Maths

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.

Two different vectors having same magnitude: -

(i) $2\hat{i}+\hat{j}+3\hat{k}$

(ii) $\hat{i}+3\hat{j}+2\hat{k}$

Kindly go through the solution

(i) True, as vector quantity $\overrightarrow{a}$ and - $\overrightarrow{a}$ are parallel to same line.

(ii) False, as collinear vector are those vectors that are parallel to same line, but it is not necessary that they are equal also.

(iii) False, as two vectors having same magnitude may have different directions, so they are not collinear.

(iv) False, as two collinear vectors having same magnitude are not equal whey they are opposite in direction.

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.

(a) Vector $\overrightarrow{a}$ and $\overrightarrow{d}$ are co initial same initial point.

(b) $\overrightarrow{b}$ and $\stackrel{?}{d}$ same magnitude & direction.

(c) $\overrightarrow{a}$ and $\overrightarrow{c}$ are collinear but not equal they are parallels their direction are not same.

95.

$\begin{array}{l}P\left(A\right)+P\left(B\right)-P\left(AandB\right)=P\left(A\right)\\ \Rightarrow P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=P\left(A\right)\\ \Rightarrow P\left(B\right)-P\left(A\cap B\right)=0\\ \Rightarrow P\left(A\cap B\right)=P\left(B\right)\\ \therefore P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(B\right)}{P\left(B\right)}=1\end{array}$

Therefore, option (B) is correct.

(i) Time period involves only magnitude. So, it is scalar quantity.

(ii) Distance involves only magnitude. So, it is scalar quantity.

(iii) Force involves both magnitude and direction. So, it is vector quantity.

(iv) Velocity involves both magnitude and direction. So, it is vector quantity.

(v) Work done involves only magnitude. So, it is scalar quantity.

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.