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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
Question
The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for x = 2. in order to make f(x) continuous at x = 2, here f(2) should be defined as:
  1. 0
  2. 1
  3. 2
  4. 3
✓

Answer

  1. 0

Solution:

Here,

x3 + x2 - 16x + 20

= x3 - 2x2 + 3x2 - 6x - 10x + 20

= x2(x - 2) + 3x(x - 2) - 10(x - 2)

= (x - 2)(x2 + 3x - 10)

= (x - 2)(x - 2) (x - 5)

= (x - 2)2(x + 5)

So, the given function can be rewritten as 

$\text{f(x)}=\frac{(\text{x}-2)^2(\text{x}+5)}{\text{x}-2}$

$\Rightarrow\text{f(x)}=(\text{x}-2)(\text{x}+5)$

If f(x) is continuous at x = 2, then

$\lim\limits_{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$

$\Rightarrow\lim\limits_{\text{x}\rightarrow2}(\text{x}-2)(\text{x}+5)=\text{f}(2)$

$\Rightarrow\text{f}(2)=0$

Hence, in order to make f(x) continuous at x = 2, f(2) should be defined as 0.

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