If
Question: If $P(n): \sqrt{n}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}, n \in N$, then $P(n)$ is true for all $n \geq$ ________________ Solution: $P(n): \sqrt{n}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}} ; n \in N$ forn= 1, $P(1): 1\frac{1}{\sqrt{1}}=1$ which is a false statement. forn= 2, $P(2): \sqrt{2}\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$ which is true forn= 3 $P(3): \sqrt{3}1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}$ which is again t...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 50 andh= 10. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=40, \sum f_{i} u_{j}=-2, h=10$ and $A=50$. Putting the values in the above formula, we have $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=50+10\left(\frac{1}{40} \times(-2)\right)$ $=50-\frac{20}{40}$ $=50-0.5$ $=49.5$ Hence, the mean is 49.5....
Read More →If P(n) :
Question: IfP(n) :n2 2n,nN, thenP(n) is true for alln _____________. Solution: P(n) :n2 2n,nN forn =1, P(1) : 1 2 which is true statement forn =2, P(2) : 22 22 which is not true/false statement forn =3, P(3) : (3)2 23 i.e 9 8 which is a false statement forn =4, P(4) : (4)2 24 i.e 16 4 4 = 16 which is a false statement forn =5, P(5) : (5)2 25 i.e 25 32 which is a true statement forn =6, P(6) : (36) 26 i.e 36 4 4 4 = 64 Which is a true statement for $n=7, P(7): 492^{7}=128$ which is again true sta...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 60 andh= 20. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=50, \sum f_{i} u_{i}=14, h=20$ and $A=60$. Putting the values in the above formula, we have $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=60+20\left(\frac{1}{50} \times(14)\right)$ $=60+\frac{280}{50}$ $=60+5.6$ $=65.6$ Hence, the mean is 65.6....
Read More →If P(n) :
Question: IfP(n) :n3nis divisible by 6,nN, thenP(n) is true for alln _____________. Solution: P(n) :n3nis divisible by 6;nN for n= 1, P(1) : (1)3 1 = 0 which is divisible by 6 forn= 2, P(2) : (2)3 2 = 8 2 = 6 which is divisible by 6 forn= 3, P(3) : (3)3 3 = 27 3 = 24; which is divisible by 6 Hence,P(n) is true forn 2...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 20 andh= 8. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=20, \sum f_{i} u_{i}=-9, h=8$ and $A=20$. Putting the values in the above formula, we get $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=20+8\left(\frac{1}{20} \times(-9)\right)$ $=20-\frac{72}{20}$ $=20-3.6$ $=16.4$ Hence, the mean is 16.4....
Read More →If P(n) :
Question: IfP(n) :n! 2n 1,nN, thenP(n) is true for alln _____________. Solution: P(n) :n! 2n 1;n forn= 1, P(1) : 1! 211 i.e 1 2= 1 i.e 1 1 which is false a statement forn= 2 P(2) : 2! 221 i.e 2 21 i.e 2 2 which is again a false statement. forn= 3 P(3) : 3! 231 i.e 6 22= 4 i.e 6 4 which is true forn= 4 P(4) : 4! 241 i.e 24 23= 8 which is true Hence, P(n) :n! 2n 1is true forn 2...
Read More →Prove that the relation R on Z defined by
Question: Prove that the relationRonZdefined by $(a, b) \in R \Leftrightarrow a-b$ is divisible by 5 is an equivalence relation onZ. Solution: We observe the following properties of relationR. Reflexivity: Let $a$ be an arbitrary element of $R$. Then, $\Rightarrow a-a=0=0 \times 5$ $\Rightarrow a-a$ is divisible by 5 $\Rightarrow(a, a) \in R$ for all $a \in Z$ So, $R$ is reflexive on $Z$. Symmetry: Let $(a, b) \in R$ $\Rightarrow a-b$ is divisible by 5 $\Rightarrow a-b=5 p$ for some $p \in Z$ $\...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 20 andh= 8. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=40, \sum f_{i} u_{i}=5, h=8$ and $A=20$. Putting the values in the above formula, we have $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=20+8\left(\frac{1}{40} \times(5)\right)$ $=20+\frac{40}{40}$ $=20+1$ $=21$ Hence, the mean is 21....
Read More →For each n ∈ N,
Question: For eachnN, 102n 1+ 1 is divisible by _____________. Solution: For eachnN, LetP(n) : 102n1+1 forn= 1 L.H.S = 102(1)1+1 = 101+ 1 = 10 + 1 = 11 i.e P(1) = 11 AssumeP(n) is true forn= 2, $P(2): 10^{2(2)-1}+1$ $=10^{4-1}+1$ $=10^{3}+1$ $=1000+1$ $=1001$ $P(2)=11(91)$ Similarly, assume thatP(k) is divisible by 11. $P(k+1): 10^{2(k+1)-1}+1$ $=10^{2 k+2-1}+1$ $=10^{2 k+1}+1$ $=10^{(2 k-1)+2}+1$ $=10^{2 k-1} \cdot 10^{2}+1 \quad\left[\right.$ since $10^{2 k-1}+1=11 \mathrm{~m}$ i. e $\left.10^...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 25 andh= 10. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=60, \sum f_{i} u_{i}=8, h=10$ and $A=25$. Putting the values in the above formula, we have $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=25+10\left(\frac{1}{60} \times(8)\right)$ $=25+\frac{80}{60}$ $=25+1.333$ $=26.333$ Hence, the mean is 26.333....
Read More →For each n ∈ N,
Question: For eachnN, 102n 1+ 1 is divisible by _____________. Solution: For eachnN, LetP(n) : 102n1+1 forn= 1 L.H.S = 102(1)1+1 = 101+ 1 = 10 + 1 = 11 i.e P(1) = 11 AssumeP(n) is true forn= 2, $P(2): 10^{2(2)-1}+1$ $=10^{4-1}+1$ $=10^{3}+1$ $=1000+1$ $=1001$ $P(2)=11(91)$ Similarly, assume thatP(k) is divisible by 11. $P(k+1): 10^{2(k+1)-1}+1$ $=10^{2 k+2-1}+1$ $=10^{2 k+1}+1$ $=10^{(2 k-1)+2}+1$ $=10^{2 k-1} \cdot 10^{2}+1 \quad\left[\right.$ since $10^{2 k-1}+1=11 \mathrm{~m}$ i. e $\left.10^...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 15 andh= 6. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=40, \sum f_{i} u_{i}=5, h=6$ and $A=15$. Putting the values in the above formula, we get $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=15+6\left(\frac{1}{40} \times(5)\right)$ $=15+\frac{30}{40}$ $=15+0.75$ $=15.75$ Hence, the mean is 15.75....
Read More →Match the following columns:
Question: Match the following columns: (a) ....... (b) ........ (c) ........ (d) ........ Solution: (a) Because it is a non-terminating and repeating decimal, it is a rational number. (b) $\pi$ is an irrational number. (c) $\frac{1}{7}=.142857142857 \ldots$ Hence, its period is 6. (d) $x^{2}+\frac{1}{x^{2}}$ $=(2-\sqrt{3})^{2}+\frac{1}{(2-\sqrt{3})^{2}}$ $=\left(2^{2}+(\sqrt{3})^{3}-2 \times 2 \times \sqrt{3}\right)+\frac{1}{\left(2^{2}+(\sqrt{3})^{3}-2 \times 2 \times \sqrt{3}\right)}$ $=(4+3-4...
Read More →If P(n) :
Question: IfP(n) : 2nn!,nN, thenP(n) is true for alln _____________. Solution: GivenP(n) : 2nn! ;nN, forn= 1, P(1) : 2' 1! i.e 2 1 Which is not true for n= 2, P(2) : 22= 4 2! i.e 4 2 Which is not true forn= 3, P(3) : 23 3! i.e $81 \times 2 \times 3$ i.e $86$ Which is again not true. forn= 4, P(4) i.e 24 4! i.e 16 24 i.e a true statement. P(5) : 25 5! i.e 32 1 2 3 4 5 i.e 32 120 Which is also true $\therefore F(n): 2^{n}n !$ is true for $n3$ $F(n): 2^{n}n !$ is true for $n3$...
Read More →If P(n) :
Question: IfP(n) : 2nn!,nN, thenP(n) is true for alln _____________. Solution: GivenP(n) : 2nn! ;nN, forn= 1, P(1) : 2' 1! i.e 2 1 Which is not true for n= 2, P(2) : 22= 4 2! i.e 4 2 Which is not true forn= 3, P(3) : 23 3! i.e $81 \times 2 \times 3$ i.e $86$ Which is again not true. Which is again not true. forn= 4, P(4) i.e 24 4! i.e 16 24 i.e a true statement. P(5) : 25 5! i.e 32 1 2 3 4 5 i.e 32 120 Which is also true $\therefore F(n): 2^{n}n !$ is true for $n3$ $F(n): 2^{n}n !$ is true for $...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 20 andh= 8. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=40, \sum f_{i} u_{i}=7, h=8$ and $A=20$. Putting the values in the above formula, we get $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=20+8\left(\frac{1}{40} \times(7)\right)$ $=20+\frac{56}{40}$ $=20+1.4$ $=21.4$ Hence, the mean is 21.4....
Read More →Show that the relation R on the set Z of integers, given by
Question: Show that the relationRon the setZof integers, given byR= {(a, b) : 2 dividesa b}, is an equivalence relation. Solution: We observe the following properties of relationR. Reflexivity: Let $a$ be an arbitrary element of the set $Z$. Then, $a \in R$ $\Rightarrow a-a=0=0 \times 2$ $\Rightarrow 2$ divides $a-a$ $\Rightarrow(a, a) \in R$ for all $a \in Z$ So, $R$ is reflexive on $Z$. Symmetry: Let $(a, b) \in R$ $\Rightarrow 2$ divides $a-b$ $\Rightarrow \frac{a-b}{2}=p$ for some $p \in Z$ ...
Read More →If P(n) : 2n < n!, n ∈ N,
Question: IfP(n) : 2nn!,nN, thenP(n) is true for alln _____________. Solution: GivenP(n) : 2nn! ;n N for n= 1, P(1) : 2' 1! i.e 2 1 Which is not true forn= 2, P(2) : 22= 4 2! i.e 4 2 Which is not true forn= 3, P(3) : 23 3! i.e $81 \times 2 \times 3$ i.e $86$ Which is again not true forn= 4, P(4) i.e 24 4! i.e 16 24 i.e a true statement. P(5) : 25 5! i.e 32 1 2 3 4 5 i.e 32 120 Which is also true P(n) : 2nn! is true forn 3 P(n) : 2nn! is true i.e forn 4...
Read More →If P(n) : 2n < n!, n ∈ N,
Question: IfP(n) : 2nn!,nN, thenP(n) is true for alln _____________. Solution: GivenP(n) : 2nn! ;n N for n= 1, P(1) : 2' 1! i.e 2 1 Which is not true forn= 2, P(2) : 22= 4 2! i.e 4 2 Which is not true forn= 3, P(3) : 23 3! i.e $81 \times 2 \times 3$ i.e $86$ Which is again not true forn= 4, P(4) i.e 24 4! i.e 16 24 i.e a true statement. P(5) : 25 5! i.e 32 1 2 3 4 5 i.e 32 120 Which is also true P(n) : 2nn! is true forn 3 P(n) : 2nn! is true i.e forn 4...
Read More →Assertion:
Question: Assertion: $\sqrt{3}$ is an irrational number. Reason:The sum of rational number and an irrational number is an irrational number.(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.(c) Assertion is true and Reasom is false.(d) Assertion is false and Reasom is true. Solution: (b) Both Assertion and Reason are true, but Reason is not a correct explan...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 100 andh= 20. We know that mean $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now we have $N=\sum f_{i}=100, \sum f_{i} u_{i}=61, h=20$ and $A=100$ Putting the values in the above formula, we get $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=100+20\left(\frac{1}{100} \times(61)\right)$ $=100+\frac{1220}{100}$ $=100+12.20$ $=112.20$ Hence, the mean is 112.20....
Read More →If P(n) :
Question: If $P(n): " 2 \times 4^{2 n+1}+3^{3 n+1}$ is divisible by $\lambda$ for all $n \in N^{\prime \prime}$ is true, then the value of $\lambda$ is _____________ Solution: If $P(n)=2 \times 4^{2 n+1}+3^{3 n+1}$ is divisible by $\lambda \forall n \in \mathbf{N}$ is true for $n=1$, $P(1): 2 \times 4^{2(1)+1}+3^{3(1)+1}$ $=2 \times 4^{3}+3^{4}$ $=2 \times 64+81$ $=128+81$ $P(1)=209$ for $n=2$, $P(2): 2^{3} \times 4^{2(2)+1}+3^{3(2)+1}$ $=2^{3} \times 4^{5}+3^{7}$ $=2^{3} \times 256+2187$ $=2048...
Read More →Find the mean of each of the following frequency distributions :
Question: Find the mean of each of the following frequency distributions : Solution: Let the assumed mean beA= 15 andh= 6. We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ Now, we have $N=\sum f_{i}=40, \sum f_{i} u_{i}=3, h=6$ and $A=15$. Putting the values in the above formula, we get $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$ $=15+6\left(\frac{1}{40} \times(3)\right)$ $=15+\frac{18}{40}$ $=15+0.45$ $=15.45$ Hence, the mean is 15.45....
Read More →Assertion: e is irrational number.
Question: Assertion:eis irrational number.Reason:is an irrational number.(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.(c) Assertion is true and Reasom is false.(d) Assertion is false and Reasom is true. Solution: (b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.It is known thateand are irrational numbers, but...
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