Question

Let f(x) =

$|x|=\{\begin{array}{cc}\frac{3\times (x-1)}{x^2-3x+2}& \text{text{for } ~~~ xneq 1,2}\\ -3& \text{text{for } ~~~ x=1}\\ 4& \text{text{for} ~~~ x=2}\end{array}$. Then f(x) is continuous

• A. Except at x = 1
• B. except at x = 2
• C. except at x = 0, 1 or 2
• D. At each real number

Answer 1

The correct option is B

except at x = 2

By observing the function, we see that the possibilities of discontinuities are there for x=1 and x=2. So

we go on to test the continuity at these points.

Given function is,

$f\left(x\right)=\frac{3x(x-1)}{(x-1)(x-2)}whenx\ne 1,2$

$=\frac{3x}{x-2}$ when $x\ne 1,2;$

For x=1

${lim}_{x\to 1}f\left(x\right)={lim}_{x\to 1}\frac{3x}{x-2}$

$=\frac{3}{-1}$=-3

That is, $li{m}_{x\to 1}f\left(x\right)=f\left(1\right)=-3$. Hence the function is continuous at x=1

Now for x=2

${lim}_{x\to 2}f\left(x\right)={lim}_{x\to 2}\frac{3x}{x-2}$=indefinite.

Hence the limit doesn't exist at x=2 for function f(x) and is continuous at x=2

Therefore right answer is (b)