If a relation R on the set {1, 2, 3}
Question: If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is (A) reflexive (B) transitive (C) symmetric (D) none of these Solution: (D) none of these R on the set {1, 2, 3} be defined by R = {(1, 2)} Hence, its clear that R is not reflexive, transitive and symmetric....
Read More →The maximum number of equivalence
Question: The maximum number of equivalence relations on the set A = {1, 2, 3} are (A) 1 (B) 2 (C) 3 (D) 5 Solution: (D) 5 Given, set A = {1, 2, 3} Now, the number of equivalence relations as follows R1= {(1, 1), (2, 2), (3, 3)} R2= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} R3= {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)} R4= {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)} R5= {(1, 2, 3) ⇔ A x A = A2} Thus, maximum number of equivalence relation is 5....
Read More →Consider the non-empty set consisting of children
Question: Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is (A) symmetric but not transitive (B) transitive but not symmetric (C) neither symmetric nor transitive (D) both symmetric and transitive Solution: (B) transitive but not symmetric aRb ⇒ a is brother ofb. This does not meanbis also a brother of a asbcan be a sister ofa. Thus, R is not symmetric. aRb ⇒ a is brother ofb. and bRc ⇒bis brother ofc. So, a is brother o...
Read More →Let T be the set of all triangles in the Euclidean plane,
Question: Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b a, b T. Then R is (A) reflexive but not transitive (B) transitive but not symmetric (C) equivalence (D) none of these Solution: (C) equivalence Given aRb, if a is congruent to b, a, b T. Then, we have aRa ⇒ a is congruent to a; which is always true. So, R is reflexive. Let aRb ⇒ a ~ b b ~ a bRa So, R is symmetric. Let aRb and bRc a ~ b and b ~ c a ~ c aRc So, R i...
Read More →Let * be binary operation defined
Question: Let * be binary operation defined on R by a * b = 1 + ab, a, b R. Then the operation * is (i) commutative but not associative (ii) associative but not commutative (iii) neither commutative nor associative (iv) both commutative and associative Solution: (i) Given that * is a binary operation defined on R by a * b = 1 + ab, a, b R So, we have a * b = ab + 1 = b * a So, * is a commutative binary operation. Now, a * (b * c) = a * (1 + bc) = 1 + a (1 + bc) = 1 + a + abc Also, (a * b) * c = ...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} \frac{1}{3}$ Solution: To Prove: $\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} \frac{1}{3}$ Formula Used: $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ where $x y-1$ Proof: $\mathrm{LHS}=\tan ^{-1} 2-\tan ^{-1} 1$ $=\tan ^{-1}\left(\frac{2-1}{1+2}\right)$ $=\tan ^{-1}\left(\frac{1}{3}\right)$ $=\mathrm{RHS}$ Therefore LHS = RHS Hence proved....
Read More →Let * be the binary operation defined on Q.
Question: Let * be the binary operation defined on Q. Find which of the following binary operations are commutative (i) a * b = a b a, b Q (ii) a * b = a2+ b2 a, b Q (iii) a * b = a + ab a, b Q (iv) a * b = (a b)2 a, b Q Solution: Given that * is a binary operation defined on Q. (i) a * b = a b, a, b Q and b * a = b a So, a * b b * a Thus, * is not commutative. (ii) a * b = a2+ b2 b * a = b2+ a2 Thus, * is commutative. (iii) a * b = a + ab b * a = b + ab So clearly, a + ab b + ab Thus, * is not ...
Read More →Functions f , g : R → R are defined,
Question: Functions f , g : R R are defined, respectively, byf(x) = x2+ 3x + 1,g(x) = 2x 3, find (i)f o g(ii)g o f(iii)f o f(iv)g o g Solution: Given,f(x) = x2+ 3x + 1,g(x) = 2x 3 (i) fog = f(g(x)) = f(2x 3) = (2x 3)2+ 3(2x 3) + 1 = 4x2+ 9 12x + 6x 9 + 1 = 4x2 6x + 1 (ii) gof = g(f(x)) = g(x2+ 3x + 1) = 2(x2+ 3x + 1) 3 = 2x2+ 6x 1 (iii) fof = f(f(x)) = f(x2+ 3x + 1) = (x2+ 3x + 1)2+ 3(x2+ 3x + 1) + 1 = x4+ 9x2+ 1 + 6x3+ 6x + 2x2+ 3x2+ 9x + 3 + 1 = x4+ 6x3+ 14x2+ 15x + 5 (iv) gog = g(g(x)) = g(2x...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x}{(x+1)\left(x^{2}+1\right)} d x$ Solution: $I=\int \frac{x}{(x+1)\left(x^{2}+1\right)}$ $\frac{\mathrm{x}}{(\mathrm{x}+1)\left(\mathrm{x}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+1}$ $x=A\left(x^{2}+1\right)+(B x+C)(x+1)$ Equating constants $0=A+C$ Equating coefficients of $x$ $1=B+C$ Equating coefficients of $x^{2}$ $0=A+B$ Solving, we get $\mathrm{A}=-\frac{1}{2} \mathrm{~B}=\frac{1}{2} \...
Read More →Using the definition,
Question: Using the definition, prove that the function f : A B is invertible if and only if f is both one-one and onto. Solution: Let f: A B be many-one function. Let f(a) = p and f(b) = p So, for inverse function we will have f-1(p) = a and f-1(p) = b Thus, in this case inverse function is not defined as we have two images a and b for one pre-image p. But for f to be invertible it must be one-one. Now, let f: A B is not onto function. Let B = {p, q, r} and range of f be {p, q}. Here image r ha...
Read More →Prove that:
Question: Prove that: $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$ Solution: To Prove: $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof: $\mathrm{LHS}=2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$ $=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$ $=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\left(\frac{1}{3} \times \frac{1}{3}\right)}\right)+\t...
Read More →Let A = {1, 2, 3, … 9} and R be the relation
Question: Let A = {1, 2, 3, 9} and R be the relation in A A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)]. Solution: Given, A = {1, 2, 3, 9} and (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) A A. Let (a, b) R(a, b) So, a + b = b + a, a, b A which is true for any a, b A. Thus, R is reflexive. Let (a, b) R(c, d) Then, a + d = b + c c + b = d + a (c, d) R(a, b) Thus, R is symmetric. ...
Read More →Each of the following defines a relation on N:
Question: Each of the following defines a relation on N: (i) x is greater than y, x, y N (ii) x + y = 10, x, y N (iii) x y is square of an integer x, y N (iv) x + 4y = 10 x, y N. Determine which of the above relations are reflexive, symmetric and transitive. Solution: (i) Given, x is greater than y; x, y N If (x, x) R, then x x, which is not true for any x N. Thus, R is not reflexive. Let (x, y) R ⇒ xRy ⇒ x y So, y x is not true for any x, y N Hence, R is not symmetric. Let xRy and yRz ⇒ x y and...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{5}{\left(x^{2}+1\right)(x+2)} d x$ Solution: $I=\int \frac{5}{\left(x^{2}+1\right)(x+2)}$ $\frac{5}{\left(x^{2}+1\right)(x+2)}=\frac{A x+B}{x^{2}+1}+\frac{C}{x+2}$ $5=(A x+B)(x+2)+C\left(x^{2}+1\right)$ Equating constants $5=2 B+C$ Equating coefficients of $x$ $0=2 A+B$ Equating coefficients of $x^{2}$ $0=A+C$ Solving, we get $A=-1, B=2, C=1$ Thus $I=\int \frac{-x+2}{x^{2}+1} d x+\int \frac{d x}{x+2}$ $=\int \frac{-\mathrm{xdx}}{\mathrm{x}^{...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$ Solution: To Prove: $\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof: $\mathrm{LHS}=\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$ $=\tan ^{-1} 1+\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\left(\frac{1}{2} \times \frac{1}{3}\right)}\right)$ $=\tan ^{-1} 1+\tan ^{-1}\l...
Read More →Let A = [–1, 1]. Then, discuss whether the following
Question: Let A = [1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective: (i) f(x) = x/2 (ii) g(x) = |x| (iii) h(x) = x|x| (iv) k(x) = x2 Solution: Given, A = [1, 1] (i) f: [-1, 1] [-1, 1], f (x) = x/2 Let f (x1) = f(x2) x1/ 2 = x2 So, f (x) is one-one. Also x [-1, 1] x/2 = f (x) = [-1/2, 1/2] Hence, the range is a subset of co-domain A So, f (x) is not onto. Therefore, f (x) is not bijective. (ii) g (x) = |x| Let g (x1) = g (x2) |x1| = |x2| x1= x2 So, ...
Read More →Let A = R – {3}, B = R – {1}. Let f : A → B be defined
Question: Let A = R {3}, B = R {1}. Let f : A B be defined by f (x) = x 2/ x 3 x A . Then show that f is bijective. Solution: Given, A = R {3}, B = R {1} And, f : A B be defined by f (x) = x 2/ x 3 x A Hence, f (x) = (x 3 + 1)/ (x 3) = 1 + 1/ (x 3) Let f(x1) = f (x2) $1+\frac{1}{x_{1}-3}=1+\frac{1}{x_{2}-3}$ $\frac{1}{x_{1}-3}=\frac{1}{x_{2}-3}$ $x_{1}=x_{2}$ So, f (x) is an injective function. Now let y = (x 2)/ (x -3) x 2 = xy 3y x(1 y) = 2 3y x = (3y 2)/ (y 1) y R {1} = B Thus, f (x) is onto ...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$ Solution: To Prove: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof: $\mathrm{LHS}=\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}$ $=\tan ^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\left(\frac{2}{11} \times \frac{7}{24}\right)}\right)$ $=\tan ^{-1}\left(\frac{48+77}{264-14}\right)$ $=\t...
Read More →Give an example of a map
Question: Give an example of a map (i) which is one-one but not onto (ii) which is not one-one but onto (iii) which is neither one-one nor onto. Solution: (i) Let f: N N, be a mapping defined by f (x) = x2 For f (x1) = f (x2) Then, x12= x22 x1= x2(Since x1+ x2= 0 is not possible) Further f is not onto, as for 1 N, there does not exist any x in N such that f (x) = 2x + 1. (ii) Let f: R [0, ), be a mapping defined by f(x) = |x| Then, its clearly seen that f (x) is not one-one as f (2) = f (-2). Bu...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{18}{(x+2)\left(x^{2}+4\right)} d x$ Solution: $I=\int \frac{18}{(x+2)\left(x^{2}+4\right)}$ $\frac{18}{(x+2)\left(x^{2}+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+4}$ $18=A\left(x^{2}+4\right)+(B x+C)(x+2)$ Equating constants $18=4 \mathrm{~A}+2 \mathrm{C}$ Equating coefficients of $x$ $0=2 B+C$ Equating coefficients of $x^{2}$ $0=A+B$ Solving, we get $A=\frac{9}{4}, \quad B=-\frac{9}{4}, \quad C=\frac{9}{2}$ Thus, $I=\frac{9}{4} \int \frac{...
Read More →Given A = {2, 3, 4}, B = {2, 5, 6, 7}.
Question: Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following: (a) an injective mapping from A to B (b) a mapping from A to B which is not injective (c) a mapping from B to A. Solution: Given, A = {2, 3, 4}, B = {2, 5, 6, 7} (i) Let f: A B denote a mapping f = {(x, y): y = x + 3} or f = {(2, 5), (3, 6), (4, 7)}, which is an injective mapping. (ii) Let g: A B denote a mapping such that g = {(2, 2), (3, 2), (4, 5)}, which is not an injective mapping. (iii) Let h: B...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$ Solution: To Prove: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof: $\mathrm{LHS}=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}$ $=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{2}{11}}{1-\left(\frac{1}{2} \times \frac{2}{11}\right)}\right)$ $=\tan ^{-1}\left(\frac{11+4}{22-2}\right)$ $=\tan ^{-1}...
Read More →Let R be relation defined on the set of natural number N as follows:
Question: Let R be relation defined on the set of natural number N as follows: R = {(x, y): x N, y N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive. Solution: Given function: R = {(x, y): x N, y N, 2x + y = 41}. So, the domain = {1, 2, 3, .., 20} [Since, y N ] Finding the range, we have R = {(1, 39), (2, 37), (3, 35), ., (19, 3), (20, 1)} Thus, Range of the function = {1, 3, 5, .., 39} R is not reflexive as (2, 2) R as 2 x...
Read More →If A = {1, 2, 3, 4 }, define relations on
Question: If A = {1, 2, 3, 4 }, define relations on A which have properties of being: (a) reflexive, transitive but not symmetric (b) symmetric but neither reflexive nor transitive (c) reflexive, symmetric and transitive. Solution: Given that, A = {1, 2, 3}. (i) Let R1= {(1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3)} R1is reflexive as (1, 1), (2, 2) and (3, 3) lie is R1. R1is transitive as (1, 2) R1, (2, 3) R1⇒ (1, 3) R1 Now, (1, 2) R1⇒ (2, 1) R1. (ii) Let R2= {(1, 2), (2, 1)} Now, (1, ...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x$ Solution: $I=\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x}=\int \frac{5 x^{2}+20 x+6}{x(x+1)^{2}}$ $\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$ $5 x^{2}+20 x+6=A(x+1)^{2}+B x(x+1)+C x$ Equating constants $6=A$ Equating coefficients of $x^{2}$ $5=A+B$ $B=-1$ Equating coefficients of $\mathrm{x}$ $20=2 A+B+C$ $20=12-1+C$ $C=9$ $I=\int \frac{6 d x}{x}-\int \frac{d x}{x+1}+...
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