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The function  $f\left ( x \right )=\dfrac{x^{2}-1}{x-1}$ at $x=1$ is :

1. Continuous and differentiable
2. Continuous but not differentiable
3. Differentiable but not continuous
4. Neither continuous nor differentiable

### 1 comment

When a function is differentiable it is also continuous. But a function can be continuous but not differentiable

f(x)=x+1 is both continuous and differential

option A is right.

graph of the given function

by

### 1 comment

what value you are getting for the function when $x=1 ?$

If it is $x \neq 1$ and $f(x)$ is defined for all reals i.e. $\mathbb{R} \backslash \{1\}$ then $f(x)$ must be continuous or if you write $f(x) = \frac{x^2 – 1}{x-1}, x \neq 1$ and $f(x) = 2, x=1$ then you can say $f(x)$ is continuous for all real values of $x$ or when your function is defined as $f(x) = x+1$ for all $x \in \mathbb{R}$ then also you can say $f(x)$ is continuous.

But here in the question domain of the function is not defined. if the domain is $x \in \mathbb{R}$ If we have to check the continuity for $x =1$ then the given function is not continuous because $f(1)$ is undefined.

If function f(x) is not continuous than it is not differentiable so Option D is true

edited

But in official answer key, Ques.70(sec B ques 10) it is option instead D.

f(x)=  (x²-1)/(x-1) = [(x+1)(x-1)]/(x-1) = x+1 (differentiable and continuous)

yes, option A is right.

graph of the given function

1 vote