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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
✓

Answer

Here f(x) = 5x - 3
  1. $\ \ \ \text{Lt}\ \ \ \ \ \ \text{f(x)}\\ \text{x} \rightarrow 0$ $= \ \ \ \text{Lt}\ \ \ \ \ \ (5\text{x}-3) = 5(0) - 3 = 0 - 3 = -3\\ \ \ \ \ \text{x} \rightarrow 0$
Now f is defined at x = 0

and f(0) = 5(0) - 3 = 0 - 3 = -3

$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(0) = -3\\ \ \ \ \text{x}\rightarrow0$

$\therefore$ f is continous at x = 0
  1. $\ \ \ \text{Lt}\ \ \ \ \ \ \text{f(x)}\\ \text{x} \rightarrow -3$$= \ \ \ \text{Lt}\ \ \ \ \ \ (5\text{x}-3) = 5(-3) - 3 = -15- 3 = -18\\ \ \ \ \ \text{x} \rightarrow -3$
Now f is defined at x = -3

and f(-3) = 5(-3) - 3 = -15 - 3 = -18

$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(-3) = -18\\ \ \ \ \text{x}\rightarrow-3$

$\therefore$ f is continous at x = -3
  1. $\ \ \ \text{Lt}\ \ \ \ \ \ \text{f(x)}\\ \text{x} \rightarrow 5$$= \ \ \ \text{Lt}\ \ \ \ \ \ (5\text{x}-3) = 5(5) - 3 = 25- 3 = 22\\ \ \ \ \ \text{x} \rightarrow 5$
Now f is defined at x = 5

and f(5) = 5(5) - 3 = 25 - 3 = 22

$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(5) = 22\\ \ \ \ \text{x}\rightarrow5$

$\therefore$ f is continous at x = 5

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