Ncert Solutions Maths class 11th

45. The given system of inequality is

5x+4y≤ 20 - (1)

x≥ 1 - (2)

and  y≥ 2 - (3)

The equation of inequality (1) is 5x+4y=20.

Putting (x, y)= (0,0) in inequality (1) we get,

5 * 0+4 * 0 ≤ 20   => 0 ≤ 20 which is true.

So, the solution region of inequality (1) includes the plane with origin (0,0).

∴ The shaded region indicates the solution of the given system of inequality.

44. The given system of inequality is

x+y≤ 9- (1)

y>x- (2)

x≥ 0 - (3)

The corresponding equation of (1) is x+y=9 and (2) is y=x

Substituting (x, y)= (0,0) in (1),

0+0 ≤ 9  => 0 ≤ 9 which is true.

And putting (1,0) in (2)

0> 1which is false.

So, solution region of inequality (1) includes origin (0,0) and solution region of inequality (2) excludes plane having (1,0).

? Solution of region of given system of inequality is the shaded region.

43. Given system of inequality is

2x+y≥ 8- (1)

x+2y≥ 10- (2)

The corresponding equations are

2x + y = 8

and x + 2y = 10

Now, putting (x, y)= (0,0) in inequality (1) and (2),

2 * 0+8 ≥ 8

0 ≥ 8 which is not true.

and 0+2 * 0 ≥ 10

0 ≥ 10 which is not true.

So, solution of plane of inequality (1) and (2) does not include the origin (0,0)

? The required solution of the given system of inequality is the shaded region.

42. The given system of inequality is

x+y≤ 6 - (1)

x+y≥ 4- (2)

So the corresponding equations are

x+y=6

and x + y = 4

Putting (x, y)= (0,0) in equality (1) and (2),

0+0 ≤ 6 and 0 + 0 ≥ 4

0 ≤ 6 is true.  => 0 ≥ 4 is false.

So, solution of plane of inequality (1) includes the origin and inequality (2) does not includes the origin.

? The reqd solution of the given system of inequality is the shaded region.

41. The given system of inequality is

2x – y> 1 - (1)

x – 2y< 1- (1)

So the corresponding equations are

2x – y=1

and x – 2y= –1

Putting (x, y)= (0,0) in (1) and (2) to cheek the inequality

2 * 0 – 0 > 1

0 > 1 which is not true.

and 0 – 2 * 0< 1

0< 1 which is not true.

So, the solution of plane of inequality (1)and (2) does not include the plane with point (0,0) or origin.

? The reqd. solution of the given system of inequality is the shaded region.

40. The given system of inequalities is

x + y ≥ 4.- (1)

2x – y< 0.- (2)

The corresponding equations are x+y=4 and 2x – y=0.

and

Put (x, y)= (1,1) in (1) and (2).

So, 1+1 ≥ 4

2 ≥ 4 which is not true.

and 2 * 1 – 1<0

1<0 which is not true.

So solution of plane of inequality (1) and (2) does not include the plane with point (1,1).

? The reqd. solution of the given system of inequality is the shaded portion.

38. 

2. The given system of inequalities are

3x+2y≤ 12- (1)

x≥ 1- (2)

y≥ 2- (3)

We draws the graphs of the lines 3x+2y=12 using points and as 3 * 0 + 2 * 0 ≤ 12

The solution is plane which includes the origin (0, 0).

0 ≤ 12

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 6\end{array}|\begin{array}{l}4\\ 0\end{array}|$

and x = 1 and y = 2.

The inequality (1), (2) and (3) represents the region between these three lines including the points on the respective lines. So, every point on the shaded region in first quadrant represents a solution of the given system of inequalities.

7. Here,

r= length of pendulum.

r= 75 cm.

(i) Arc of length, l = 10 cm

Ø= $\frac{l}{r}$ = $\frac{10cm}{75cm}=\frac{2}{15}$ radian.

(ii) Arc of length, l = 15 cm.

So, Ø= $\frac{l}{r}$ = $\frac{15cm}{75cm}=\frac{1}{5}$ radian.

(iii) Arc for length, l= 21 cm.

So, Ø= $\frac{l}{r}=\frac{21cm}{75cm}=\frac{7}{25}$ radian.

6. Let r₁ and r₂ be the radii of two circles.

Then using relation