मराठी

Find the points of discontinuity, if any, of the following functions: f ( x ) = { x 10 − 1 , if x ≤ 1 x 2 , if x > 1 - Mathematics

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प्रश्न

Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}x^{10} - 1, & \text{ if }  x \leq 1 \\ x^2 , & \text{ if } x > 1\end{cases}\]

बेरीज

उत्तर

Given: `f(x)[{[x^10- 1 ,  \text{ if } ≤ 1],[x^2,\text{ if }    x>1]]`

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

if `c< 1, ` then `f(c)=c^10-1` and `lim_(x-> c) f(x)=lim_(x->c)(x^10  -1 )=c^10  -1`

`∴ lim_(x->c)f(x)=f(c)`

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

`lim_(x->c)f(x)=lim_(x->c)(x^10  -1)=1^10  -1 =1-1=0`

The right hand limit of f at = 1 is,

`lim_(x->1)f(x)=lim_(x->1)(x^2)=1^2=1`

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

`if c>1, " then "  f(c)=c^2`

`lim_(x->c)f(x)=lim_(x->c)f(c)=c^2`

`∴lim_(x->c)f(x)=f(c)`

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

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पाठ 9: Continuity - Exercise 9.2 [पृष्ठ ३४]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 9 Continuity
Exercise 9.2 | Q 3.1 | पृष्ठ ३४

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