5 Marks Questions

Question 15 Marks

Let $A$ and $B$ be two sets. Show that $f : A \times B \rightarrow B \times A$ such that $f ( a , b )=( b , a )$ is
$(i)$ injective
$(ii)$ bijective

Answer

i. Let $\left(a_1 b_1\right)$ and $\left(a_2, b_2\right) \in A \times B$ such that
$f\left(a_1, b_1\right)=f\left(a_2, b_2\right)$
$\Rightarrow\left(a_1, b_1\right)=\left(a_2, b_2\right)$
$\Rightarrow a_1=a_2 \text { and } b_1=b_2$
$\Rightarrow\left(a_1, b_1\right)=\left(a_2, b_2\right)$
Therefore, $f$ is injective.
ii. Let $(b, a)$ be an arbitrary
Element of $B \times A$, then $b \in B$ and $a \in A$
$\Rightarrow( a , b )) \in( A \times B )$
Thus for all $(b, a) \in B \times A$ their exists $(a, b)) \in(A \times B)$
such that
$f (a, b) = (b, a)$
So $f: A \times B \rightarrow B \times A$
is an onto function.
Hence $f$ is bijective.

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Question 25 Marks

Find perpendicular distance of the point $(1, 0, 0)$ from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ .Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Answer

Suppose the point $(1,0,0)$ be $P$ and the point through which the line passes be $Q(1,-1,-10)$. The line is parallel to the vector $\vec{b}=2 \hat{i}-3 \hat{j}+8 \hat{k}$
Now,$\begin{array}{l}\overrightarrow{P Q}=0 \hat{i}-\hat{j}-10 \hat{k} \\\therefore \vec{b} \times \overrightarrow{P Q}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\2 & -3 & 8 \\0 & -1 & -10\end{array}\right|\end{array}$
$=38 \hat{i}+20 \hat{j}-2 \hat{k}$
$\Rightarrow|\vec{b} \times \overrightarrow{P Q}|=\sqrt{38^2+20^2+2^2}$
$=\sqrt{1444+400+4}$
$=\sqrt{1848}$
$d=\frac{\mid \vec{b} \times \overrightarrow{P Q \mid}}{|\vec{b}|}$
$=\frac{\sqrt{1848}}{\sqrt{77}}$
$=\sqrt{24}$
$=2 \sqrt{6}$
Suppose $L$ be the foot of the perpendicular drawn from the point $P(1,0,0)$ to the given line$-$

The coordinates of a general point on the line
$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8} \text { are given by }$
$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda$
$\Rightarrow x=2 \lambda+1$
$y=-3 \lambda-1$
$z=8 \lambda-10$
Suppose the coordinates of $L$ be
$(2 \lambda+1,-3 \lambda-1,8 \lambda-10)$
Since, The direction ratios of $PL$ are proportional to,
$2 \lambda+1-1,-3 \lambda-1-0,8 \lambda-10-0$, i.e., $2 \lambda,-3 \lambda-1,8 \lambda-10$
Since, The direction ratios of the given line are proportional to $2, -3, 8,$ but $PL$ is perpendicular to the given line.
$\therefore 2(2 \lambda)-3(-3 \lambda-1)+8(8 \lambda-10)=0$
$\Rightarrow \lambda=1$ Substituting $\lambda=1$ in $(2 \lambda+1,-3 \lambda-1,8 \lambda-10)$
we get the coordinates of $L$ as $(3,-4,-2)$. Equation of the line $PL$ is given by
$\frac{x-1}{3-1}=\frac{y-0}{-4-0}=\frac{z-0}{-2-0}$
$=\frac{z-1}{1}=\frac{y}{-2}=\frac{z}{-1}$
$\Rightarrow \vec{r}=\hat{i}+\lambda(\hat{i}-2 \hat{j}-\hat{k})$

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Question 35 Marks

The cost of 4kg onion, 3kg wheat and 2kg rice is Rs. 60. The cost of 2kg onion, 4kg wheat and 6kg rice is Rs. 90. The cost of 6kg onion 2kg wheat and 3kg rice is Rs. 70. Find the cost of each item per kg by matrix method.

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Question 45 Marks

Let L be the set of all lines in xy plane and R be the relation in L. define as $s R =\left\{\left( L _1, L_2\right): L _1 \| L _2\right\}$ Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4

Answer

$L _1 \| L _1$ i.e $\left( L _1, L_1\right) \in R$ Hence reflexive
Let $\left(L_1, L_2\right) \in R$, then
$L _1 \| L _2$ which implies $L _2 \| L 1$
$\Rightarrow\left( L _2, L_1\right) \in R$ Hence symmetric
We know the
$L _1 \| L _2$ and $L _2 \| L _3$
Then $L _1 \| L _3$
Therefore, $\left(L_1, L_2\right) \in R$ and $\left(L_2, L_3\right) \in R$ implies $\left(L_1, L_3\right) \in R$
Hence Transitive Hence, R is an equivalence relation.
Any line parallel to y = 2x + 4 is of the form y = 2x + K where k is a real number.
Therefore, set of all lines parallel to y = 2x + 4 is {y : y = 2x + k , k is a real number}

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Question 55 Marks

$\overrightarrow{A B}=3 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{C D}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6 \hat{i}+7 \hat{j}+4 \hat{k}$ and $-9 \hat{j}+2 \hat{k}$, respectively. Find the position vector of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{P Q}$ is perpendicular to $\overrightarrow{A B}$ and $\overrightarrow{C D}$ both.

Answer

We have, $\overrightarrow{A B}=3 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{C D}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$
Also, the position vectors of $A$ and $C$ are $6 \hat{i}+7 \hat{j}+4 \hat{k}$ and $-9 \hat{j}+2 \hat{k}$, respectively. Since, $\overrightarrow{P Q}$ is perpendicular to both $\overrightarrow{A B}$ and $\overrightarrow{C D}$.
So, $P$ and $Q$ will be foot of perpendicular to both the lines through $A$ and $C.$
Now, equation of the line through A and parallel to the vector $\overrightarrow{A B}$ is,
$\vec{r}=(6 \hat{i}+7 \hat{j}+4 \hat{k})+\lambda(3 \hat{i}-\hat{j}+\hat{k})$
And the line through C and parallel to the vector $\overrightarrow{C D}$ is given by
$\vec{r}=-9 \hat{j}+2 \hat{k}+\mu(-3 \hat{i}+2 \hat{j}+4 \hat{k})$
Let $\vec{r}=(6 i+7 j+4 \hat{k})+\lambda(3 \hat{i}-\hat{j}+\hat{k})$
and $\vec{r}=-9 \hat{j}+2 \hat{k}+\mu(-3 \hat{i}+2 \hat{j}+4 \hat{k}) \ldots (ii)$
Let $P(6+3 \lambda, 7-\lambda, 4+\lambda)$ is any point on the first line and Q be any point on second line is given by
$(-3 \mu,-9+2 \mu, 2+4 \mu)$
$\therefore \overrightarrow{P Q}=(-3 \mu-6-3 \lambda) \hat{i}+(-9+2 \mu-7+\lambda) \hat{j}+(2+4 \mu-4-\lambda) \hat{k}$
$=(-3 \mu-6-3 \lambda) \hat{i}+(2 \mu+\lambda-16) \hat{j}+(4 \mu-\lambda-2) \hat{k}$
If $\overrightarrow{P Q}$ is perpendicular to the first line, then
$3(-3 \mu-6-3 \lambda)-(2 \mu+\lambda-16)+(4 \mu-\lambda-2)=0$
$\Rightarrow-9 \mu-18-9 \lambda-2 \mu-\lambda+16+4 \mu-\lambda-2=0$
$\Rightarrow-7 \mu-11 \lambda-4=0 \ldots . \text { (iii) }$
If $\overrightarrow{P Q}$ is perpendicular to the second line, then
$-3(-3 \mu-6-3 \lambda)+(2 \mu+\lambda-16)+(4 \mu-\lambda+2)=0$
$\Rightarrow 9 \mu+18+9 \lambda+4 \mu+2 \lambda-32+16 \mu-4 \lambda-8=0$
$\Rightarrow 29 \mu+7 \lambda-22=0 \ldots \ldots \text { (iv) }$
On solving Eqs. $(iii)$ and $(iv)$, we get
$-49 \mu-77 \lambda-28=0$
$\Rightarrow 319 \mu+77 \lambda-242=0$
$\Rightarrow 270 \mu-270=0$
$\Rightarrow \mu=1$
Using $\mu$ in Eq. $(iii)$, we get
$-7(1)=-11 \lambda-4=0$
$\Rightarrow-7-11 \lambda-4=0$
$\Rightarrow-11-11 \lambda=0$
$\Rightarrow \lambda=-1$
$\therefore \overrightarrow{P Q}=(-3(1)-6-3(-1)) \hat{i}+(2(1)+(-1)-16) \hat{j}+(4(1)-(-1)-2) \hat{k}$
$=-6 \hat{i}-15 \hat{j}+3 \hat{k}$

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Question 65 Marks

Find the area of the region enclosed by the parabola $x^2=y$ the line $y=x+2$ and $x-$ axis.

Answer

Equation of parabola is $x^2=y \ldots (i)$
Equation of line is $y=x+2 \ldots (ii)$
Here the two points of intersections of parabola $(i)$ and line $(ii)$ are $A (-1,1)$ and $B (2,4)$.
Area $\text{ALODBM =}$ Area bounded by parabola $(i)$ and $x -$ axis
$=\left|\int_{-1}^2 x^2 d x\right|=\left(\frac{x^3}{3}\right)_{-1}^2$
$=\frac{8}{3}-\left(\frac{-1}{3}\right)$
$=\frac{8}{3}+\frac{1}{3}=\frac{9}{3}=3$ sq units

Also Area of trapezium $\text{ALMB =}$ Area bounded by line $(ii)$ and $x-$ axis
$=\left|\int_{-1}^2(x+2) d x\right|=\left(\frac{x^2}{2}+2 x\right)_{-1}^2$
$=2+4-\left(\frac{1}{2}-2\right)$
$=6-\frac{1}{2}+2$
$=\frac{15}{2} \text { sq. units }$
Now Required area = Area of trapezium $\text{ALMB -}$ Area $\text{ALODBM}$
$=\frac{15}{2}-3=\frac{9}{2}$ sq. units

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