Class 11th

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

So, radius, r = $\frac{40}{2}$ cm = 20 cm

Length of chord (AB) = 20cm

In OAB

OA = OB=AB=20 cm

Hence, AOAB is equilateral triangle and end of the angle is 60°

:. Ø =60° = 60 *
$\frac{\mathrm{}\mathrm{\pi }}{180}$ radian =$\frac{\mathrm{\pi }}{3}$ radian

Hence, length of minor are of the chord, l=rØ.

l = 20 * $\frac{\mathrm{\pi }}{3}$ cm

l = $\frac{2\mathrm{0\pi }}{3}$ cm.

4. Here l = 22cm.

r =100cm.

Ø =?

Hence by r = $\frac{1}{\mathrm{Ø}}$

= Ø = $\frac{l}{r}$ = $\frac{22}{100}$ radian

= $\frac{22}{100}$ * $\mathrm{180°}$/π

$\frac{\mathrm{}\mathrm{}}{}$$\frac{}{}$= $\frac{22}{100}$ * 180° * $\frac{7}{22}$

= ${\left(\frac{63}{5}\right)}^0$

=12 $\frac{3°}{5}$ = 12° $\frac{3*60\text{'}}{5}=12°36\text{'}$

3. Given that a wheel makes 360 revolutions in one minute

Then, number of revolutions in one second = $\frac{360}{60}$ =6.

In 1 complete revolution the wheel turns 360°= 2π radian.

So, In 6 revolution, the wheel will turns 6*2π radian = 12π radian.

Hence, in one second the wheel will turn an angle of 12π radian.