Ncert Solutions Maths class 12th

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.

13. $$Given, f(x) = $\{\begin{array}{ccc}\hfill 3 \pi 0\le x\le 1\hfill & \hfill 4 \pi 1<x<3\hfill & \hfill 5 \pi 3\le x\le 10.\hfill \end{array}$

For x = c such that $0\le c<1$

f(c) = 3

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 3 = 3 = f(c)

So, f is continuous in [0, 1].

For x = c = 1,

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

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

∴ L.H.L $\cancel{=}$ R.H.L.

f is discontinuity at x = 1

for x = c such that $1<c<3.$

f(c) = 4

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

So, f is continuous in $x\in (1,3)$

For x = c = 3

L.H.L. $\underset{x\to 3^{-}}{lim}$ f(x) = $\underset{x\to 3^{-}}{lim}$ 4 = 4

R.H.L. $\underset{x\to 3^{+}}{lim}$ f(x) = $\underset{x\to 3^{+}}{lim}$ 5 = 5.

So, f is discontinuous at x = 3.

For x = c such that $3<c\le 10$

f (c) = 5.

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 5 = 5 = f(c)

So, f is continuous in $x\in (3,10]$

12. Given, f(x) = $\{\begin{array}{ll}x+5 if x\left|?\right|1\hfill & x-5 if x>1.\hfill \end{array}$

For x = c < 1.

F (c) = c + 5

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ f x + 5 = c + 5

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

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

For x = c > 1

F (c) = c 5

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ x 5 = c 5.

$\underset{x\to c}{lim}$ f(x) = f(c)

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

For x = 1

L.H.L. = $\underset{x\to 1^{-}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ x + 5 = 1 + 5 = 6.

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f(x) = $\underset{x\to 1^{-}}{lim}$ x 5 = 1 5 = 4.

L.H.L. $\cancel{=}$ R.H.L.

f is not continuous at x = 1

So, point of discontinuity of f is at x = 1.

Discuss the continuity of the function $f$ , where $f$ is defined by

11. Given, f (x) = $\{\begin{array}{ll}{x}^{10}-1, if x\le 1\hfill & x^2, if x>1.\hfill \end{array}$

For x = c < 1.

f (c) = $\underset{x\to c}{lim}$ f (x) = c10 1.

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

For x = c > 1.

f (c) = $\underset{x\to c}{lim}$ f (x) = c2

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

For x = c = 1,

L.H.L = $\underset{x\to 1^{-}}{lim}$ f (x) = $\underset{x\to 1^{-}}{lim}$ x10 1 110 1 = 0.

R.H.L. = $\underset{x\to 1^{+}}{lim}$ f (x) = $\underset{x\to 1^{+}}{lim}$ x2 = 12 = 1.

∴ L.H.L $\cancel{=}$ R.H.L.

So, f is not continuous at x = 1.

Hence, f has point of discontinuity at x = 1.

10. Given f (x) = $\{\begin{array}{cc}\hfill x^3-3if x\left|?\right|2\hfill & \hfill x^2+1 if x>2\hfill \end{array}$

For x = c < 2,

f (c) = c3 3

$\underset{x\to c}{lim}$ f (x) = $\underset{x\to c}{lim}$ x3 3 = c3 3.

So f is continuous at x $\left|<\right|$ 2.

For x = c > 2

f (c) = x2 + 1 = c2 + 1

$\underset{x\to 2}{lim}$ f (x) = $\underset{x\to 2}{lim}$ x2 + 1 = c2 + 1 = f (c)

So, f is continuous at x $\left|>\right|$ 2.

For x = c = 2, f (2) = 23 3 = 8 3 = 5.

L.H.L. $\underset{x\to 2^{-}}{lim}$ f (x) = $\underset{x\to 2^{-}}{lim}$ x3 3 = 23 3 = 5.

R.H.L. $\underset{x\to 2^{+}}{lim}$ f (x) = $\underset{x\to 2^{+}}{lim}$ x2 + 1 = 22 + 1 = 5

∴ R.H.L. = L.H.L. = f (2).

So, f is continuous at x = 2

Hence f has no point of discontinuity.

9. Given, f (x) = $\{\begin{array}{cc}\hfill x+1,\pi x\ge 1\hfill & \hfill x^2+1, \pi x<1.\hfill \end{array}$

For x = c < 1,

$\underset{x\to c}{lim}$ f (x) = $\underset{x\to c}{lim}$ x2 + 1 = c2 + 1

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

So f is continuous at x = c < 1.

For x = c > 1,

F (c) = c + 1

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

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

So, f is continuous at x = c > 1.

For x = c = 1, + (1) = 1 + 1 = 2

L.H.L. = $\underset{x\to 1^{-}}{lim}$ f (x) = $\underset{x\to 1^{-}}{lim}$ x2 + 1 = 12 + 1 = 2.

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

∴ L.H.L = R.H.L. = f (1)

So, f is continuous at x = 1.Hence f has no point of discontinuity.

8. Given, f(x) = $fxxx$

For x = c < 0,

f(c) = 1

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

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

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

For x = c > 0,

F (c) = 1

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

∴f(c) = $\underset{x\to c}{lim}$ f(x)

f is continuous at x > 0.

For x = c 0.

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

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

∴ L.H.L. $\cancel{=}$ R.H.L.

is now continuous at x = 0, point of discontinuity of f is at x = 0.