Question

The function f defined by,
2x, if x is rational
f(x)=
-x+27, if x is irrational
continuous only at,
(a) x = 0
(b) x = 3
(C) x = 6
(d) x = 9
ОА
Answer​

Answer 1

Answer:

1. x is 12
2. x is 15
3. x is 18 OA ans

Answer 2

Concept:

To solve this question we first recall the concept of continous function.

Graphically, A function is said to be continuous if its graph does not have any break in between the curve.

Mathematically , a function  is continuous if its left hand limit (LHL) is equal to right hand limit (RHL).

Given:

The function f defined by :

$f(x)=\left \{ {{2x \; ,\; \; if\; x\; is\; rational} \atop {-x+27\; ,\;\;}if \;x\;is\;irrational\;} \right.$

To find:

The pont at which the given function f(x) is continuous.

Solution:

For option (a) at x=0

LHL =  $\lim_{x \to 0^{-}} f(x)$

       = $\lim_{x \to 0^{-}}2x$

       = 0

RHL = $\lim_{x \to 0^{+}} f(x)$

       = $\lim_{x \to 0^{+}}(-x+27)$

       = 27

So, LHL ≠ RHL

So, f(x) is not continuous at x=0.

For option (b) at x=3

LHL =  $\lim_{x \to 3^{-}} f(x)$

       = $\lim_{x \to 3^{-}}2x$

       = 6

RHL = $\lim_{x \to 3^{+}} f(x)$

       = $\lim_{x \to 3^{+}}(-x+27)$

       = 24

So, LHL ≠ RHL

So, f(x) is not continuous at x=3.

For option (c) at x=6

LHL =  $\lim_{x \to 6^{-}} f(x)$

       = $\lim_{x \to 6^{-}}2x$

       = 12

RHL = $\lim_{x \to 6^{+}} f(x)$

       = $\lim_{x \to 6^{+}}(-x+27)$

       = 21

So, LHL ≠ RHL

So, f(x) is not continuous at x=6.

For option (d) at x=9

LHL =  $\lim_{x \to 9^{-}} f(x)$

       = $\lim_{x \to 9^{-}}2x$

       = 18

RHL = $\lim_{x \to 9^{+}} f(x)$

       = $\lim_{x \to 9^{+}}(-x+27)$

       = 18

So, LHL = RHL

So, f(x) is continuous at x=9.

Hence, option (d) is correct choice.