NIELIT 2017 DEC Scientific Assistant A - Section B: 10

Question

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

• A. Continuous and differentiable
• B. Continuous but not differentiable
• C. Differentiable but not continuous
• D. Neither continuous nor differentiable

Comments

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

Answer 1

option A is right.

graph of the given function

Comments

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.

Answer 2

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

Comments

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

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

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yes, option A is right.

graph of the given function

Answer 3

I AM 100% sure that this function is not continuous .....come on even by watching question we can tell it

1.(x^2-1)/(x-1) here the value of x can't be equal to 1 so that denominator never become zero and function not become unefined 

2. so that's why here 1 can't be part of its domain the 1 must be omitted from the DOMAIN and by intentionally they were asking at x=1   so they fixing a trap  here they think we are dumb and actually we are dumb that we can'tt see it

OPTION : D is true only

alert: dont check on desmos without putting the value of x subscript there

even  if you guys check the f(a) where a=1 we got infinity value and we know that function is not defined at x=1 so how this function can be continuous