 1

and for any

x, y \in R, f (x y + 1) = f (x) f (y) - f (y) - x + 2

. Then,

f

is