NIELIT2017 STA-set-c-119

Question

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

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

Comments

D) => As f(x) limit does exist at x=1 but value at point x=1 is not given.So function will be discontinuous at x=1.

If a function is discontinuous at some point then it is non differentiable at that point also(because we cannot make unique tangent at that point).

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@Dheeraj Pant NIELIT changed its answer key to "A) Both continuous and differentiable". I'm confused because of that change.

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yes , Option A is right

f(x) = (x-1)* (x+1)/(x-1)

     = x+1

y =  x+1  [form a straight line ]

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A) Both continuous and differentiable may be correct.

Because we can also write f(x) as:

$f(x) =\frac{x^{2}-1}{x - 1} = \frac{(x - 1)(x + 1)}{(x - 1)} = x + 1$

Now I think f(x) should be both.

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I think answer should be D, if additional information would have been given like f(1)=2 then i can say that it will be continuous. And  we can  cancel (x-1) factor from numerator and denominator only after assuming that x!=1. And if i assume that then what about value of f(x) at x=1??

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only option D is correct...

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At x=1, LHL = RHL =f(1)=2,so option A is right.

Answer 1

Given $f(x)=\frac{x^2 - 1}{x-1}$
Using $a^2 - b^2 = (a+b)(a-b)$, we get $f(x) = \frac{(x+1)(x-1)}{(x-1)}$
$\rightarrow f(x)=x+1$.
Since we divided out numerator and denominator, we have a hole at $x = 1$, so the function is discontinuous at $x=1$
(A function which is not continuous isn't differentiable either).

Option D.

(You can refer to the video to know why this is discontinuous): 

Comments

you mean option d.

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@LiteYagami yeah sorry. Corrected it now.