Let A = {1, 2, 3} and R
Question: Let $\mathrm{A}=\{1,2,3\}$ and $R=\left\{(a, b):\left|a^{2}-b^{2}\right| \leq 5, a, b \in A\right\}$. Then write $\mathrm{R}$ as set of ordered pairs. Solution: Given: A = {1, 2, 3} $R=\left\{(a, b):\left|a^{2}-b^{2}\right| \leq 5, a, b \in A\right\}$ We know that $\left|1^{2}-1^{2}\right| \leq 5$ $\left|2^{2}-2^{2}\right| \leq 5$ $\left|3^{2}-3^{2}\right| \leq 5$, $\left|1^{2}-2^{2}\right| \leq 5$, $\left|2^{2}-1^{2}\right| \leq 5$ $\left|2^{2}-3^{2}\right| \leq 5$ $\left|3^{2}-2^{2}\...
Read More →Let A = {1, 2, 3} and R
Question: Let $\mathrm{A}=\{1,2,3\}$ and $R=\left\{(a, b):\left|a^{2}-b^{2}\right| \leq 5, a, b \in A\right\}$. Then write $\mathrm{R}$ as set of ordered pairs. Solution: Given: A = {1, 2, 3} $R=\left\{(a, b):\left|a^{2}-b^{2}\right| \leq 5, a, b \in A\right\}$ We know that $\left|1^{2}-1^{2}\right| \leq 5$ $\left|2^{2}-2^{2}\right| \leq 5$ $\left|3^{2}-3^{2}\right| \leq 5$, $\left|1^{2}-2^{2}\right| \leq 5$, $\left|2^{2}-1^{2}\right| \leq 5$ $\left|2^{2}-3^{2}\right| \leq 5$ $\left|3^{2}-2^{2}\...
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Question: $\frac{\cos x}{(1-\sin x)(2-\sin x)}$ [Hint: Put $\left.\sin x=f\right]$ Solution: $\frac{\cos x}{(1-\sin x)(2-\sin x)}$ Let $\sin x=t \Rightarrow \cos x d x=d t$ $\therefore \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}$ Let $\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}$ $1=A(2-t)+B(1-t)$ ...(1) Substitutingt= 2 and thent= 1 in equation (1), we obtain A= 1 andB= 1 $\therefore \frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}$ $\Rightarrow \int \fr...
Read More →Solve each of the following systems of equations by the method of cross-multiplication :
Question: Solve each of the following systems of equations by the method of cross-multiplication : $2(a x-b y)+a+4 b=0$ $2(b x+a y)+b-4 a=0$ Solution: GIVEN: $2(a x-b y)+a+4 b=0$ $2(b x+a y)+b-4 a=0$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $2(a x-b y)+a+4 b=0$ $2(b x+a y)+b-4 a=0$ After rewriting equations $2 a x-2 b y+(a+4 b)=0$ $2 b x+2 a y+(b-4 a)=0$ By cross multiplication method we get $\frac{x}{(...
Read More →If R is a relation from set A
Question: If R is a relation from set A = (11, 12, 13) to set B = (8, 10, 12) defined byy=x 3, then write R1. Solution: Given: A = (11, 12, 13) and B = (8, 10, 12) R is defined by (y=x 3) from A to B. We know: $8=11-3$ $10=13-3$ $\therefore R=\{(11,8),(13,10)\}$ Or, $\mathrm{R}^{-1}=\{(8,11),(10,13)\}$...
Read More →If R is a relation from set A
Question: If R is a relation from set A = (11, 12, 13) to set B = (8, 10, 12) defined byy=x 3, then write R1. Solution: Given: A = (11, 12, 13) and B = (8, 10, 12) R is defined by (y=x 3) from A to B. We know: $8=11-3$ $10=13-3$ $\therefore R=\{(11,8),(13,10)\}$ Or, $\mathrm{R}^{-1}=\{(8,11),(10,13)\}$...
Read More →Solve each of the following systems of equations by the method of cross-multiplication :
Question: Solve each of the following systems of equations by the method of cross-multiplication : $\frac{57}{x+y}+\frac{6}{x-y}=5$ $\frac{38}{x+y}+\frac{21}{x-y}=9$ Solution: GIVEN: $\frac{57}{x+y}+\frac{6}{x-y}=5$ $\frac{38}{x+y}+\frac{21}{x-y}=9$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{57}{x+y}+\frac{6}{x-y}-5=0$ $\frac{38}{x+y}+\frac{21}{x-y}-9=0$ let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ ...
Read More →If R = {(x, y) :
Question: If R = {(x,y) :x,y Z,x2+y2 4} is a relation defined on the set Z of integers, then write domain of R. Solution: Given: R = {(x,y) :x,y Z,x2+y2 4} We know: $(-2)^{2}+0^{2} \leq 4$ $(2)^{2}+0^{2} \leq 4$ $(-1)^{2}+0^{2} \leq 4$ $(1)^{2}+0^{2} \leq 4$ $(-1)^{2}+(1)^{2} \leq 4$ $0^{2}+0^{2} \leq 4$ $(1)^{2}+(1)^{2} \leq 4$ $(-1)^{2}+(-1)^{2} \leq 4$ $\therefore$ Domain $(R)=\{-\overline{2},-1,0,1,2\}$...
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Question: $\frac{1}{x\left(x^{n}+1\right)}$ [Hint: multiply numerator and denominator by $x^{n-1}$ and put $\left.x^{n}=t\right]$ Solution: $\frac{1}{x\left(x^{n}+1\right)}$ Multiplying numerator and denominator by $x^{n-1}$, we obtain $\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n-1} x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)}$ Let $x^{n}=t \Rightarrow x^{n-1} d x=d t$ $\therefore \int \frac{1}{x\left(x^{n}+1\right)} d x=\int \frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} ...
Read More →Determine the measure of each of the equal angles of a right-angled isosceles triangle. OR
Question: Determine the measure of each of the equal angles of a right-angled isosceles triangle. OR ABC is a right-angled triangle in which A = 90 and AB = AC. Find B and C. Solution: ABC is a right angled triangle Consider on a right - angled isosceles triangle ABC such that A = 90and AB = AC Since, AB = AC ⟹C =B .... (i) [Angles opposite to equal sides are equal] Now, Sum of angles in a triangle = 180 A +B +C =180 ⟹ 90 +B+B = 180 ⟹2B = 90 ⟹B = 45 B = 45,C = 45 Hence, the measure of each of th...
Read More →In Fig. (10).2(5) AB = AC and DB = DC, find the ratio ∠ABD: ∠ACD.
Question: In Fig. (10).2(5) AB = AC and DB = DC, find the ratio ABD: ACD. Solution: Consider the figure Given, AB = AC, DB = DC and given to find the ratio ABD =ACD Now,ΔABC andΔDBC are isosceles triangles since AB = AC and DB = DC respectively ABC =ACB andDBC =DCB [Angles opposite to equal sides are equal] Now consider, ABD :ACD (ABC -DBC): (ACB -DCB) (ABC -DBC): (ABC -DBC) [ABC =ACB andDBC =DCB] 1: 1 ABD: ACD = 1: 1...
Read More →If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Question: If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other. Solution: ED is a straight line segment and B and C an points on it. EBC =BCD = straight angle = 180 EBA +ABC =ACB +ACD EBA =ACD +ACB -ABC EBA =ACD [From (i) ABC = ACD] ABE =ACD Hence proved...
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Question: $\frac{1}{x^{4}-1}$ Solution: $\frac{1}{\left(x^{4}-1\right)}=\frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)}=\frac{1}{(x+1)(x-1)\left(1+x^{2}\right)}$ Let $\frac{1}{(x+1)(x-1)\left(1+x^{2}\right)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C x+D}{\left(x^{2}+1\right)}$ $1=A(x-1)\left(x^{2}+1\right)+B(x+1)\left(x^{2}+1\right)+(C x+D)\left(x^{2}-1\right)$ $1=A\left(x^{3}+x-x^{2}-1\right)+B\left(x^{3}+x+x^{2}+1\right)+C x^{3}+D x^{2}-C x-D$ $1=(A+B+C) x^{3}+(-A+B+D) x^{2}+(A+B-C) x+(-A+B-D...
Read More →Find the measure of each exterior angle of an equilateral triangle.
Question: Find the measure of each exterior angle of an equilateral triangle. Solution: Given to find the measure of each exterior angle of an equilateral triangle consider an equilateral triangle ABC. We know that for an equilateral triangle AB = BC = CA andABC =BCA = CAB =180/3= 60.... (i) Now, Extend side BC to D, CA to E and AB to F. Here BCD is a straight line segment BCD = Straight angle =180 BCA +ACD = 180 [From (i)] 60 +ACD = 180 ACD = 120 Similarly, we can findFAB andFBC also as 120 bec...
Read More →In a ∆ABC = AC and ∠ACD = (10)5°. Find ∠BAC.
Question: In a ∆ABC = AC and ACD = (10)5. Find BAC. Solution: We have, AB = AC andACD = (10)5 Since,BCD = 180 = Straight angle BCA +ACD = 180 BCA + (10)5 = 180 BCA = l80 - (10)5 BCA = 75 And also, ΔABCis an isosceles triangle [AB = AC] ABC =ACB [Angles opposite to equal sides are equal] From (i), we have ACB = 75 ABC =ACB = 75 And also, Sum of Interior angles of a triangle = 180 ABC =BCA +CAB = 180 75 + 75 +CAB =180 150 +BAC = 180 BAC = 180 - 150 = 30 BAC = 30...
Read More →The vertical angle of an isosceles triangle is (10)0°. Find its base angles.
Question: The vertical angle of an isosceles triangle is (10)0. Find its base angles. Solution: Consider an isoscelesΔABC such that AB = AC Given that vertical angle A is (10)0 To find the base angles SinceΔABC is isosceles B =C [Angles opposite to equal sides are equal] And also, Sum of interior angles of a triangle = 180 A +B + C = 180 (10)0+BB = 180 2B = 180∘- (10)0 B = 40 B = C = 40...
Read More →In a ΔABC, if AB = AC and ∠B = 70°. Find ∠A.
Question: In a ΔABC, if AB = AC and B = 70. Find A. Solution: Consider aΔABC, if AB = AC andB = 70 Since, AB = ACΔABC is an isosceles triangle B =C [Angles opposite to equal sides are equal] B =C = 70∘ And also, Sum of angles in a triangle = 180 A +B + C = 180 A + 70+ 70= 180 A = 180- 140 A = 40...
Read More →In a ΔABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Question: In a ΔABC, if A = 120 and AB = AC. Find B and C. Solution: Consider aΔABC Given Mat A = 120 and AB = AC and given to find B and C. We can observe thatΔABC is an isosceles triangle since AB = AC B =C (i) [Angles opposite to equal sides are equal] We know that sum of angles in a triangle is equal to 180 ⟹A +B +C = 180 [From (i)] ⟹A +B +B = 180 ⟹ 120+ 2B = 180 ⟹ 2B = 180- 120 ⟹ B = C = 30...
Read More →Prove that the medians of an equilateral triangle are equal.
Question: Prove that the medians of an equilateral triangle are equal. Solution: To prove the medians of an equilateral triangle are equal. Median: The line Joining the vertex and midpoint of opposite side. Now, consider an equilateral triangle ABC. Let D, E, F are midpoints of BC, CA and AB. Then, AD, BE and CF are medians of ABC. Now, D Is midpoint of BC ⟹ BD = DC =BC/2 Similarly, CE = EA =AC/2 AF = FB =AB/2 SinceΔABC is an equilateral triangle ⟹ AB = BC = CA... (i) ⟹BD = DC = CE = EA = AF = F...
Read More →If R is a relation defined on the set Z of integers by the rule (x, y)
Question: If R is a relation defined on the set Z of integers by the rule (x,y) R ⇔x2+y2= 9, then write domain of R. Solution: We need to find (x,y) R such thatx2+y2= 9. Now, $(3)^{2}+0^{2}=9$ $\Rightarrow(-3)^{2}+0^{2}=9$ $x$ can take values $-3,0$ and $3 .$ $\therefore$ Domain $(R)=\{-3,0,3\}$...
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Question: $\frac{3 x-1}{(x+2)^{2}}$ Solution: Let $\frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}$ $\Rightarrow 3 x-1=A(x+2)+B$ Equating the coefficient ofxand constant term, we obtain A= 3 2A+B= 1 ⇒B= 7 $\therefore \frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}$ $\Rightarrow \int \frac{3 x-1}{(x+2)^{2}} d x=3 \int \frac{1}{(x+2)} d x-7 \int \frac{x}{(x+2)^{2}} d x$ $=3 \log |x+2|-7\left(\frac{-1}{(x+2)}\right)+\mathrm{C}$ $=3 \log |x+2|+\frac{7}{(x+2)}+\mathrm{C}$...
Read More →If n(A) = 3, n(B) = 4,
Question: Ifn(A) = 3,n(B) = 4, then writen(A A B). Solution: Given: n(A) = 3 andn(B) = 4 Now, we have: $n(A \times A \times B)=n(A \times A) \times n(B)=3 \times 3 \times 4=36$...
Read More →In fig. (10).23, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT = QT (ii) ∠TQR = 15°
Question: In fig. (10).23, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT = QT (ii) TQR = 15 Solution: Given that PQRS is a square and SRT is an equilateral triangle. And given to prove that (i) PT = QT and (ii)TQR = 15 Now, PQRS is a square PQ = QR = RS = SP ... (i) AndSPQ =PQR =QRS =RSP = 90= right angle And also, SRT is an equilateral triangle. SR = RT = TS ... (ii) AndTSR =SRT =RTS = 60 From (i) and (ii) PQ = QR = SP = SR = RT = TS .... (iii) And also, TSP =TSR +RSP =...
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Question: $\frac{2}{(1-x)\left(1+x^{2}\right)}$ Solution: Let $\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{(1-x)}+\frac{B x+C}{\left(1+x^{2}\right)}$ $2=A\left(1+x^{2}\right)+(B x+C)(1-x)$ $2=A+A x^{2}+B x-B x^{2}+C-C x$ Equating the coefficient of $x^{2}, x$, and constant term, we obtain $A-B=0$ $B-C=0$ $A+C=2$ On solving these equations, we obtain $A=1, B=1$, and $C=1$ $\therefore \frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^{2}}$ $\Rightarrow \int \frac{2}{(1-x)\left(1+x^...
Read More →Solve each of the following systems of equations by the method of cross-multiplication :
Question: Solve each of the following systems of equations by the method of cross-multiplication : $a^{2} x+b^{2} y=c^{2}$ $b^{2} x+a^{2} y=d^{2}$ Solution: GIVEN: $a^{2} x+b^{2} y=c^{2}$ $b^{2} x+a^{2} y=d^{2}$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $a^{2} x+b^{2} y-c^{2}=0$ $b^{2} x+a^{2} y-d^{2}=0$ By cross multiplication method we get $\frac{x}{\left(-d^{2} b^{2}\right)-\left(-c^{2} a^{2}\right)}=...
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