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Directional Continuity and Continuity over an Interval - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • Directional Continuity and Continuity over an Interval is considered one of the most asked concept.

  • 53 Questions around this concept.

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The function \mathrm{f(x)}  is discontinuous only at \mathrm{x=0} such that - \mathrm{f(x)=1 \vee x \in R}.The total number of such functions is:

\mathrm{f(x)=\left\{\begin{array}{ll}\left(x^{2}+e^{\frac{1}{2-1}}\right)^{-1}, & x=2 \\ 1, & x=2\end{array}\right.} is continuous from right at the point \mathrm{x=2}, then \mathrm{k} equals

 The number of points of discontinuity of \mathrm{f(x)=[2 x]^{2}-(2 x)^{2}} (where [ ] denotes the greatest integer function and \mathrm{\left \{ \right \}} is fractional part of \mathrm{x} )  in the interval \mathrm{(-2,2)},is

The function \mathrm{f(x)=\frac{x^{3}}{8}-\sin \pi x+4} in \mathrm{[-4,4]} does not take the value

Let \mathrm{f(x)} be a continuous function defined for \mathrm{1 \leq x \leq 3}. If \mathrm{f(x)} takes rational values for all \mathrm{x} and \mathrm{f(2)=10}, then the value of \mathrm{f(1.5)} is:

Let f is a continuous function in [a, b], g is a continuous function in [b, c]. A function h(x) is defined as
\mathrm{h(x)=f(x) \text { for all } x \in[a, b)=g(x) \text { for } x \in(b, c] }
\mathrm{ \text { if } f(b)=g(b) \text { then : } }
 

\mathrm{\operatorname{Let} f(x)=[\tan x \mid \cot x] \mid . x \in\left[\frac{\pi}{12}, \frac{\pi}{2}\right) } ,( where \mathrm{[\cdot] \text { denotes }} the greatest integer less than or equal to \mathrm{x}). Then the number of proints, where \mathrm{f(x)} is discontinuous is

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Let \mathrm{f(x)=\left\{\begin{array}{cc}-3+|x|, & -\infty<x<1 \\ a+|2-x|, & 1 \leq x<\infty\end{array}\right.} and
\mathrm{g(x)=\left\{\begin{array}{cc}2-|-x| . & -\infty<x<2 \\ -b+\operatorname{sgn}(x), & 2 \leq x<\infty\end{array}\right.}

where \mathrm{\operatorname{sgn}(x)} denotes signum function of \mathrm{x}. If \mathrm{h(x)=f(x)+g(x)} is discontinuous at exactly one point, then which of the following is not possible?

A discontinuous function \mathrm{y=f(x)} satisfying \mathrm{x^2+y^2=4} is given by \mathrm{f(x)=} ________ for \mathrm{0 \leq x \leq 2}.

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Let \mathrm{f(x)=\left[2 x^3-6\right]}, where [x] is the greatest integer less than or equal to x. Then the number of points in (1,2) where f is discontinuous is:

Concepts Covered - 2

Directional Continuity and Continuity over an Interval

Directional Continuity and Continuity over an Interval

A function may happen to be continuous in only one direction, either from the “left” or from the “right”.  

A function y = f(x) is left - continuous at x = a if \\\mathrm{\lim_{x\rightarrow a^-}\;f(x)=f(a)\;\;\;\;or\;\;\;\lim_{h\rightarrow 0^+}\;f(a-h)=f(a)}\;\;\;\;or \;\;\;\;LHL=f(a)

A function y = f(x) is right - continuous at x = a if \\\mathrm{\lim_{x\rightarrow a^+}\;f(x)=f(a)\;\;\;\;or\;\;\;\lim_{h\rightarrow 0^+}\;f(a+h)=f(a)}\;\;\;\;or\;\;\;\;RHL=f(a)

For example,

\\\mathrm{f(x)=y=[x],\;\;\;\left (where,[.]\;is\;G.I.F \right ).\;\;is \;right\;continuous\;at\;x=2}\\\\f(2)=\lim_{x\rightarrow 2^+}[x]=2

But it is NOT left continuous at x = 2 as LHL = 1 but f(2) = 2. Hence f(2) does not equal LHL at x = 2.

Continuity over an Interval

Over an open interval (a, b)

A function f(x) is continuous over an open interval (a, b)  if f(x) is continuous at every point in the interval. 

For any c ∈ (a, b), f(x) is continuous if

\lim_{x\rightarrow c^-}\;f(x)=\lim_{x\rightarrow c^+}\;f(x)=f(c)

 

Over a closed interval [a, b]

A function f(x) is continuous over a closed interval of the form [a, b] if

  • it is continuous at every point in (a, b) and
  • is right-continuous at x = a and
  • is left-continuous at x = b.

i.e.At x = a, we need to check f(a)=\lim _{x \rightarrow a^{+}} f(x)\,\,\,(=\lim _{h \rightarrow 0^+} f(a+h)=\mathrm{R.H.L.}). L.H.L. should not be evaluated to check continuity x = a

And at x = b, we need to check f(b)=\lim _{x \rightarrow b^{-}} f(x) \,\,\,(=\lim _{h \rightarrow 0^+} f(b-h)=\mathrm{L.H.L.}). R.H.L. should not be evaluated to check continuity x = b

 

Consider one example,

f(x) = [ x ], prove that this function is not continuous in [2, 3],

Sol.

Condition 1

For continuity in (2,3)

        At any point x = c lying in (2,3),

        f(c) = [c] = 2 ( as c lies in (2,3))

        LHL at x = c : \lim _{x \rightarrow c^{-}} [x]=2 (as in close left neighbourhood of x = c, the function equals 2)

        RHL at x = c: \lim _{x \rightarrow c^{+}} [x]=2  (as in close right neighbourhood of x = c, the function equals 2)

        So function is continuous for any c lying in (2,3). Hence the function is continuous in (2,3)

Condition 2

Right continuity at x = 2

f(2) = 2

\\\mathrm{\lim_{x\rightarrow 2^+}f(x)=\lim_{x\rightarrow 2^+}[x]=\lim_{h\rightarrow 0^+}[2+h]=2}

So f(x) is left continuous at x = 2

Condition 3

Left continuity at x = 3

f(3) = 3 and 

\lim_{x\rightarrow 3^-}[x]=\lim_{h\rightarrow 0^+}[3-h]=2

(as in left neghbourhood of 3, f(x) = 2)

So f(3) does not equal LHL at x = 3

hence f(x) is not left continuous at x = 3

So third condition is not satisfies and hence f(x) is not continuous in [2,3]

Discontinuity and Removable Types Discontinuity

Discontinuity and Removable Types Discontinuity

A function that is non-continuous, is said to be a discontinuous function.

There are various kinds of discontinuity at a point, which are classified as shown below:

1. Removable Discontinuity   

2. Non-Removable Discontinuity

  • Finite Type

  • Infinite Type

  • Oscillatory

 

Removable Discontinuity

In this type of discontinuity, the limit of the function  \lim_{x\rightarrow a}\;f(x) necessarily exists but it is either not equal to f(a) or f(a) is not defined. However, it is possible to redefine the function at x = a in such a way that the limit of the function at x = a is equal to f(a), i.e. \lim_{x\rightarrow a}\;f(x)=f(a).

Again  removable discontinuity can be classified into missing point discontinuity and isolated point discontinuity

\\\text{Consider the function }f(x)=\frac{x^2-4}{x-2},\;\;\text{where, }x\neq2\\\therefore \;\;\;\;\;\;f(x)=\frac{(x+2)(x-2)}{(x-2)}=(x+2),\;x\neq2

In the above graph, observe that the graph has a hole (missing point) at x = 2, which makes it discontinuous at x = 2.

Here LHL = RHL = 4

Here function f(x) is not defined at x = 2, i.e. f(2) is not defined. However, we can redefine the function as

f(x)=\left\{\begin{matrix} \frac{x^2-4}{x-2}, & \;\;x\neq2\\ 4, & \;\;x=2 \end{matrix}\right. 

It makes the function continuous as now LHL = RHL = f(2)

Study it with Videos

Directional Continuity and Continuity over an Interval
Discontinuity and Removable Types Discontinuity

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Books

Reference Books

Directional Continuity and Continuity over an Interval

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 4.5

Line : 11

Discontinuity and Removable Types Discontinuity

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 4.3

Line : 4

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