Maths

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 + $\frac{2}{3}.$

15. Given, f(x) = $\{\begin{array}{ccc}\hfill -2,\hfill & \hfill if x\le -1\hfill & \hfill 2x,\hfill \\ \hfill if -1<x\le 1\hfill & \hfill 2,\hfill & \hfill if x>1.\hfill \end{array}$

For x = c < 1,

f(c) = 2

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ ( 2) = 2 = f(c)

So, f is continuous at x< 1.

For x = c > 1,

f(c) = 2

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ . 2 = 2 = f(c)

So, f is continuous at x $\left|>\right|$ 1.

For x = 1,

L.H.L. = $\underset{x\to -1^{-}}{lim}$ f(x) = $\underset{x\to -1^{-}}{lim}$ 2 = 2

R.H.L. = $\underset{x\to -1^{+}}{lim}$ f(x) = $\underset{x\to -1^{+}}{lim}$ . 2x = 2 ( 1) = 2

and f( 1) = 2

So, L.H.L. = R.H.L. = f( 1)

∴f is continuous at x = 1.

For x = 1,

L.H.L. = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ . 2x = 2.1 = 2

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ . 2 = 2.

f(1) = 2

f(1) = L.H.L = R.H.L.

So, f is continuous at x = 1.

14. Given f(x) = $\{\begin{array}{ccc}\hfill 2x, if x<0\hfill & \hfill 0, if 0\le x\le 1\hfill & \hfill 4x, if x>1.\hfill \end{array}$

For (c) = c < 0,

f(c) = 2c.

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 2x = 2c = f(c)

So, f is continuous at x $\left|<\right|$ 0

For x = c > 1,

f(c) = 4c

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 4x = 4c = f(c)

So, f is continuous at x> 1.

For x = 0

L.H.L. = $\underset{x\to 0^{-}}{lim}$ f(x) = $\underset{x\to 0^{-}}{lim}$ . 2x = 2 (0) = 0

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

f(0) = 0.

∴ L.H.L. = R.H.L. = f(0).

So, f is continuous at x = 0.

For x = 1.

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

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{+}}{lim}$ . 4x = 4 (1) = 4.

∴ L.H.L. $\cancel{=}$ R.H.L.

So, f is discontinuous at x = 1.