Three relations R1, R2 and R3 are defined on a set A = {a, b, c} as follows:
Question: Three relationsR1,R2andR3are defined on a setA= {a, b, c} as follows:R1= {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}R2= {(a, a)}R3= {(b, c)}R4= {(a, b), (b, c), (c, a)}. Find whether or not each of the relationsR1,R2,R3,R4onAis (i) reflexive (ii) symmetric and (iii) transitive. Symmetric: Clearly $(a, a) \in R \Rightarrow(a, a) \in R .$ So, $\mathrm{R}_{2}$ is symmetric. Transitive: $R_{2}$ is clearly a transitive relation, since there is only one element in it. So...
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Question: $(125)^{-1 / 3}=?$ (a) 5 (b) $-5$ (C) $\frac{1}{5}$ (d) $-\frac{1}{5}$ Solution: $(125)^{-\frac{1}{3}}$ $=\left(5^{3}\right)^{-\frac{1}{3}}$ $=5^{3 \times\left(-\frac{1}{3}\right)} \quad\left[\left(x^{a}\right)^{b}=x^{a b}\right]$ $=5^{-1}$ $=\frac{1}{5} \quad\left(x^{-a}=\frac{1}{x^{a}}\right)$ Hence, the correct answer is option (c)....
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Question: 72n+ 23n3. 3n1is divisible by 25 for allnN. Solution: Let P(n) be the given statement. Now, $P(n): 7^{2 n}+2^{3 n-3} \cdot 3^{n-1}$ is divisible by 25 Step 1: $P(1): 7^{2}+2^{3-3} \cdot 3^{1-1}=49+1=50$ It is divisible by 25 . Thus, $P(1)$ is true Step 2: Let $P(m)$ be true. Now, $7^{2 m}+2^{3 m-3} \cdot 3^{m-1}$ is divisible by 25 . Suppose : $7^{2 m}+2^{3 m-3} \cdot 3^{m-1}=25 \lambda \quad \ldots(1)$ We have to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=7^{2 m+...
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Question: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$ (a) $\sqrt{2}$ (b) 2 (c) 4 (d) 8 Solution: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ $=\frac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}}$ $=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}} \quad[\sqrt{a b}=\sqrt{a} \times \sqrt{b}]$ $=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}$ $=\frac{4}{2}$ $=2$ Hence, the correct answer is option (b)....
Read More →n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
Question: n(n+ 1) (n+ 5) is a multiple of 3 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): n(n+1)(n+5)$ is a multiple of 3 . Step 1: $P(1): 1(1+1)(1+5)=12$ It is a multiple of 3 . Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $m(m+1)(m+5)$ is a multiple of 3 . Suppose $m(m+1)(m+5)=3 \lambda$, where $\lambda \in N$. We have to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=(m+1)(m+2)(m+6)$ $=m(m+1)(m+6)+2(m+1)(m+6)$ $=m(m+1)(m+5+1)+2(m+1)(m+6)$ $=m(...
Read More →The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
Question: The value of $(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)$ is (a) 1(b) 2(c) 4(d) 0 Solution: The given expression is $(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)$ Simplifying the given expression, we have $(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)$ $=\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)$ $=\frac{\sin \thet...
Read More →n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
Question: n(n+ 1) (n+ 5) is a multiple of 3 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): n(n+1)(n+5)$ is a multiple of 3 . Step 1: $P(1): 1(1+1)(1+5)=12$ It is a multiple of 3 . Hence, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $m(m+1)(m+5)$ is a multiple of 3 . Suppose $m(m+1)(m+5)=3 \lambda$, where $\lambda \in N$. We have to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=(m+1)(m+2)(m+6)$ $=m(m+1)(m+6)+2(m+1)(m+6)$ $=m(m+1)(m+5+1)+2(m+1)(m+6)$ $=m(...
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Question: $\sqrt{10} \times \sqrt{15}=?$ (i) $\sqrt{25}$ (ii) $5 \sqrt{6}$ (iii) $6 \sqrt{5}$ (iv) None of these Solution: $\sqrt{10} \times \sqrt{15}$ $=\sqrt{2 \times 5} \times \sqrt{3 \times 5}$ $=\sqrt{2} \times \sqrt{5} \times \sqrt{3} \times \sqrt{5} \quad[\sqrt{a b}=\sqrt{a} \times \sqrt{b}]$ $=5 \times \sqrt{2 \times 3}$ $=5 \sqrt{6}$ Hence, the correct answer is option (ii)....
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Question: (ab)n=anbnfor allnN. Solution: LetP(n) be the given statement. Now, $P(n):(a b)^{n}=a^{n} b^{n}$ for all $n \in N$. Step 1: $P(1):(a b)^{1}=a^{1} b^{1}=a b$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $(a b)^{m}=a^{m} b^{m}$ We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1):(a b)^{m+1}=(a b)^{m} \cdot a b$ $=a^{m} b^{m} \cdot a b$ $=a^{m} a \cdot b^{m} b$ $=a^{m+1} b^{m+1}$ Hence, $P(m+1)$ is true. By the $p$ rinciple of mathematical $i$ nduction,...
Read More →Let A be the set of all human beings in a town at a particular time.
Question: LetAbe the set of all human beings in a town at a particular time. Determine whether each of the following relations are reflexive, symmetric and transitive: (i) $R=\{(x, y): x$ and $y$ work at the same place $\}$ (ii) $R=\{(x, y)$ : $x$ and $y$ live in the same locality $\}$ (iii) $R=\{(x, y): x$ is wife of $y\}$ (iv) $R=\{(x, y): x$ is father of and $y\}$ Solution: (i) Reflexivity: Let $x$ be an arbitrary element of $R$. Then, $x \in R$ $\Rightarrow x$ and $x$ work at the same place ...
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Question: (ab)n=anbnfor allnN. Solution: LetP(n) be the given statement. Now, $P(n):(a b)^{n}=a^{n} b^{n}$ for all $n \in N$. Step 1: $P(1):(a b)^{1}=a^{1} b^{1}=a b$ Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $(a b)^{m}=a^{m} b^{m}$ We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1):(a b)^{m+1}=(a b)^{m} \cdot a b$ $=a^{m} b^{m} \cdot a b$ $=a^{m} a \cdot b^{m} b$ $=a^{m+1} b^{m+1}$ Hence, $P(m+1)$ is true. By the $p$ rinciple of mathematical $i$ nduction,...
Read More →The value of
Question: The value of $\frac{4 \sqrt{12}}{12 \sqrt{27}}$ is (a) $\frac{1}{9}$ (b) $\frac{2}{9}$ (c) $\frac{4}{9}$ (d) $\frac{8}{9}$ Solution: $\frac{4 \sqrt{12}}{12 \sqrt{27}}$ $=\frac{4 \sqrt{2 \times 2 \times 3}}{12 \sqrt{3 \times 3 \times 3}}$ $=\frac{4 \times 2 \sqrt{3}}{12 \times 3 \sqrt{3}}$ $=\frac{2}{9}$ Hence, the correct answer is option (b)....
Read More →sin θ1−cot θ+cos θ1−tan θ is equal to
Question: $\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$ is equal to (a) 0 (b) 1 (c) $\sin \theta+\cos \theta$ (d) $\sin \theta-\cos \theta$ Solution: The given expression is $\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$. Simplifying the given expression, we have $\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$ $=\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}$ $=\frac{...
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Question: 32n+28n 9 is divisible by 8 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 3^{2 n+2}-8 n-9$ is divisible by 8 for all $n \in N$. Step :1 $P(1)=3^{2+2}-8-9=81-17=64$ It is divisible by $8 .$ Thus, $P(1)$ is true. $\operatorname{Step}(2)$ : Let $P(m)$ be true. Then, $3^{2 m+2}-8 m-9$ is divisible by 8 . Let: $3^{2 m+2}-8 m-9=8 \lambda$ where $\lambda \in N \quad \ldots(1)$ We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=3^{2 m+4}-8(m+1)-17$ $...
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Question: The value of $\sqrt{20} \times \sqrt{5}$ is (a) 10 (b) $2 \sqrt{5}$ (c) $20 \sqrt{5}$ (d) $4 \sqrt{5}$ Solution: $\sqrt{20} \times \sqrt{5}$ $=\sqrt{2 \times 2 \times 5} \times \sqrt{5}$ $=2 \sqrt{5} \times \sqrt{5}$ $=2 \times 5$ $=10$ Hence, the correct answer is option (a)....
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Question: 32n+28n 9 is divisible by 8 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 3^{2 n+2}-8 n-9$ is divisible by 8 for all $n \in N$. Step :1 $P(1)=3^{2+2}-8-9=81-17=64$ It is divisible by $8 .$ Thus, $P(1)$ is true. $\operatorname{Step}(2)$ : Let $P(m)$ be true. Then, $3^{2 m+2}-8 m-9$ is divisible by 8 . Let: $3^{2 m+2}-8 m-9=8 \lambda$ where $\lambda \in N \quad \ldots(1)$ We need to show that $P(m+1)$ is true whenever $P(m)$ is true. Now, $P(m+1)=3^{2 m+4}-8(m+1)-17$ $...
Read More →the quotient is
Question: When $15 \sqrt{15}$ is divided by $3 \sqrt{3}$, the quotient is (a) $5 \sqrt{3}$ (b) $3 \sqrt{5}$ (c) $5 \sqrt{5}$ (d) $3 \sqrt{3}$ Solution: $15 \sqrt{15} \div 3 \sqrt{3}$ $=\frac{15 \sqrt{5 \times 3}}{3 \sqrt{3}}$ $=\frac{15 \times \sqrt{5} \times \sqrt{3}}{3 \sqrt{3}} \quad[\sqrt{a b}=\sqrt{a} \times \sqrt{b}]$ $=5 \sqrt{5}$ Hence, the correct answer is option (c)....
Read More →sin θ1+cos θis equal to
Question: $\frac{\sin \theta}{1+\cos \theta}$ is equal to (a) $\frac{1+\cos \theta}{\sin \theta}$ (b) $\frac{1-\cos \theta}{\cos \theta}$ (c) $\frac{1-\cos \theta}{\sin \theta}$ (d) $\frac{1-\sin \theta}{\cos \theta}$ Solution: The given expression is $\frac{\sin \theta}{1+\cos \theta}$. Multiplying both the numerator and denominator under the root by $(1-\cos \theta)$, we have $\frac{\sin \theta}{1+\cos \theta}$ $=\frac{\sin \theta(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}$ $=\frac{\sin \...
Read More →when simplified is
Question: $(6+\sqrt{27})-(3+\sqrt{3})+(1-2 \sqrt{3})$ when simplified is (a) positive and irrational (b) positive and rational (c) negative and irrational (d) neqative and rational Solution: $(6+\sqrt{27})-(3+\sqrt{3})+(1-2 \sqrt{3})$ $=6+\sqrt{3 \times 3 \times 3}-(3+\sqrt{3})+(1-2 \sqrt{3})$ $=6+3 \sqrt{3}-3-\sqrt{3}+1-2 \sqrt{3}$ $=4$ Thus, the given expression when simplified is positive and rational.Hence, the correct answer is option (b)....
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Question: 52n+224n 25 is divisible by 576 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 5^{2 n+2}-24 n-25$ is divisible by 576 for all $n \in N$. Step 1: $P(1)=5^{2+2}-24-25=625-49=576$ It is divisible by 576 . Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $5^{2 m+2}-24 m-25$ is divisible by 576 Let $5^{2 m+2}-24 m-25=576 \lambda$, where $\lambda \in N$. We need to show that $P(m+1)$ is true wheneve $r P(m)$ is true. Now, $P(m+1)=5^{2 m+4}-24(m+1)-25$ $=5^{2} \times...
Read More →cos4 A − sin4 A is equal to
Question: $\cos ^{4} A-\sin ^{4} A$ is equal to (a) $2 \cos ^{2} A+1$ (b) $2 \cos ^{2} A-1$ (c) $2 \sin ^{2} A-1$ (d) $2 \sin ^{2} A+1$ Solution: The given expression is $\cos ^{4} A-\sin ^{4} A$. Factorising the given expression, we have $\cos ^{4} A-\sin ^{4} A$ $=\left(\cos ^{2} A+\sin ^{2} A\right) \times\left(\cos ^{2} A-\sin ^{2} A\right)$ $=1 \times\left(\cos ^{2} A-\sin ^{2} A\right)$ $=\cos ^{2} A-\sin ^{2} A$ $=\cos ^{2} A-\left(1-\cos ^{2} A\right)$ $=\cos ^{2} A-1+\cos ^{2} A$ $=2 \c...
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Question: 52n+224n 25 is divisible by 576 for allnN. Solution: LetP(n) be the given statement. Now, $P(n): 5^{2 n+2}-24 n-25$ is divisible by 576 for all $n \in N$. Step 1: $P(1)=5^{2+2}-24-25=625-49=576$ It is divisible by 576 . Thus, $P(1)$ is true. Step 2 : Let $P(m)$ be true. Then, $5^{2 m+2}-24 m-25$ is divisible by 576 Let $5^{2 m+2}-24 m-25=576 \lambda$, where $\lambda \in N$. We need to show that $P(m+1)$ is true wheneve $r P(m)$ is true. Now, $P(m+1)=5^{2 m+4}-24(m+1)-25$ $=5^{2} \times...
Read More →when simplified is
Question: $(-2-\sqrt{3})(-2+\sqrt{3})$ when simplified is (a) positive and irrational(b) positive and rational(c) negative and irrational(d) negative and rational Solution: $(-2-\sqrt{3})(-2+\sqrt{3})$ $=-(2+\sqrt{3}) \times[-(2-\sqrt{3})]$ $=(2+\sqrt{3})(2-\sqrt{3})$ $=2^{2}-(\sqrt{3})^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$ $=4-3$ $=1$ Thus, the given expression when simplified is positive and rational.Hence, the correct answer is option (b)....
Read More →sec4 A − sec2 A is equal to
Question: $\sec ^{4} A-\sec ^{2} A$ is equal to (a) $\tan ^{2} \mathrm{~A}-\tan ^{4} \mathrm{~A}$ (b) $\tan ^{4} A-\tan ^{2} A$ (c) $\tan ^{4} A+\tan ^{2} A$ (d) $\tan ^{2} \mathrm{~A}+\tan ^{4} \mathrm{~A}$ Solution: The given expression is $\sec ^{4} \mathrm{~A}-\sec ^{2} \mathrm{~A}$. Taking common $\sec ^{2} \mathrm{~A}$ from both the terms, we have $\sec ^{4} A-\sec ^{2} A$ $=\sec ^{2} \mathrm{~A}\left(\sec ^{2} \mathrm{~A}-1\right)$ $=\left(1+\tan ^{2} \mathrm{~A}\right) \tan ^{2} \mathrm{...
Read More →Which of the following is the value of
Question: Which of the following is the value of $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7}) ?$ (a) $-4$ (b) 4 (c) $\sqrt{11}$ (d) $\sqrt{7}$ Solution: $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})$ $=(\sqrt{11})^{2}-(\sqrt{7})^{2} \quad\left[(a-b)(a+b)=a^{2}-b^{2}\right]$ $=11-7$ $=4$ Hence, the correct answer is option (b)....
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