Q.Show that $f(x)=|x-1|$ is continuous at $x=1$ but not differentiable at $x=1$.

The expression is $\tfrac00$ at $x=2$; differentiate top and bottom (L'Hôpital) to get $f(2)-2f'(2)=-4$.

Evaluating a $\tfrac00$ limit that encodes a derivative is a CBSE/NCERT Class 12 limits and differentiability skill.

At $x=2$ the numerator is $2f(2)-2f(2)=0$ and the denominator is $0$, so it is $\tfrac00$. Differentiate numerator and denominator (with $f(2)=4$ a constant):

$\lim_{x\to2}\frac{x f(2) - 2f(x)}{x-2} = \lim_{x\to2}\frac{f(2) - 2f'(x)}{1} = f(2) - 2f'(2).$

Substitute $f(2)=4,\ f'(2)=4$:

$= 4 - 2\cdot 4 = -4.$

Match value and slope of the two pieces at $x=1$: continuity gives $a+b=0$, differentiability gives $a=-1$; hence $a-b=-2$.

Continuity and differentiability of a piecewise function is a CBSE/NCERT Class 12 continuity and differentiability topic.

Continuity at $x=1$: the left piece value equals the right piece value:

$1^2 + a\cdot1 + b = 1 \Rightarrow 1 + a + b = 1 \Rightarrow a + b = 0.$

Differentiability at $x=1$: the left derivative $\dfrac{d}{dx}(x^2+ax+b)=2x+a$ at $x=1$ is $2+a$; the right derivative $\dfrac{d}{dx}(x)=1$. Equate:

$2 + a = 1 \Rightarrow a = -1.$

Then $b = -a = 1$. Therefore

$a - b = -1 - 1 = -2.$

This limit is exactly the definition of the derivative $G'(1)$.

By definition, $\displaystyle\lim_{x\to 1}\frac{G(x)-G(1)}{x-1} = G'(1)$.

Given $G(x) = -\sqrt{25-x^2} = -(25-x^2)^{1/2}$, differentiate using the chain rule:

$G'(x) = -\frac{1}{2}(25-x^2)^{-1/2}\cdot(-2x) = \frac{x}{\sqrt{25-x^2}}$

At $x=1$: $G'(1) = \dfrac{1}{\sqrt{25-1}} = \dfrac{1}{\sqrt{24}}$.