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) = $\underset{x\to 1^{-}}{lim}$ 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.

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) = $\underset{x\to 2^{-}}{lim}$ .2x + 3 = 2. 2 + 3 = 4 + 3 = 7.

R.H.L. = $\underset{x\to 2^{+}}{lim}$ f(x) = $\underset{x\to 2^{+}}{lim}$ 2x 3 = 2. 2 3 = 4 3 = 1.

∴ LHL $\cancel{=}$ RHL

∴ f is not continuous at x = 2.i e, point of discontinuity

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 continuous at x = 1.

At x = 2,

f (2) = 5.

$\underset{x\to 1}{lim}$ f (x) = $\underset{x\to 2}{lim}$ 5 = 5

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

So f is continuous at x = 2.

Find all points of discontinuity of f, where f is defined by

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 the corner points of the feasible region.

The values of z at these corner points are given below:

The maximum value of z is 16000 at (800, 400).

Hence, 800 and 400 dolls of type A and type B should be produced, respectively, to get the maximum profit of? 16000.

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 feasible region.

The values of z at these corner points are given below:

The maximum value of z is 595 at (140, 50).

Hence, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.

Thus, the maximum amount of nitrogen added to the garden is 595 kg.