Ncert Solutions Maths class 12th

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) = $\underset{x\to 1^{-}}{lim}$&nbs

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 =

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) =&

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. = $\underset{x\to 1^{-}}{lim}$&n

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

7. Given, f(x) = $\{\begin{array}{lll}\left|x\right|+3 if x\le -3\hfill & -2x if -3<x<3\hfill & 6x+2 if x\ge 3\hfill \end{array}$

For x = $?c<-3,$

f ( 3) = e + 3 (∴x< 3, $\left|x\right|=-x$ )

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ $\left|x\right|+3=-a+3.$

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

So, f is continuous at x = c < 3.

For x = c > 3

f(3) = 6.3 + 2 = 18 + 2 = 20

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 6x + 2 = 18 + 2 = 20

∴ $\underset{x\to c}{lim}$ f(x) = f(c).So f is continuous at x = c > 3.

For. C = 3,

f ( 3

6. Given f(x) = $\{\begin{array}{cc}\hfill 2x+3 if x\le 2\hfill & \hfill 2x-3 if x>2.\hfill \end{array}$

For x = c < 2,

F (c) = 2c + 3

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 2x + 3 = 2c + 3

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

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

For x = c > 2.

F (c) = 2c 3

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ 2x 3 = 2c 3

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

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

For x = c = 2,

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

5. Given, f (a) = $\{\begin{array}{ll}x, if x\le 1\hfill & 5, if x>1.\hfill \end{array}$

At x = 0,

(0) = 0

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

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

So, f is continuous at x = 0.

At x = 1,

Left hand limit,

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

Right hand limit,

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

L.H.L. $\cancel{=}$ R.H.L.

So, f is not contin

Let x and y be the number of dolls of type A and B, respectively, that are produced in a week.

The given problem can be formulated as given below:

$Maximisez=12x+16y\dots \dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+y \le 1200\dots \dots \dots \dots \left(ii\right)$

$y\le x/2orx\ge 2y\dots \dots \dots .(ii\mathrm{i)}$

$x–3y\le 600\dots \dots \dots \dots .\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots \dots \left(v\right)$

The feasible region determined by the system of constraints is given below:

A (600, 0), B (1050, 150) and C (800, 400) are th

Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.

The problem can be formulated as given below:

$Maximisez=3x+3.5y\dots \dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y \ge 240\dots \dots \dots ..\left(ii\right)$

$x+0.5y\ge 90\dots \dots \dots ..\left(iii\right)$

$1.5x+2y\le 310\dots \dots \dots ..\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots .\left(v\right)$

The feasible region determined by the system of constraints is given below:

A (140, 50), B (20, 140) and C (40, 100) are the corner points of the