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Question: $\frac{1}{\left(e^{x}-1\right)}$ [Hint: Put $\left.e^{x}=t\right]$ Solution: $\frac{1}{\left(e^{x}-1\right)}$ Let $e^{x}=t \Rightarrow e^{x} d x=d t$ $\Rightarrow \int \frac{1}{e^{x}-1} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t$ Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$ $1=A(t-1)+B t$ ...(1) Substitutingt= 1 andt= 0 in equation (1), we obtain A= 1 andB= 1 $\therefore \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$ $\begin{aligned} \Rightarrow \int \frac{...
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 : $\frac{b}{a} x+\frac{a}{b} y=a^{2}+b^{2}$ $x+y=2 a b$ Solution: GIVEN: $\frac{b}{a} x+\frac{a}{b} y=a^{2}+b^{2}$ $x+y=2 a b$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{b}{a} x+\frac{a}{b} y-\left(a^{2}+b^{2}\right)=0$ $x+y-2 a b=0$ By cross multiplication method we get $\frac{x}{(-2 a b)\left(\frac{a}...
Read More →In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Question: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle. Solution: LetΔABC be isosceles such that AB = AC. B =C Given that vertex angle A is twice the sum of the base angles B and C. i.e.,A= 2(B +C) A = 2(B +B) A = 2(2B) A = 4(B) Now, We know that sum of angles in a triangle = 180 A +B +C =180 4B +B +B = 180 6B =180 B = 30 Since,B =C B =C = 30 AndA = 4B A = 4 30 = 120 Therefore, angles of the given triangle are 120, 30 and...
Read More →If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Question: If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles. Solution: Given that the bisector of the exterior vertical angle of a triangle is parallel to the base and we have to prove that the triangle is isosceles. Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle EAC and AD∥BC Let EAD = (i), DAC = (ii), ABC = (iii) and ACB = (iv) We have, (i) = (ii) [AD is a bisector ofEAC] (i) = (iii) ...
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 : $\frac{a x}{b}-\frac{b y}{a}=a+b$ $a x-b y=2 a b$ Solution: GIVEN: $\frac{a x}{b}-\frac{b y}{a}=a+b$ $a x-b y=2 a b$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{a x}{b}-\frac{b y}{a}-(a+b)=0$ $a x-b y-2 a b=0$ By cross multiplication method we get $\frac{x}{(-2 a b)\left(-\frac{b}{a}\right)-(-b)(-(a+b)...
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Question: $\frac{1}{x\left(x^{4}-1\right)}$ Solution: $\frac{1}{x\left(x^{4}-1\right)}$ Multiplying numerator and denominator by $x^{3}$, we obtain $\frac{1}{x\left(x^{4}-1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}-1\right)}$ $\therefore \int \frac{1}{x\left(x^{4}-1\right)} d x=\int \frac{x^{3}}{x^{4}\left(x^{4}-1\right)} d x$ Let $x^{4}=t \Rightarrow 4 x^{3} d x=d t$ Iet $x^{4}=t \Rightarrow \Delta x^{3} d x=d t$ $\therefore \int \frac{1}{x\left(x^{4}-1\right)} d x=\frac{1}{4} \int \frac{d t}{t(t-1...
Read More →Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(x, y) :
Question: Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(x,y) :xyis odd}. Write R in roster form. Solution: Given: A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x,y) :xyis odd} Since $1-4=-3$ is odd, we have: $1-6=-5$ is odd $2-9=-7$ is odd $3-4=-1$ is odd $3-6=-3$ is odd $5-4=1$ is odd $5-6=-1$ is odd R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}...
Read More →Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(x, y) :
Question: Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(x,y) :xyis odd}. Write R in roster form. Solution: Given: A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x,y) :xyis odd} Since $1-4=-3$ is odd, we have: $1-6=-5$ is odd $2-9=-7$ is odd $3-4=-1$ is odd $3-6=-3$ is odd $5-4=1$ is odd $5-6=-1$ is odd R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}...
Read More →In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Question: In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent. Solution: Given that, in two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other. We have to prove that the triangles are congruent. Let us consider two right triangles such that B =C = 90.... (i) AB = DE.... (ii) C =F From(iii) Now observe the two triangles ABC and DEF C =F [Fro...
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 : $m x-n y=m^{2}+n^{2}$ $x+y=2 m$ Solution: GIVEN: $m x-n y=m^{2}+n^{2}$ $x+y=2 m$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $m x-n y-\left(m^{2}+n^{2}\right)=0$ $x+y-2 m=0$ By cross multiplication method we get $\frac{x}{(-2 m)(-n)-\left(-\left(m^{2}+n^{2}\right)\right)}=\frac{-y}{(-2 m)(m)-\left(-\left(m^{2...
Read More →Let A and B be two sets such that n(A) = 3 and n(B) = 2.
Question: Let A and B be two sets such thatn(A) = 3 andn(B) = 2. If (x, 1), (y, 2), (z, 1) are in A B, write A and B. Solution: Given: (x, 1), (y, 2), (z, 1) are in A B n(A) = 3 andn(B) = 2 $(x, 1) \in A \times B \Rightarrow x \in A, 1 \in B$ Similarly, $y \in A, 2 \in B$ and $z \in A, 1 \in B$ So, A = {x,y,z} and B = {1,2}...
Read More →If R = [(x, y) : x, y ∈ W, 2x + y = 8],
Question: If R = [(x,y) :x,y W, 2x+y= 8], then write the domain and range of R. Solution: R = {(x,y) :x,y W, 2x+y= 8} For $\mathrm{x}=0, \mathrm{y}=8$ For $\mathrm{x}=1, \mathrm{y}=6$ For $\mathrm{x}=2, \mathrm{y}=4$ For $\mathrm{x}=3, \mathrm{y}=2$ For $x=4, y=0$ For $\mathrm{x}=5, \mathrm{y}0$ So, $\mathrm{y}0$ for all $\mathrm{x}5$ Domain (R) = {0,1,2,3,4} and Range (R) = {0,2,4,6,8}...
Read More →If R = [(x, y) : x, y ∈ W, 2x + y = 8],
Question: If R = [(x,y) :x,y W, 2x+y= 8], then write the domain and range of R. Solution: R = {(x,y) :x,y W, 2x+y= 8} For $\mathrm{x}=0, \mathrm{y}=8$ For $\mathrm{x}=1, \mathrm{y}=6$ For $\mathrm{x}=2, \mathrm{y}=4$ For $\mathrm{x}=3, \mathrm{y}=2$ For $x=4, y=0$ For $\mathrm{x}=5, \mathrm{y}0$ So, $\mathrm{y}0$ for all $\mathrm{x}5$ Domain (R) = {0,1,2,3,4} and Range (R) = {0,2,4,6,8}...
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Question: $\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}$ Solution: $\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}$ Let $x^{2}=t \Rightarrow 2 x d x=d t$ $\therefore \int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x=\int \frac{d t}{(t+1)(t+3)}$ ...(1) Let $\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}$ $1=A(t+3)+B(t+1)$ ...(1) Substitutingt= 3 andt= 1 in equation (1), we obtain $\Rightarrow \int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x=\int\left\{...
Read More →BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC
Question: BD and CE are bisectors of B and C of an isosceles ΔABC with AB = AC. Prove that BD = CE Solution: Given thatΔABC is isosceles with AB = AC and BD and CE are bisectors ofB andC We have to prove BD = CE Since AB = AC ⟹ΔABC =ΔACB ... (i) [Angles opposite to equal sides are equal] Since BD and CE are bisectors ofB andC ABD =DBC =BCE = ECA =B/2=C/2 Now, Consider ΔEBC = ΔDCB EBC =DCB [B =C] [From (i)] BC = BC [Common side] BCE =CBD [From (ii)] So, by ASA congruence criterion, we haveΔEBCΔDC...
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 : $\frac{a^{2}}{x}-\frac{b^{2}}{y}=0$ $\frac{a^{2} b}{x}+\frac{b^{2} a}{y}=a+b, x, y \neq 0$ Solution: GIVEN: $\frac{a^{2}}{x}-\frac{b^{2}}{y}=0$ $\frac{a^{2} b}{x}+\frac{a b^{2}}{y}=a+b$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{a^{2}}{x}-\frac{b^{2}}{y}=0$ $\frac{a^{2} b}{x}+\frac{a b^{2}}{y}-(a+b)=0...
Read More →Two lines AB and CD intersect at O such that BC is equal and parallel to AD.
Question: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O. Solution: Given that lines AB and CD Intersect at O. Such that BC∥AD and BC = AD... (i) We have to prove that AB and CD bisect at O. To prove this first we have to prove that ΔAOD ΔBOC...
Read More →If A = [1, 3, 5] and B = [2, 4], list of elements of R, if
Question: If A = [1, 3, 5] and B = [2, 4], list of elements of R, if R = {(x,y) :x,y A B andxy} Solution: Given: A = {1, 3, 5} and B = {2, 4} R = {(x,y) :x,y A B andxy} A B = {(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)} As 3 2, 5 2 and 5 4, we have R = {(3,2),(5,2),(5,4)}...
Read More →Given that RT = TS, ∠1 = 2, ∠2 and 4 = 2 ∠(3) Prove that ΔRBT ≅ ΔSAT.
Question: Given that RT $=T S, \angle 1=2, \angle 2$ and $4=2 \angle(3)$ Prove that $\triangle R B T \cong \triangle S A T$. Solution: In the figure, given that RT = TS.... (i) 1 = 22.... (ii) And4 = 23.... (iii) To prove thatΔRBT ΔSAT. Let the point of intersection RB and SA be denoted by O Since RB and SA intersect at O AOR =BOS [Vertically opposite angles] 1 =4 22 = 23 [From (ii) and (iii)] 2 =3 .(iv) Now we have RT = TS inΔTRS ΔTRS is an isosceles triangle TRS =TSR .... (v) But we have TRS =...
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Question: $\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$ Solution: $\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$ Let $\frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+3\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ $4 x^{2}+10=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+3\right)$ $4 x^{2}+10=A x^...
Read More →If R = {(2, 1), (4, 7), (1, −2), ...},
Question: If R = {(2, 1), (4, 7), (1, 2), ...}, then write the linear relation between the components of the ordered pairs of the relation R. Solution: Given: R = {(2, 1), (4, 7), (1, 2), ...} We can observe that $1=3 \times 2-5$ $7=3 \times 4-5$ $-2=3 \times 1-5$ Thus, the linear relation between the components of the ordered pairs of the relation R isy= 3x-5....
Read More →Let R = [(x, y) :
Question: Let R = [(x,y) :x,y Z,y= 2x 4]. If (a, -2) and (4,b2) R, then write the values ofaandb. Solution: R = [(x,y) :x,y Z,y= 2x 4] $(a,-2)$ and $\left(4, b^{2}\right) \in \mathrm{R}$ So, $-2=2($ a) $-4$ $\Rightarrow 2=2 a$ $\Rightarrow a=1$ Also, $b^{2}=2(4)-4$ $\Rightarrow b^{2}=4$ $\Rightarrow b=\pm 2$...
Read More →Let R = [(x, y) :
Question: Let R = [(x,y) :x,y Z,y= 2x 4]. If (a, -2) and (4,b2) R, then write the values ofaandb. Solution: R = [(x,y) :x,y Z,y= 2x 4] $(a,-2)$ and $\left(4, b^{2}\right) \in \mathrm{R}$ So, $-2=2($ a) $-4$ $\Rightarrow 2=2 a$ $\Rightarrow a=1$ Also, $b^{2}=2(4)-4$ $\Rightarrow b^{2}=4$ $\Rightarrow b=\pm 2$...
Read More →AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from
Question: AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. (10).26). Show that the line PQ is perpendicular bisector of AB. Solution: Consider the figure. We have AB is a line segment and P, Q are points on opposite sides of AB such that AP = BP ... (i) AQ = BQ ... (ii) We have to prove that PQ is perpendicular bisector of AB. Now considerΔPAQ andΔPBQ, We have AP = BP [From (i)] AQ = BQ [From (ii)] And PQ - P...
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 : $6(a x+b y)=3 a+2 b$ $6(b x-a y)=3 b-2 a$ Solution: GIVEN: $6(a x+b y)=3 a+2 b$ $6(b x-a y)=3 b-2 a$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation, after rewriting equations $6 a x+6 b y-(3 a+2 b)=0$ $6 b x-6 a y-(3 b-2 a)=0$ By cross multiplication method we get $\frac{x}{-(3 b-2 a)(6 b)-((-6 a) \times-(3 a+2 ...
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