A relation R is defined from a set A = [2, 3, 4, 5] to a set B = [3, 6, 7, 10] as follows:
Question: A relation R is defined from a set A = [2, 3, 4, 5] to a set B = [3, 6, 7, 10] as follows: (x,y) R ⇔xis relatively prime toy Express R as a set of ordered pairs and determine its domain and range. Solution: Given: (x,y) R ⇔xis relatively prime to y. Here, 2 is co-prime to 3 and 7. 3 is co-prime to 7 and 10. 4 is co-prime to 3 and 7. 5 is co-prime to 3, 6 and 7. Thus, we get: R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)} Domain of R = {2, 3, 4, 5} Range o...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{44}{x+y}+\frac{30}{x-y}=10$ $\frac{55}{x+y}+\frac{40}{x-y}=13$ Solution: The given equations are: $\frac{44}{x+y}+\frac{30}{x-y}=10$ $\frac{55}{x+y}+\frac{40}{x-y}=13$ Let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ then equations are $44 u+30 v=10 \ldots(i)$ $55 u+40 v=13 \ldots($ ii $)$ Multiply equation $(i)$ by 4 and equation $(i i)$ by 3 add both equations, we get Put the value of $u$ in equation $(i)$, we get $44 \times \frac{1}{11}+30...
Read More →In the given fig, find the values of x, y and z.
Question: In the given fig, find the values of x, y and z. Solution: From the given figure y = 25 [Vertically opposite angles are equal] Now x + y = 180 [Linear pair of angles] x = 180 - 25 x = 155 Also, z = x = 155 [Vertically opposite angles] y = 25 z = 155...
Read More →If A = [1, 2, 3], B = [4, 5, 6],
Question: If A = [1, 2, 3], B = [4, 5, 6], which of the following are relations from A to B? Give reasons in support of your answer.(i) [(1, 6), (3, 4), (5, 2)] (ii) [(1, 5), (2, 6), (3, 4), (3, 6)] (iii) [(4, 2), (4, 3), (5, 1)] (iv) A B. Solution: Given:A = {1, 2, 3} and B = {4, 5, 6}Thus, we have: A B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}(i) {(1, 6), (3, 4), (5, 2)} Since it is not a subset of A B, it is not a relation from A to B. (ii) {(1, 5), (2, 6), (3...
Read More →In the below fig. three coplanar lines intersect at a point O, forming angles as shown in the figure.
Question: In the below fig. three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u. Solution: Vertically opposite angles are equal So SOD = z = 90 DOF = y = 50 Now, x + y + z = 180 [Linear pair] x + y + z = 180 90 + 50 + x = 180 x = 180 - 140 x = 40 Hence values of x, y, z and u are 40, 50, 90 and 40 respectively in degrees....
Read More →If A × B
Question: IfABCDandAB ϕ, prove thatACandBD. Solution: Let: $(x, y) \in(A \times B)$ $\therefore x \in A, y \in B$ Now, $\because(A \times B) \subseteq(C \times D)$ $\therefore(x, y) \in(C \times D)$ Or $x \in C$ and $y \in D$ Thus, we have : $A \subseteq C \ B \subseteq D$...
Read More →In the below fig, lines l1, and l2 intersect at O, forming angles as shown in the figure. If x = 45. Find the values of x,
Question: In the below fig, lines l1, and l2intersect at O, forming angles as shown in the figure. If x = 45. Find the values of x, y. z and u. Solution: Given that X = 45, Y = ?, Z = ?, u = ? Vertically opposite angles are equal Therefore z = x = 45 z and u are angles that are a linear pair Therefore, z + u = 180 z = 180 - u u = 180 - x u = 180 - 45 u = 135 x and y angles are a linear pair x+ y = 180 y = 180 - x y =180 - 45 y = 135 Hence, x = 45, y = 135, z = 135 and u = 45...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $x-y+z=4$ $x+y+z=2$ $2 x+y-3 z=0$ Solution: The given equations are: $x-y+z=4 \ldots(i)$ $x+y+z=2 \quad \ldots($ ii $)$ $2 x+y-3 z=0 \quad \ldots($ iii $)$ First of all we find the value of $x$ $x=4+y-z$ Put the value of $x$ in equation $(i)$, we get $4+y-z+y+z=2$ $\Rightarrow 2 y=-2$ $\Rightarrow y=-1$ Put the value of $x$ and $y$ in equation in $(i i i)$ we get $2(4+y-z)+y-3 z=0$ $\Rightarrow 8-2-2 z-1-3 z=0$ $\Rightarrow-5 z=-5$ $\Rightarrow...
Read More →Prove
Question: $\frac{4 x+1}{\sqrt{2 x^{2}+x-3}}$ Solution: Let $4 x+1=A \frac{d}{d x}\left(2 x^{2}+x-3\right)+B$ $\Rightarrow 4 x+1=A(4 x+1)+B$ $\Rightarrow 4 x+1=4 A x+A+B$ Equating the coefficients ofxand constant term on both sides, we obtain $4 A=4 \Rightarrow A=1$ $A+B=1 \Rightarrow B=0$ Let $2 x^{2}+x-3=t$ $\therefore(4 x+1) d x=d t$ $\Rightarrow \int \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x=\int \frac{1}{\sqrt{t}} d t$ $=2 \sqrt{t}+\mathrm{C}$ $=2 \sqrt{2 x^{2}+x-3}+\mathrm{C}$...
Read More →Prove that:
Question: Prove that: (i) (AB) C= (AC) (BC) (ii) (A B) C = (AC) (BC) Solution: (i) (AB) C= (AC) (BC) Let(a, b) be an arbitrary element of (AB) C. Thus, we have: $(a, b) \in(A \cup B) \times C$ $\Rightarrow a \in(A \cup B)$ and $b \in C$ $\Rightarrow(a \in A$ or $a \in B)$ and $b \in C$ $\Rightarrow(a \in A$ and $b \in C)$ or $(a \in B$ and $b \in C)$ $\Rightarrow(a, b) \in(A \times C)$ or $(a, b) \in(B \times C)$ $\Rightarrow(a, b) \in(A \times C) \cup(B \times C)$ $\therefore(A \cup B) \times C...
Read More →In the below fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS − ∠POS).
Question: In the below fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = 1/2(QOS POS). Solution: Given that OR perpendicular POR = 90 POS + SOR = 90 [ POR = POS + SOR] ROS = 90 POS... (1) QOR= 90 (∵ OR PQ) QOS ROS = 90 ROS = QOS 90 By adding (1) and (2) equations, we get ROS = QOS POS ROS = 1/2(QOS POS)...
Read More →Prove that:
Question: Prove that: (i) (AB) C= (AC) (BC) (ii) (A B) C = (AC) (BC) Solution: (i) (AB) C= (AC) (BC) Let(a, b) be an arbitrary element of (AB) C. Thus, we have: $(a, b) \in(A \cup B) \times C$ $\Rightarrow a \in(A \cup B)$ and $b \in C$ $\Rightarrow(a \in A$ or $a \in B)$ and $b \in C$ $\Rightarrow(a \in A$ and $b \in C)$ or $(a \in B$ and $b \in C)$ $\Rightarrow(a, b) \in(A \times C)$ or $(a, b) \in(B \times C)$ $\Rightarrow(a, b) \in(A \times C) \cup(B \times C)$ $\therefore(A \cup B) \times C...
Read More →Prove
Question: $\frac{1}{\sqrt{(x-a)(x-b)}}$ Solution: $(x-a)(x-b)$ can be written as $x^{2}-(a+b) x+a b$. Therefore $x^{2}-(a+b) x+a b$ $=x^{2}-(a+b) x+\frac{(a+b)^{2}}{4}-\frac{(a+b)^{2}}{4}+a b$ $=\left[x-\left(\frac{a+b}{2}\right)\right]^{2}-\frac{(a-b)^{2}}{4}$ $\Rightarrow \int \frac{1}{\sqrt{(x-a)(x-b)}} d x=\int \frac{1}{\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\left(\frac{a-b}{2}\right)^{2}}} d x$ Let $x-\left(\frac{a+b}{2}\right)=t$ $\therefore d x=d t$ $\Rightarrow \int \frac{...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $x-y+z=4$ $x-2 y-2 z=9$ $2 x+y+3 z=1$ Solution: The given equations are: $x-y+z=4 \ldots(i)$ $x-2 y-2 z=9 \ldots(i i)$ $2 x+y+3 z=1 \ldots(i i i)$ First of all we find the value of $x$ $x=4+y-z$ Put the value of $x$ in equation $(i i)$, we get $4+y-z-2 y-2 z=9$ $\Rightarrow-3 z-y=5 \ldots(i v)$ Put the value of $x$ and $y$ in equation in $($ iii $)$ we get $2(4+y-z)+y+3 z=1$ $\Rightarrow 8+2 y-2 z+y+3 z=1$ $\Rightarrow 3 y+z=-7 \quad \ldots(v)$...
Read More →In the below fig, lines PQ and RS intersect each other at point O.
Question: In the below fig, lines PQ and RS intersect each other at point O. If POR: ROQ = 5: 7. Find all the angles. Solution: Given POR and ROPis linear pair POR + ROP = 180 Given that POR: ROQ = 5:7 Hence, POR = (5/12) 180 = 75 Similarly ROQ = (7/7 + 5) 180 = 105 Now POS = ROQ = 105 [Vertically opposite angles] Also, SOQ = POR = 75 [Vertically opposite angles]...
Read More →In below fig, ray OS stand on a line POQ.
Question: In below fig, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of POS and SOQ respectively. If POS = x, find ROT? Solution: Given, Ray OS stand on a line POQ Ray OR and Ray OT are angle bisectors ofPOS and SOQrespectively POS = x POS and SOQis linear pair POS + QOS = 180 x + QOS = 180 QOS = 180 - x Now, ray or bisector POS ROS = 1/2 POS x/2 ROS = x/2 [Since POS = x] Similarly ray OT bisector QOS TOS = 1/2 QOS = (180 - x)/2 [QOS = 180 - x] = 90 - x/2 Hence,ROT = ROS + R...
Read More →Prove
Question: $\frac{1}{\sqrt{8+3 x-x^{2}}}$ Solution: $8+3 x-x^{2}$ can be written as $8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$ Therefore, $8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$ $=\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}$ $\Rightarrow \int \frac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x$ Let $x-\frac{3}{2}=t$ $\therefore d x=d t$ $\Rightarrow \int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x=\int \fra...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $23 x-29 y=98$ $29 x-23 y=110$ Solution: The given equations are: $23 x-29 y=98 \ldots(i)$ $29 x-23 y=110 \ldots($ ii $)$ Multiply equation $(i)$ by 23 and equation $(i i)$ by 29 and subtract (ii) from (i) we get Put the value of $x$ in equation $(i)$, we get $23 \times 3-29 y=98$ $\Rightarrow-29 y=29$ $\Rightarrow y=-1$ Hence the value of $x=3$ and $y=-1$...
Read More →In below fig. OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Question: In below fig. OP, OQ, OR and OS are four rays. Prove that: POQ + QOR + SOR + POS = 360 Solution: Given that OP, OQ, OR and OS are four rays You need to produce any of the ray OP, OQ, OR and OS backwards to a point in the figure. Let us produce ray OQ backwards to a point T So that TOQ is a line Ray OP stands on the TOQ SinceTOP, POQis a linear pair TOP + POQ = 180... (1) Similarly, Ray OS stands on the line TOQ TOS + SOQ = 180... (2) ButSOQ = SOR + QOR ... (3) So, eqn (2) becomes TOS +...
Read More →If A = {1, 2, 3}, B = {3, 4}
Question: If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find(i) A (B C) (ii) (A B) (A C) (iii) A (B C) (iv) (A B) (A C) Solution: Given: A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6} (i) A (B C) Now,(B C) = {4} A (B C) = {(1, 4), (2, 4), (3, 4)}(ii) (A B) (A C) Now, (A B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} And, (A C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} (A B) (A C) = {(1, 4), (2, 4), (3, 4)} (iii) A (B C) Now, (B C) = {3, 4, 5, 6} A (B C) = {...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $99 x+101 y=499$ $101 x+99 y=501$ Solution: The given equations are: $99 x+101 y=499 \ldots(i)$ $101 x+99 y=501 \ldots(i i)$ Multiply equation $(i)$ by 99 and equation $(i i)$ by 101 , and subtract (ii) from (i) we get Put the value of $x$ in equation $(i)$, we get $99 \times 3+101 y=499$ $\Rightarrow 101 y=202$ $\Rightarrow y=2$ Hence the value of $x=3$ and $y=2$...
Read More →If the figure below forms a linear pair,
Question: If the figure below forms a linear pair, EOB = FOC = 90 and DOC = FOG = AOB = 30 Find the measure ofFOE, COB and DOE Name all the right angles Name three pairs of adjacent complementary angles Name three pairs of adjacent supplementary angles Name three pairs of adjacent angles Solution: (i)FOE = x, DOE = y and BOC = z SinceAOF, FOGis a linear pair AOF+ 30 = 180 AOF= 180 - 30 AOF= 150 AOB + BOC + COD + DOE + EOF = 150 30 + z + 30 + y + x = 150 x + y + z =150 - 30 - 30 x + y + z = 90 .....
Read More →Prove
Question: $\frac{1}{\sqrt{(x-1)(x-2)}}$ Solution: $(x-1)(x-2)$ can be written as $x^{2}-3 x+2$. Therefore, $x^{2}-3 x+2$ $=x^{2}-3 x+\frac{9}{4}-\frac{9}{4}+2$ $=\left(x-\frac{3}{2}\right)^{2}-\frac{1}{4}$ $=\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{3}\right)^{2}$ $\therefore \int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x$ Let $x-\frac{3}{2}=t$ $\therefore d x=d t$ $\Rightarrow \int \frac{1}{\sqrt{\left(x-\frac{3}{2}...
Read More →Solve the following systems of equations:
Question: Solve the following systems of equations: $\frac{4}{x}+3 y=14$ $\frac{3}{x}-4 y=23$ Solution: The given equations are: $\frac{4}{x}+3 y=14 \ldots$(i) $\frac{3}{x}-4 y=23 \ldots(i i)$ Multiply equation $(i)$ by 4 and equation $(i i)$ by 3, add both equations, we get $\frac{16}{x}+12 y=56$ Put the value of $x$ in equation $(i)$, we get $\frac{4}{\frac{1}{5}}+3 y=14$ $\Rightarrow 3 y=-6$ $\Rightarrow y=-2$ Hence the value of $x=\frac{1}{5}$ and $y=-2$....
Read More →Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that:
Question: LetA= {1, 2},B= {1, 2, 3, 4},C= {5, 6} andD= {5, 6, 7, 8}. Verify that:(i)ACBD (ii)A (BC) = (AB) (AC) Solution: Given: A= {1, 2},B= {1, 2, 3, 4},C= {5, 6} andD= {5, 6, 7, 8} (i)ACBD LHS:AC ={(1, 5), (1, 6), (2, 5), (2, 6)} RHS:BD ={(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)} ACBD (ii)A (BC) = (AB) (AC) We have: $(B \cap C)=\phi$ LHS: $A \times(B \cap C)=\phi$ Now, (AB) = {(1, 1), (1, 2), (1, 3), (1, 4),...
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