Question

The function defined by $f\left(x\right)=\{\begin{array}{cc}|x-3|;& x\ge 1\\ \frac{1}{4}{x}^2-\frac{3}{2}x+\frac{13}{4};& x<1\end{array}$ is

• A. Continuous at x = 1
• B. Continuous at x = 3
• C. Differentiable at x = 1
• D. All the above

Answer 1

The correct option is D

All the above

Since |x - 3| = x - 3, if x $\ge$ 3
= -x +3 , if x < 3
$\therefore$ The given function can be defined as
$f\left(x\right)=\{\begin{array}{cc}\frac{1}{4}{x}^2-\frac{3}{2}x+\frac{13}{4},& x<1\\ 3-x,& 1\le x<3\\ x-3,& x\ge 3\end{array}$

Now we can check the continuity of the function at x = 1 & x = 3
At x = 1
${lim}_{h\to 0}f(1+h)=3-(1+h)=2-h=2$
${lim}_{h\to 0}f(1-h)=\frac{1}{4}(1-h{)}^2-\frac{3}{2}(1-h)+\frac{13}{4}$
$=\frac{1}{4}[1+h^2-2h]-\frac{3}{2}+\frac{3h}{2}+\frac{13}{4}$
$=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}-\frac{h}{2}+\frac{3h}{2}+\frac{h^2}{4}$
$=2+h+\frac{h^2}{4}$
= 2 (Since h $\to$ 0)
& f(1) = 3 - 1 = 2
Since f(1+h) = f(1-h) = f(1)
We can say f(x) is continuous at x = 1.

Let's check the continuity at x = 3
${lim}_{h\to 0}$ f(3+h) = (3+h)-3 = h = 0
${lim}_{h\to 0}$ f(3-h) = 3-(3-h) = h = 0

f(3) = 3 - 3 = 0
Here also we can see that f(3+h) = f(3-h) = f(3)
So, f(x) is continuous at x = 3

Let's check the differentiability at x = 1
We are checking differentiability only at x = 1 as only that much we are asked to find, refer options.

L.H.D = ${lim}_{h\to 0}\frac{f\left(1\mu h\right)-f\left(1\right)}{-h}$
= $\frac{\frac{1}{4}(1-h{)}^2-\frac{3}{2}(1-h+\frac{13}{4}-2)}{-h}$
= $\frac{h+\frac{h^2}{4}}{-h}$
= $-1-\frac{h}{4}$
= -1

R.H.D = ${lim}_{h\to 0}\frac{f(1+h)-f\left(1\right)}{h}$
= $\frac{3-(1+h)-2}{h}$
= $\frac{-h}{h}$
= -1
Since, L.H.D = R.H.D at x = 1, we can say function is differentiable at x = 1