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Let $f(x)=|x-3|+|x+5|$ and
$A=\left\{a \in \mathbb{R} / \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\right.$ exists $\}$
Then the number of real numbers which are in $(-\infty,-3)$ $\cup(5, \infty)$ but not in $A$ is
Options:
$A=\left\{a \in \mathbb{R} / \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\right.$ exists $\}$
Then the number of real numbers which are in $(-\infty,-3)$ $\cup(5, \infty)$ but not in $A$ is
Solution:
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Verified Answer
The correct answer is:
1
$A=I R /\{3,-5\}$
Now, real numbers which are in $(-\infty,-3) \cup(5, \infty)$ but not in $A$ is only 3 .
$\therefore$ No. of required real no. is 1 .
Now, real numbers which are in $(-\infty,-3) \cup(5, \infty)$ but not in $A$ is only 3 .
$\therefore$ No. of required real no. is 1 .
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