If A = {1, 2, 4},
Question: If A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, write (A C) (B C). Solution: A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5} Now, (A C) = {1, 4} (B C) = {4} Thus, we have: (A C) (B C) = {(1, 4), (4, 4)}...
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Question: $\frac{x^{3}+x+1}{x^{2}-1}$ Solution: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left(x^{3}+x+1\right)$ by $x^{2}-1$, we obtain $\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1}$ Let $\frac{2 x+1}{x^{2}-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$ $2 x+1=A(x-1)+B(x+1)$ ...(1) Substitutingx= 1 and 1 in equation (1), we obtain $A=\frac{1}{2}$ and $B=\frac{3}{2}$ $\therefore \frac{x^{3}+x+1}{x^{2}-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$ $\Rightarrow...
Read More →If A = {1, 2, 3, 4, 5, 6),
Question: IfA= {1, 2, 3, 4, 5, 6), then the number of sub sets ofAcontaining elements 2, 3 and 5 is _______ . Solution: IfA= {1, 2, 3, 4, 5, 6) i.e n(A) = 6 Then number of subsets ofAcontaining 2, 3, 5 i. e 3 elements is ${ }^{6} \mathrm{C}_{3}=\frac{6 !}{3 ! 3 !}$ $=\frac{6 \times 5 \times 4 \times 3 !}{3 ! 3 !}$ $=\frac{6 \times 5 \times 4}{3 \times 2}$ Then number of subsets ofA Containing 3 elements = 20...
Read More →In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively.
Question: In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN. Solution: Given that, In PQR, PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP respectively and given to prove that LN = MN Here we can observe that PQR is an isosceles triangle PQ = QR andQPR =QRP .... (i) And also, L and M are midpoints of PQ and QR respectively PL = LQ = QM = MR =PQ/2=QR/2 And also, PQ = QR Now, considerΔLPN andΔMRN, LP = MR [From - (2)] LP...
Read More →If n (A × B) = 200 and n(A) = 50,
Question: Ifn(AB) = 200 andn(A) = 50, then the number of elements inP(B) is _______ . Solution: Givenn(AB) = 200 $n(A)=50$ Sincen(AB) =n(A)n(B) ⇒ 200 =n(A)n(B) i.e $n(B)=\frac{200}{5}=4$ i.e number of elements ofBare 4 i.e number of elements inP(B) = 24...
Read More →In Fig. (10).22, the sides BA and CA have been produced such that: BA = AD and CA = AE. Prove that segment DE ∥ BC.
Question: In Fig. (10).22, the sides BA and CA have been produced such that: BA = AD and CA = AE. Prove that segment DE ∥ BC. Solution: Given that, the sides BA and CA have been produced such that BA = AD and CA = AE and given to prove DE ∥ BC Consider triangle BAC and DAE, We have BA = AD and CA= AE [given in the data] And alsoBAC=DAE [vertically opposite angles] So, by SAS congruence criterion, we have BAC ≃ DAE BC = DE andDEA = BCA, EDA = CBA [Corresponding parts of congruent triangles are eq...
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Question: $\frac{5 x}{(x+1)\left(x^{2}-4\right)}$ Solution: $\frac{5 x}{(x+1)\left(x^{2}-4\right)}=\frac{5 x}{(x+1)(x+2)(x-2)}$ Let $\frac{5 x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}$ $5 x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)$ ...(1) Substitutingx= 1, 2, and 2 respectively in equation (1), we obtain $A=\frac{5}{3}, B=-\frac{5}{2}$, and $C=\frac{5}{6}$ $\therefore \frac{5 x}{(x+1)(x+2)(x-2)}=\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}$ $\Rightarrow \int \frac{5 x}{...
Read More →If A = {3, 5, 6, 9) and R is a relation in A defined as R
Question: IfA= {3, 5, 6, 9) andRis a relation inAdefined asR= {(x,y) :x+y 18), thenRin roster form is ______ . Solution: A= {3, 5, 6, 9) R= {(x,y) :x+y 18 } Then roaster form forRis collection of elements of type {(x,y)} satisfyingR. forx= 3, possible values ofyare 3, 5, 6, 9 forx= 5, possible values ofyare 3, 5, 6, 9 forx= 6, possible values ofyare 3, 5, 6, 9 forx= 9, possible values ofy are 3, 5, 6. $\therefore R=\{(3,3)(3,5)(3,6)(3,9)(5,3)(5,5)(5,6)(5,9)(6,3)(6,5)(6,6)(6,9)(9,3)(9,5)(9,6)\}$ ...
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 : $(a-b) x+(a+b) y=2 a^{2}-2 b^{2}$ $(a+b)(x+y)=4 a b$ Solution: GIVEN: $(a-b) x+(a+b) y=2 a^{2}-2 b^{2}$ $(a+b)(x+y)=4 a b$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $(a-b) x+(a+b) y-2 a^{2}+2 b^{2}=0$ $(a+b) x+(a+b) y-4 a b=0$ By cross multiplication method we get $\frac{x}{(-4 a b)(a+b)-(a+b)\left(-2 a^{2}...
Read More →If n(A ∩ B′) = 9,
Question: Ifn(AB) = 9,n(A'B) = 10 andn(AB) = 24, thenn(AB) = ___________ . Solution: $\left.\begin{array}{rl}n\left(A \cap B^{\prime}\right) =9 \\ \text { If } n\left(A^{\prime} \cap B\right) =10 \\ n(A \cup B) =24\end{array}\right\}$ given n(A) =n(A⋃B) n(A' B) n(B) =n(A⋃B)n(AB') i.en(A) = 24 10 =14 $n(B)=24-9=15$ $\therefore n(A \times B)=n(A) \cdot n(B)$ n(AB) = 14 15 = 210...
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Question: $\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}$ Solution: $\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}=\frac{2 x-3}{(x+1)(x-1)(2 x+3)}$ Let $\frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2 x+3)}$ $\Rightarrow(2 x-3)=A(x-1)(2 x+3)+B(x+1)(2 x+3)+C(x+1)(x-1)$ $\Rightarrow(2 x-3)=A\left(2 x^{2}+x-3\right)+B\left(2 x^{2}+5 x+3\right)+C\left(x^{2}-1\right)$ $\Rightarrow(2 x-3)=(2 A+2 B+C) x^{2}+(A+5 B) x+(-3 A+3 B-C)$ Equating the coefficients of $x^{2}$ and $x$. we obt...
Read More →If R = {(x, y) :
Question: IfR= {(x,y) :wherex R and 5x 5} is a relation, then range (R) = _______ . Solution: R= {(x,y) : wherexRand 5x 5} Range R = ?? (Correction; NO information as y is given )...
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Question: $\frac{3 x+5}{x^{3}-x^{2}-x+1}$ Solution: $\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$ Let $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)}$ $3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}$ $3 x+5=A\left(x^{2}-1\right)+B(x+1)+C\left(x^{2}+1-2 x\right)$ ...(1) Substitutingx= 1 in equation (1), we obtain B= 4 Equating the coefficients of $x^{2}$ and $x$, we obtain A+C= 0 B 2C= 3 On solving, we obtain $A=-\frac{1}{2}$ and $C=\frac{1}{2}$ $\therefore \...
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 : $b x+c y=a+b$ $a x\left(\frac{1}{a-b}-\frac{1}{a+b}\right)+c y\left(\frac{1}{b-a}-\frac{1}{b+a}\right)=\frac{2 a}{a+b}$ Solution: GIVEN: To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation By cross multiplication method we get $\frac{x}{\left(-\frac{2 a c}{(a+b)}\right)-\left(-(a+b) \times\left(\frac{c}{(b-a)}-\frac{...
Read More →Fill in the blanks to make the following statements true:
Question: Fill in the blanks to make the following statements true: (i) Sum of the angles of a triangle is _______ . (ii) An exterior angle of a triangle is equal to the two ________ opposite angles. (iii) An exterior angle of a triangle is always ________ than either of the interior opposite angles. (iv) A triangle cannot have more than _______ right angles. (v) A triangles cannot have more than _______ obtuse angles. Solution: (i) 180 (ii) Interior (iii) Greater (iv) One (v) One...
Read More →If A ={(x : x ∈ W, x < 2},
Question: IfA={(x:xW,x 2},B= {x:xN, 1 x 5} andC= 3, 5, thenA (B C) = _________. Solution: A= {x:xW;x 2y= {0, 1} B= {x:xN ; 1 x 5}= {2, 3, 4} C= {3, 5}ThenBC= {3} $\therefore A \times(B \cap C)=\{(0,3),(1,3)\}$...
Read More →Which of the following statements are true (T) and which are false (F):
Question: Which of the following statements are true (T) and which are false (F): (i) Sum of the three angles of a triangle is 180. (ii) A triangle can have two right angles. (iii) All the angles of a triangle can be less than 60. (iv) All the angles of a triangle can be greater than 60. (v) All the angles of a triangle can be equal to 60. (vi) A triangle can have two obtuse angles. (vii) A triangle can have at most one obtuse angles. (viii) If one angle of a triangle is obtuse, then it cannot b...
Read More →In figure 9.39, AB ∥ DE. Find ∠ACD.
Question: In figure 9.39, AB ∥ DE. Find ACD. Solution: SinceAB ∥ DE ABC = CDE = 40 [Alternate angles] ACB = 180 ABC BAC =180 40 30 = 110 ACD = 180 110 [Linear pair] = 70...
Read More →The relation R
Question: The relationR= {(x,y) :x,yZ,x2+y2= 64} is __________. Solution: R= {(x,y) :x,yZ:x2+y2= 64}Since x2+y2= 64 y2= 64x2 The only possibilities forxare 0, 8 and 8, and foryare 0, 8 and 8 . i.eR= {(0, 8), (8, 0), (8, 0) (0, 8)} HenceR= { (0, 8) (8, 0) (8, 0) (0, 8)}...
Read More →In figure 9.38, AE bisects ∠CAD and ∠B = ∠C. Prove that AE ∥ BC.
Question: In figure 9.38, AE bisects CAD and B = C. Prove that AE ∥ BC. Solution: Let B = C = x Then, CAD = B + C = 2x (exterior angle) ⇒1/2CAD = x ⇒ EAC = x ⇒ EAC = C These are alternate interior angles for the lines AE and BC AE ∥ BC...
Read More →If R =
Question: IfR= {(x,y) :x,y Z,x2+y2= 25}, then Domain (R) = .................... and Range (R) = __________. Solution: IfR= { (x,y) :x,y Z ,x2+y2= 25} Sincex2+y2=25 $\therefore$ only possibilities for $x$ are $0,3,4$ and 5 and similarly for $y$ are $0,3,4$ and 5 i.eR= {(0, 5) , (3, 4) , (4, 3), (5, 0)} $\left\{\because\right.$ all above satisfy $\left.x^{2}+y^{2}=25\right\}$ $\therefore$ Domain $(R)=\{0,3,4,5\}$ and Range $(R)=\{0,3,4,5\}$...
Read More →In triangle ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove
Question: In triangle ABC, BD AC and CE AB. If BD and CE intersect at O, prove that BOC = 180 - A. Solution: In quadrilateral AEOD A + AEO + EOD + ADO = 360 ⇒ A + 90 + 90 + EOD = 360 ⇒ A + BOC = 180 [∵ EOD = BOC vertically opposite angles] ⇒ BOC = 180 A...
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Question: $\frac{x}{(x-1)^{2}(x+2)}$ Solution: Let $\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+2)}$ $x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2}$ Substitutingx= 1, we obtain $B=\frac{1}{3}$ Equating the coefficients of $x^{2}$ and constant term, we obtain $A+C=0$ $-2 A+2 B+C=0$ On solving, we obtain $A=\frac{2}{9}$ and $C=\frac{-2}{9}$ $\therefore \frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)}$ $\Rightarrow \int \frac{x}{(x-1)^{2}(x+2)} d x=\fra...
Read More →In a triangle ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
Question: In a triangle ABC, AD bisects A and C B. Prove that ADB ADC. Solution: ∵ C B [Given] ⇒ C + x B + x [Adding x on both sides] ⇒ 180 - ADC 180 - ADB ⇒ - ADC - ADB ⇒ADB ADC Hence proved....
Read More →In figure 9.37, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.
Question: In figure 9.37,AM BC and AN is the bisector of A. If B = 65 and C = 33, find MAN. Solution: Let BAN = NAC = x [∵ AN bisects A] ANM = x + 33 [Exterior angle property] In ΔAMB BAM = 90 65 = 25 [Exterior angle property] MAN = BAN BAM = (x 25) Now in ΔMAN, (x - 25) + (x + 33) + 90 = 180 [Angle sum property] ⇒ 2x + 8 = 90 ⇒ 2x = 82 ⇒ x = 41 MAN = x 25 = 41 25 = 16...
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