Continuity and Differentiability

 25. Given, f(x) = $\{\begin{array}{cc}\hfill \frac{\pi cosx}{\pi -2x}\hfill & \hfill if x\ne \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill 3\hfill & \hfill if x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \end{array}$

For continuity at $x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\underset{x\to \raisebox{1ex}{${\pi }^{-}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}f(x)=\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}f(x)=f\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right).$

$\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{\pi -2x}=\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2^{+}$}\right.}{lim}\frac{xcosx}{\pi -2x}=3.$

Take $\underset{x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{x-2x}=3$ .

Putting x = $\frac{\pi }{2}+h$ such that as $x\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,h\to 0.$

So $\underset{h\to 0}{lim}\frac{xcos(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+h)}{x-2(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+n)}=\underset{h\to 0}{lim}\frac{x(-sinx)}{-2h}$

$=\underset{h\to 0}{lim}\frac{xsinh}{2h}$

$=\frac{k}{2}\underset{h\to 0}{lim}\frac{sinh}{h}=\frac{k}{2}.$

i e, $\frac{k}{2}=3$

$\Rightarrow$ k = 6

Similarly from $\underset{x\to \raisebox{1ex}{${\pi }^{+}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{xcosx}{\pi -2x}=3$

$\Rightarrow \underset{h\to 0}{lim}\frac{hcos\left(\frac{\pi }{2}+h\right)}{\pi -2\left(\frac{\pi }{2}+h\right)}=\underset{h\to 0}{lim}\frac{-hsinh}{-2h}$

$=\frac{}{2}\underset{h\to 0}{lim}\frac{sinh}{h}$

$=\frac{x}{2}$

So, $\frac{x}{2}=3$

$\Rightarrow$ k = 6

24. Given, f(x) = $\{\begin{array}{cc}\hfill sinx-cosx,\hfill & \hfill if x\ne 0\hfill \\ \hfill -1,\hfill & \hfill if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f(c) = sin c cos c.

$\underset{x\to c}{lim}$ f (x) = $\underset{x\to c}{lim}$ (sin x cos x) = sin c cos c = f(c)

So, f is continuous at $x\ne 0$

For x = 0,

f(0) = 1

$\underset{x\to 0}{lim}$ f (x) = $\underset{x\to 0}{lim}$ (sin x cos x) = sin 0 cos 0 = 0 1 = 1

$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}(sinx-corx)=sin0-cos0=-1.$

∴ $\underset{x\to 0-}{lim}$ f(x) = $\underset{x\to 0+}{lim}$ f (x) = f (0)

So, f is continuous at x = 0.

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

23. Given f (x) = $\{\begin{array}{ll}{x}^{2}sin\frac{1}{x}, if x\ne 0.\hfill & 0 if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f (c) = $c^{2}sin\frac{1}{c}$

$\underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}{x}^{2}sin\frac{1}{x}=c^{2}sin\frac{1}{c}.$

So, f is continuous for $x\ne 0.$

For x = 0,

f (0) = 0

$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\left(x^{2}sin\frac{1}{x}\right)$

As we have sin $\frac{1}{x}\in [-1,1]$

$\underset{x\to 0}{lim}$ f (x) = 02 a where $a\in [-1,1]$

= 0 = f (0).

∴ f is also continuous at x = 0.

22. Given f(x) = $\{\begin{array}{ll}\frac{sinx}{x},\hfill & if x<0.\hfill \\ x+1,\hfill & if x\ge 0.\hfill \end{array}$

For x = c < 0,

f(c) = $\frac{sinc}{c}$

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ $\frac{sinx}{x}=\frac{sinc}{c}=f(c)$

So, f is continuous for x < 0

For x = c > 0

f(c) = c + 1

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ x + 1 = c + 1 = f(c)

So, f is continuous for x > 0.

For x = 0.

L.H.L. = $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{sinx}{x}=1.$

R.H.S. = $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}x+1=0+1=1$

And f(0) = 0 + 1 = 1

L.H.L = R.H.L. = f(0)

So, f is continuous at x = 1.

Hence, discontinuous point of x does not exit.

19. Given f (x) = x2 sin x + 5.

At x = .

$f(\pi )={\pi }^2-sin\pi +5={\pi }^2-0+5={\pi }^2+5$

$\underset{x\to \pi }{lim}$ f (x) = $\underset{x\to \pi }{lim}$  [x2 sin x + 5]

If x = $\pi +h$ then as x, h 0, so,

$\underset{x\to \pi }{lim}$ f (x) = $\underset{x\to 0}{lim}$  [ ( + h)2 sin ( + h) + 5]

= ( + 0)2 $\underset{h\to 0}{lim}$ $\begin{array}{rr}\hfill [sin\pi cosh+cos\pi \cdot sina]& \hfill +5.\end{array}$

= 2 $\underset{h\to 0}{lim}$ sin cos h $\underset{h\to 0}{lim}$ cos sin h + 5

= x2 0 * (1) ( 1) 0 + 5.

= 2 + 5 = f (x)

So, f is continuous at x = .

18. Given, g (x) = x [x].

For $n\in z,$

g (n) = n [n] = nn = 0

$\underset{x\to n^{-}}{lim}$ f (x) = $\underset{x\to n^{-}}{lim}$  (x [x]) = n [n 1] = n + 1 = 1

$\underset{x\to n^{+}}{lim}$ g (x) = $\underset{x\to n^{+}}{lim}$ x [x] = n [n] = 0

So,  $\underset{x\to n^{-}}{lim}$ g (x) $\cancel{=}$ $\underset{x\to n^{+}}{lim}$ g (x).

g (x) is d is continuous at all x $\in z.$

17. Given, f (x) = $\{\begin{array}{cc}\hfill \lambda \left(x^2-2x\right)\hfill & \hfill if x\left|?\right|0\hfill \\ \hfill 4x+1\hfill & \hfill if x>0.\hfill \end{array}$

For continuity at x = 0,

$\underset{x\to 0^{-}}{lim}$ f (x) = $\underset{x\to 0^{+}}{lim}$ f (x) = f (0).

$\underset{x\to 0^{-}}{lim}$ $\lambda \left(x^2-2x\right)$ = $\underset{x\to 0^{+}}{lim}$ 4x + 1 = $\lambda \left(0^2-2.0\right)$

0 = 1 = 0 which is not true

Hence, f is not continuous for any value of $\lambda .$

For x = 1,

$\underset{x\to 1}{lim}$ f (x) = f (1).

$\underset{x\to 1}{lim}$ 4x + 1 = 4 (1) + 1

$\Rightarrow$ 4 + 1 = 4 + 1

$\Rightarrow$ 5 = 5.

So, f is continuous at x = 1 value of $\lambda$

16. Given, f (x) = $\{\begin{array}{ll}ax+1, if x\le 3\hfill & bx+3, if x>3\hfill \end{array}$ is continuous at x = 3

So, f (3) = 3a + 1

L.H.L = $\underset{x\to 3^{-}}{lim}$ f (x) = $\underset{x\to 3^{-}}{lim}$ ax + 1 = 3a + 1

R.H.L = $\underset{x\to 3^{+}}{lim}$ f (x) = $\underset{x\to 3^{+}}{lim}$ b x + 3 = 3b + 3

for continuity at x = 3,

L.H.L = R.H.L. = f (3)

$\Rightarrow$ 3a + 1 = 3 + 3 = 3a + 1

So, 3a + 1 = 3b + 3

3a = 3b + 3 1

3a = 3b + 2.

a = b + $\frac{2}{3}.$