If R is a relation on the set A
Question: If R is a relation on the set A = [1, 2, 3, 4, 5, 6, 7, 8, 9] given byxRy⇔y= 3x, then R =(a) [(3, 1), (6, 2), (8, 2), (9, 3)] (b) [(3, 1), (6, 2), (9, 3)] (c) [(3, 1), (2, 6), (3, 9)] (d) none of these Solution: (d) none of these A = {1, 2, 3, 4, 5, 6, 7, 8, 9} xRy⇔y= 3x Forx= 1,y= 3 Forx= 2,y= 6 Forx= 3,y= 9Thus, R = {(1,3),(2,6),(3,9)}...
Read More →Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Question: Prove that the straight lines perpendicular to the same straight line are parallel to one another. Solution: Let AB and CD be drawn perpendicular to the Line MN ABD = 90 [AB is perpendicular to MN] .... (i) CON = 90 [CD is perpendicular to MN ] .... (ii) Now, ABD = CDN = 90[From (i) and (ii)] Therefore, AB ∥CD, Since corresponding angles are equal....
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 : $\frac{x}{a}+\frac{y}{b}=2$ $a x-b y=a^{2}-b^{2}$ Solution: GIVEN: $\frac{x}{a}+\frac{y}{b}=2$ $a x-b y=a^{2}-b^{2}$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{x}{a}+\frac{y}{b}-2=0$ $a x-b y-\left(a^{2}-b^{2}\right)=0$ By cross multiplication method we get $\frac{x}{\left(\left(\frac{1}{b} \times-\le...
Read More →Prove
Question: $\frac{x+2}{\sqrt{4 x-x^{2}}}$ Solution: Let $x+2=A \frac{d}{d x}\left(4 x-x^{2}\right)+B$ $\Rightarrow x+2=A(4-2 x)+B$ Equating the coefficients ofxand constant term on both sides, we obtain $-2 A=1 \Rightarrow A=-\frac{1}{2}$ $4 A+B=2 \Rightarrow B=4$ $\Rightarrow(x+2)=-\frac{1}{2}(4-2 x)+4$ $\therefore \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x$ $x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$ $=-\frac{1}{2} \int \frac{4-2 x}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-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 : $a x+b y=a^{2}$ $b x+a y=b^{2}$ Solution: GIVEN: $a x+b y=a^{2}$ $b x+a y=b^{2}$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $a x+b y-a^{2}=0$ $b x+a y-b^{2}=0$ By cross multiplication method we get $x=\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{\left(a^{2}-b^{2}\right)}\left\{\sin c e\left(a^{3}-b^{3}\right)=(a...
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 : $x+a y=b$ $a x-b y=c$ Solution: GIVEN: $x+a y=b$ $a x-b y=c$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $x+a y-b=0$ $a x-b y-c=0$ By cross multiplication method we get $x=\frac{\left(a c+b^{2}\right)}{\left(b+a^{2}\right)}$, and $y=\frac{-(-c+a b)}{\left(-b-a^{2}\right)} y=\frac{(a b-c)}{\left(a^{2}+b\right)...
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 : $a x+b y=a-b$ $b x-a y=a+b$ Solution: GIVEN: $a x+b y=a-b$ $b x-a y=a+b$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $a x+b y-(a-b)=0$ $b x-a y-(a+b)=0$ By cross multiplication method we get Therefore $x=1$ and $y=-1$ Hence we get the value of $x=1$ and $y=-1$...
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 : $\frac{x+y}{x y}=2, \frac{x-y}{x y}=6$ Solution: GIVEN: $\frac{x+y}{x y}=2$ $\frac{x-y}{x y}=6$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{x+y}{x y}=2$ $\frac{1}{x}+\frac{1}{y}=2$ $\frac{1}{x}+\frac{1}{3}-2=0$...(1) $\frac{x-y}{x y}=6$ $\frac{1}{y}-\frac{1}{x}=6$ $\frac{1}{y}-\frac{1}{x}-6=0$ ...(2) L...
Read More →If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5},
Question: If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A B) (B C) is(a) {(1, 2), (1, 5), (2, 5)} (b) [(1, 4)] (c) (1, 4) (d) none of these Solution: (b) {(1, 4)}A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5} (A B) = {1} (B C) = {4} So, (A B) (B C) = {(1,4)}...
Read More →Let R be a relation on N × N defined by
Question: Let R be a relation on N N defined by (a,b) R (c,d) ⇔a+d=b+cfor all (a,b), (c,d) N NShow that: (i) (a,b) R (a,b) for all (a,b) N N (ii) (a,b) R (c,d) ⇒ (c,d) R (a,b) for all (a,b), (c,d) N N (iii)(a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) for all (a,b), (c,d), (e,f) N N Solution: We are given , (a,b) R (c,d) ⇔a+d=b+cfor all (a,b), (c,d) N N (i) $(a, b) R(a, b)$ for all $(a, b) \in N \times N$ $\because a+b=b+a$ for all $a, b \in N$ $\therefore(a, b) R(a, b)$ for all $a, b \in N$ (...
Read More →Let R be a relation on N × N defined by
Question: Let R be a relation on N N defined by (a,b) R (c,d) ⇔a+d=b+cfor all (a,b), (c,d) N NShow that: (i) (a,b) R (a,b) for all (a,b) N N (ii) (a,b) R (c,d) ⇒ (c,d) R (a,b) for all (a,b), (c,d) N N (iii)(a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) for all (a,b), (c,d), (e,f) N N Solution: We are given , (a,b) R (c,d) ⇔a+d=b+cfor all (a,b), (c,d) N N (i) $(a, b) R(a, b)$ for all $(a, b) \in N \times N$ $\because a+b=b+a$ for all $a, b \in N$ $\therefore(a, b) R(a, b)$ for all $a, b \in N$ (...
Read More →Let R be a relation on N × N defined by
Question: Let R be a relation on N N defined by (a,b) R (c,d) ⇔a+d=b+cfor all (a,b), (c,d) N NShow that: (i) (a,b) R (a,b) for all (a,b) N N (ii) (a,b) R (c,d) ⇒ (c,d) R (a,b) for all (a,b), (c,d) N N (iii)(a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) for all (a,b), (c,d), (e,f) N N Solution: We are given , (a,b) R (c,d) ⇔a+d=b+cfor all (a,b), (c,d) N N (i) $(a, b) R(a, b)$ for all $(a, b) \in N \times N$ $\because a+b=b+a$ for all $a, b \in N$ $\therefore(a, b) R(a, b)$ for all $a, b \in N$ (...
Read More →In the below fig, if l ∥ m ∥ n and ∠1 = 60°. Find ∠2.
Question: In the below fig, if l ∥ m ∥ n and 1 = 60. Find 2. Solution: Since l parallel to m and p is the transversal Therefore, Given: l ∥ m ∥ n 1 = 60 To find2 1 = 3 = 60 [Corresponding angles] Now,3and4are linear pair of angles 3 + 4 = 180 60 +4 = 180 4 = 180 - 60 ⟹ 120 Also, m ∥ n and P is the transversal Therefore4 = 2 = 120 (Alternative interior angle] Hence 22 = 120...
Read More →In the below fig, ∠1 = 60° and ∠2 = (2/3)rd of a right angle. Prove that l ∥ m.
Question: In the below fig, 1 = 60 and 2 = (2/3)rd of a right angle. Prove that l ∥ m. Solution: Given: 1 = 60 and 2 = (2/3)rd of a right angle To Prove: Parallel Drawn to m Proof1 = 60 2 = (2/3) 90 = 60 Since1 = 1 = 60 Therefore, Parallel to m as pair of corresponding angles are equal....
Read More →For the relation
Question: For the relation R1defined on R by the rule (a,b) R1⇔ 1 +ab 0. Prove that: (a,b) R1and (b,c) R1⇒ (a,c) R1is not true for alla,b,c R. Solution: We have: (a,b) R1⇔ 1 +ab 0 Let: $a=1, b=-\frac{1}{2}$ and $c=-4$ Now, $\left(1,-\frac{1}{2}\right) \in R_{1}$ and $\left(-\frac{1}{2},-4\right) \in R_{1}$, as $1+\left(-\frac{1}{2}\right)0$ and $1+\left(-\frac{1}{2}\right)(-4)0$ But $1+1 \times(-4)0$ $\therefore(1,-4) \notin R_{1}$ And, (a,b) R1and (b,c) R1 Thus, (a,c) R1is not true for alla,b,c...
Read More →If each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Question: If each of the two lines is perpendicular to the same line, what kind of lines are they to each other?3 Solution: Let AB and CD be perpendicular to MN ABD = 90 [AB perpendicular to MN] ..... (i) CON = 90 [CO perpendicular to MN] .... (ii) Now, ABD = CDN = 90 (From (i) and (ii)) AB parallel to CD, Since corresponding angles are equal...
Read More →Let R be the relation on Z defined by
Question: Let R be the relation on Z defined by R = {(a,b) :a,b Z,abis an integer} Find the domain and range of R. Solution: R = {(a,b) :a,b Z,a bis an integer}We know: Difference of any two integers is always an integer. Thus, for alla,b Z, we getabas an integer. Domain (R) = Z And, Range (R) = Z...
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 : $2 x-y=6$ $x-y=2$ Solution: GIVEN: $2 x-y=6$ $x-y=2$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $2 x-y-6=0$ $x-y-2=0$ By cross multiplication method we get $y=\frac{-2}{-1}=2$ Hence we get the value of $x=4$ and $\mathrm{y}=2$...
Read More →Two unequal angles of a parallelogram are in the ratio 2: 3. Find all its angles in degrees.
Question: Two unequal angles of a parallelogram are in the ratio 2: 3. Find all its angles in degrees. Solution: Let A = 2x and B = 3x Now, A +B = 180 [Co-interior angles are supplementary] 2x + 3x - 180 [AD II BC and AB is the transversal) ⟹ 5x = 180 x = 180/5 x = 36 Therefore, A = 2 36 = 72 b = 3 36 = 108 Now, A = C = 72 [Opposite side angles of a parallelogram are equal) B = D = 108...
Read More →Prove
Question: $\frac{6 x+7}{\sqrt{(x-5)(x-4)}}$ Solution: $\frac{6 x+7}{\sqrt{(x-5)(x-4)}}=\frac{6 x+7}{\sqrt{x^{2}-9 x+20}}$ Let $6 x+7=A \frac{d}{d x}\left(x^{2}-9 x+20\right)+B$ $\Rightarrow 6 x+7=A(2 x-9)+B$ Equating the coefficients ofxand constant term, we obtain $2 A=6 \Rightarrow A=3$ $-9 A+B=7 \Rightarrow B=34$ $\therefore 6 x+7=3(2 x-9)+34$ $\int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}}=\int \frac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x$ $=3 \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \frac{...
Read More →The adjacent figure shows a relationship between the sets P and Q.
Question: The adjacent figure shows a relationship between the sets P and Q. Write this relation in (i) set builder form (ii) roster form. What is its domain and range? Solution: (i) We have: $5-2=3$ $6-2=4$ $7-2=5$ $\therefore \mathrm{R}=\{(x, y): y=x-2, x \in P, y \in Q\}$ (ii) R = {(5, 3), (6, 4), (7, 5)}(iii) Domain (R) = {5, 6, 7} Range (R) = {3, 4, 5}...
Read More →The adjacent figure shows a relationship between the sets P and Q.
Question: The adjacent figure shows a relationship between the sets P and Q. Write this relation in (i) set builder form (ii) roster form. What is its domain and range? Solution: (i) We have: $5-2=3$ $6-2=4$ $7-2=5$ $\therefore \mathrm{R}=\{(x, y): y=x-2, x \in P, y \in Q\}$ (ii) R = {(5, 3), (6, 4), (7, 5)}(iii) Domain (R) = {5, 6, 7} Range (R) = {3, 4, 5}...
Read More →In the below fig, AB ∥ CD and P is any point shown in the figure. Prove that:
Question: In the below fig, AB ∥ CD and P is any point shown in the figure. Prove that: ABP + BPD + CDP = 360 Solution: Through P, draw a line PM parallel to AB or CD. Now, A8 || PM ⟹ ABP + BPM = 180 And CD||PM = MPD + CDP = 180 Adding (i) and (ii), we get A8P + (BPM + MPD) CDP = 360 ⟹ ABP + BPD + CDP = 360...
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 : $2 x+y=35$ $3 x+4 y=65$ Solution: GIVEN: $2 x+y=35$ $3 x+4 y=65$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $2 x+y-35=0$ $3 x+4 y-65=0$ By cross multiplication method we get Also $\mathrm{y}=\frac{25}{5}$ Hence we get the value of $x=15$ and $y=5$...
Read More →Let A = [1, 2, 3, 4, 5, 6]. Let R be a relation on A defined by
Question: Let A = [1, 2, 3, 4, 5, 6]. Let R be a relation on A defined by {(a,b) :a,b A,bis exactly divisible bya}(i) Writer R in roster form (ii) Find the domain of R (ii) Find the range of R. Solution: A = [1, 2, 3, 4, 5, 6] R = {(a,b) :a,b A,bis exactly divisible bya}(i) Here, 2 is divisible by 1 and 2. 3 is divisible by 1 and 3. 4 is divisible by 1 and 4. 5 is divisible by 1 and 5. 6 is divisible by 1, 2, 3 and 6. R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (...
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