Continuity and Differentiability

33. Let g (x) = $x$ is continuous being a modules f x and h (x) = $x$ is also continuous being a modules $\forall x\in ?$

Then, f (x) = g (x) h (x).is also continuous for all x. E. R.

Hence, there is no point of discontinuous for f (x).

32. Given, f (x) = sin $x$

Let g (x) = sin x and h (x) = $x$ then as sine f x and modulus f x are continuous in x e R

g and h are continuous.

So, (goh) (x) = g (h (x) = g (|x|) = sin |x| = f (x)

Is a continuous f x being a competitive f x of two continuous f x.

31. Given, f (x) = $\left|cosx\right|.$

Let g (x) = cos x and h (x) = $x$

Hence, as cosine function and modulus f x are continuous $\forall x\in ?,g$ h are continuous.

Then, (hog) x = h (g (x)

= h (cos x)

$=\left|cosx\right|.$

= f (x) is also continuous being

A composites fxn of two continuous f x $\forall x\in ?$

30. Given f (x) = cos (x2)

Let g (x) = cos x is a lregononuie fa (cosine) which is continuous function

and let h (x) = x2 is a polynomial f xn which is also continuous

Hence (goh) x = g (h (x)

= g (x)2

= cos (x2)

= f (x)

is also a continuous f x being a composite fxn of how continuous f x $\forall x\in ?$

29. Given, f(x) = $\{\begin{array}{ccc}\hfill 5\hfill & \hfill if x\le 2\hfill & \hfill ax+b\hfill \\ \hfill if 2<x<10\hfill & \hfill 21\hfill & \hfill if x\ge 10\hfill \end{array}$

For continuity at x = 2,

$\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{+}}{lim}f(x)=f(2)$

$\Rightarrow \underset{x\to 2^{-}}{lim}5=\underset{x\to 2^{+}}{lim}ax+b=5.$

$\Rightarrow$ 5 = 2a + b (i)

For continuous at x = 10,

$\underset{x\to 10^{-}}{lim}f(x)=\underset{x\to 10^{+}}{lim}f(x)=f(10)$

$\Rightarrow \underset{x\to 10^{-}}{lim}ax+b=\underset{x\to 10^{+}}{lim}21=21.$

$\Rightarrow$ 10a + b = 21 (2).

So, e q (2) 5 e q (1) we get,

10a + b 5 (2a + b) = 21 5 5.

$\Rightarrow$ 10a + b 10a 5b = 21 25.

$\Rightarrow$ 4b = 4

$\Rightarrow$ b = 1.

And putting b = 1 in e q (1),

$\Rightarrow$ 2a = 5 b = 5 1 = 4

$\Rightarrow a=\frac{4}{2}=2.$

Hence, a = 2 and b = 1.

 28. Given, f (x) $\{\begin{array}{ll}kx+1,\hfill & if x\le 5\hfill \\ 3x-5,\hfill & if x>5.\hfill \end{array}$

For continuity at x = 5,

$\underset{x\to 5^{-}}{lim}f(x)=\underset{x\to 5^{-}}{lim},kx+1=5x+1.$

$\underset{x\to 5^{+}}{lim}f(x)=\underset{x\to 5^{+}}{lim}3x-5=15-5=10$

f (5) = 5k + 1

So,  $\underset{x\to 5^{-}}{lim}f(x)=\underset{x\to 5^{+}}{lim}f(x)=f(5).$

i e, 5k + 1 = 10

$\Rightarrow$ 5k = 10 1

$\Rightarrow$ k = $\frac{9}{5}.$

26. Given f (x) = $\{\begin{array}{ll}k{x}^2\hfill & if x\le 2.\hfill \\ 3\hfill & if x>2.\hfill \end{array}$

For continuous at x = 2,

f (2) = k (2)2 = 4x.

L.H.L. = $\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}{x}^2=4x$

R.H.L. = $\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}3=3$

Then, L.H.L = R.H.L. = f (2)

i e, 4x = 3

$\Rightarrow x=\frac{3}{4}.$

 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.