f (x) = |x - 3| = $\begin{cases} 3-x, & x < 3 \\ x-3, & x \ge 3 \end{cases}$ Let c be a real number. Case I: c < 3. Then f(c) = 3 – c. $\lim\limits_{x\to c}$ f (x) = $\lim\limits_{x\to c}$ (3 - x) = 3 - x Since, $\lim\limits_{x\to c}$ f (x) = f (c), f is continuous at all negative real numbers. Case II: c = 3. Then f(c) = 3 – 3 = 0 $\lim\limits_{x\to c}$ f (x) = $\lim\limits_{x\to c}$ f (x) (x - 3) = 3 - 3 = 0 Since, $\lim\limits_{x\to c}$ f (x) f (x) = f (3), f is continuous at x = 3. Case III: c > 3. Then f(c) = c – 3. $\lim\limits_{x\to c}$ f (x) = $\lim\limits_{x\to c}$ (x - 3) = x - 3 Since, $\lim\limits_{x\to c}$ f (x) = f (c), f is continuous at all positive real numbers Therefore, f is continuous function. Now, we need to show that f(x) = |x - 3|, x ∊ R is not differentiable at x = 3. Consider the left hand limit of f at x = 3 $\lim\limits_{h\to 0^{-}}$ $\frac{f(3+h)-f(3)}{h}$ = $\lim\limits_{h\to 0^{-}}$ $\frac{|3+h-3|-|3-3|}{h}$ = $\lim\limits_{h\to 0^{-}}$ $\frac{|h|-0}{h}$ = $\lim\limits_{h\to 0^{-}}$ $\frac{-h}{h}$ = - 1 h < 0 ⇒ |h| = - h Consider the right hand limit of f at x = 3 $\lim\limits_{h\to 0^{+}}$ $\frac{f(3+h)-f(3)}{h}$ $\lim\limits_{h\to 0^{+}}$ $\frac{|3+h-3|-|3-3|}{h}$ = $\lim\limits_{h\to 0^{+}}$ $\frac{|h|-0}{h}$ = $\lim\limits_{h\to 0^{+}}$ $\frac{h}{h}$ = 1 h > 0 ⇒ |h| = h Since the left and right hand limits are not equal, f is not differentiable at x = 3. OR y = a $\cos t + \log \tan\left(\frac{t}{2}\right)$ find $\frac{d^2y}{dx^2}$ ⇒ $\frac{dy}{dt}$ = a $\left| \frac{d}{dt} + \cos t + \frac{d}{dt} \left( \log \tan \frac{t}{2} \right) \right|$ = a $\left| -\sin t + \cot \frac{t}{2} \times \sec^3 \frac{t}{2} \times \frac{1}{2} \right|$ = a $\left| -\sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} \right|$ = a $\left( -\sin t + \frac{1}{\sin t} \right)$ = a $\frac{-\sin^2 t + 1}{\sin t}$ = a $\frac{\cos^2 t}{\sin t}$ x = a sin t $\frac{dx}{dt}$ = a $\frac{d}{dt}$ sin t = a cos t ∴ $\frac{dy}{dx}$ = $\frac{\frac{dy}{dx}}{\frac{dx}{dt}}$ = $\frac{a \frac{\cos^2 t}{\sin t}}{a \cos t}$ = $\frac{\cos t}{\sin t}$ = cot t $\frac{d^2y}{dx^2}$ = - $\csc^2 t \, \frac{dt}{dx}$ = - $\csc^2 t \times \frac{1}{a \cos t}$ = - $\frac{1}{a \sin^2 t \cos t}$