If
Question: If $R=\left\{(x, y): x^{2}+y^{2} \leq 4, x, y \in Z\right\}$ is a relation in $\mathrm{Z}$, then the domain of $R$ is Solution: Given: $R=\left\{(x, y): x^{2}+y^{2} \leq 4, x, y \in Z\right\}$ $R=\{(-2,0),(2,0),(0,2),(0,-2),(-1,1),(-1,-1),(1,-1),(1,1),(0,1),(1,0),(-1,0),(0,-1),(0,0)\}$ Therefore, Domain of R = {2, 1, 0, 1, 2} Hence, if $R=\left\{(x, y): x^{2}+y^{2} \leq 4, x, y \in Z\right\}$ is a relation in $Z$, then the domain of $R$ is $\{\underline{-2}, \underline{-1}, 0,1,2\}$....
Read More →If a relation R on the set (1, 2, 3) be defined by R = {(1, 2)}, then R is
Question: If a relationRon the set (1, 2, 3) be defined byR= {(1, 2)}, thenRis (a) reflexive (b) transitive (c) symmetric (d) none of these Solution: Given: ArelationRon the set {1, 2, 3} be defined byR= {(1, 2)}. R= {(1, 2)} Since, $(1,1) \notin R$ Therefore, It is not reflexive. Since, $(1,2) \in R$ but $(2,1) \notin R$ Therefore, It is not symmetric. But there is no counter example to disapprove transitive condition.Therefore, it is transitive. Hence, the correct option is (b)....
Read More →Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
Question: Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: (i) $1+i$ (ii) $\sqrt{3}+i$ (iii) $1-i$ (iv) $\frac{1-i}{1+i}$ (v) $\frac{1}{1+i}$ (vi) $\frac{1+2 i}{1-3 i}$ (vii) $\sin 120^{\circ}-i \cos 120^{\circ}$ (viii) $\frac{-16}{1+i \sqrt{3}}$ Solution: (i) $z=1+i$ $r=|z|$ $=\sqrt{1+1}$ $=\sqrt{2}$ Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ $\Rightarrow \tan \alpha=\left(\frac{1}{1}\right)$ $...
Read More →Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
Question: Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: (i) $1+i$ (ii) $\sqrt{3}+i$ (iii) $1-i$ (iv) $\frac{1-i}{1+i}$ (v) $\frac{1}{1+i}$ (vi) $\frac{1+2 i}{1-3 i}$ (vii) $\sin 120^{\circ}-i \cos 120^{\circ}$ (viii) $\frac{-16}{1+i \sqrt{3}}$ Solution: (i) $z=1+i$ $r=|z|$ $=\sqrt{1+1}$ $=\sqrt{2}$ Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ $\Rightarrow \tan \alpha=\left(\frac{1}{1}\right)$ $...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: For real numbers $x$ and $y$, define $x R y$ iff $x-y+\sqrt{2}$ is an irrational number. Then the relation $R$ is (a) reflexive (b) symmetric (c) transitive (d) none of these Solution: We have, $R=\{(x, y): x-y+\sqrt{2}$ is an irrational number; $x, y \in \mathbf{R}\}$ As, $x-x+\sqrt{2}=\sqrt{2}$, which is an irrational number $\Rightarrow(x, x) \in R$ So, $R$ is reflexive relation Since, $(\sqrt{2}, 2) \in R$ i. e. $\sqrt{2}-2+\s...
Read More →Find the mean, median and mode of the following data:
Question: Find the mean, median and mode of the following data: Solution: Consider the following data. Here, the maximum frequency is 12 so the modal class is 6080. Therefore, $l=60$ $h=20$ $f=12$ $f_{1}=10$ $f_{2}=6$ $F=24$ Median $=l+\frac{\frac{N}{2}-F}{f} \times h$ $=60+\frac{25-24}{12} \times 20$ $=60+\frac{1}{12} \times 20$ Median $=61.66$ Thus, the median of the data is 61.66. Mean $=\frac{\sum f_{i} x_{i}}{\sum f}$ $=\frac{3120}{50}$ Mean = 62.4 Thus, the mean of the data is 62.4.Mode $=...
Read More →100 surnames were randomly picked up from a local telephone directly and the frequency
Question: 100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, fund the modal size of the surnames. Solution: Consider the following table. Here, the maximum frequency is 40 so the modal class is 710. Therefore, $l=7$ $h=3$ $f=40$ $f_{1}=30$ $f_{2}=16...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Consider a non-empty set consisting of children in a family and a relationRdefined asaRbifais brother ofb. Then,Ris (a) symmetric but not transitive (b) transitive but not symmetric (c) neither symmetric nor transitive (d) both symmetric and transitive Solution: We have, $R=\{(a, b): a$ is brother of $b\}$ Let $(a, b) \in R$. Then, $a$ is brother of $b$ but $b$ is not necessary brother of $a$ (As, $b$ can be sister of $a$ ) $\Righ...
Read More →Using the remainder theorem, find the remainder,
Question: Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=6 x^{3}+13 x^{2}+3, g(x)=3 x+2$. Solution: $p(x)=6 x^{3}+13 x^{2}+3$ $g(x)=3 x+2=3\left(x+\frac{2}{3}\right)=3\left[x-\left(-\frac{2}{3}\right)\right]$ By remainder theorem, when $p(x)$ is divided by $(3 x+2)$, then the remainder $=p\left(-\frac{2}{3}\right)$. Putting $x=-\frac{2}{3}$ in $p(x)$, we get $p\left(-\frac{2}{3}\right)=6 \times\left(-\frac{2}{3}\right)^{3}+13 \times\left(-\frac{2}{...
Read More →The following frequency distribution gives the monthly consumption of electricity
Question: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Solution: The given data is shown below. Here, the maximum frequency is 20 so the modal class is 125145. Therefore, $l=125$ $h=20$ $f=20$ $f_{1}=13$ $f_{2}=14$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=125+\frac{7}{13} \times 20$ $=125+\frac{140}{13}$ Mode $=135.76$ units Thus, the mode of ...
Read More →Using the remainder theorem, find the remainder,
Question: Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=2 x^{3}+x^{2}-15 x-12, g(x)=x+2$. Solution: $p(x)=2 x^{3}+x^{2}-15 x-12$ $g(x)=x+2$ By remainder theorem, whenp(x) is divided by (x+2), then the remainder =p(2).Puttingx= 2 inp(x), we get $p(-2)=2 \times(-2)^{3}+(-2)^{2}-15 \times(-2)-12=-16+4+30-12=6$ Remainder = 6Thus, the remainder whenp(x) is divided byg(x) is 6....
Read More →Find the values of the following expressions:
Question: Find the values of the following expressions: Solution: (i) $i^{49}+i^{68}+i^{89}+i^{110}$ $=i^{4 \times 12+1}+i^{4 \times 17}+i^{4 \times 22+1}+i^{4 \times 27+2}$ $=\left\{\left(i^{4}\right)^{12} \times i\right\}+\left\{\left(i^{4}\right)^{17}\right\}+\left\{\left(i^{4}\right)^{22} \times i\right\}+\left\{\left(i^{4}\right)^{27} \times i^{2}\right\}$ $=i+1+i+i^{2} \quad\left[\because i^{4}=1\right]$ $=2 i+1-1 \quad\left[\because i^{2}=-1\right]$ $=2 i$ (ii) $i^{30}+i^{80}+i^{120}$ $=i...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $a R b$ if $a$ is congruent to $b$ for all $a$, $b \in T$. Then, $R$ is a) reflexive but not symmetric (b) transitive but not symmetric (c) equivalence (d) none of these Solution: We have, $R=\{(a, b): a$ is congruent to $b ; a, b \in T\}$ As, $a \cong a$ $\Rightarrow(a, a) \in R$ So, $R$ is reflexive relation Let $(a, b) \in \m...
Read More →Using the remainder theorem, find the remainder,
Question: Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=x^{3}-2 x^{2}-8 x-1, g(x)=x+1 .$ Solution: $p(x)=x^{3}-2 x^{2}-8 x-1$ $g(x)=x+1$ By remainder theorem, whenp(x) is divided by (x+1), then the remainder =p(1).Puttingx= 1 inp(x), we get $p(-1)=(-1)^{3}-2 \times(-1)^{2}-8 \times(-1)-1=-1-2+8-1=4$ Remainder = 4Thus, the remainder whenp(x) is divided byg(x) is 4....
Read More →A student noted the number of cars passing through a spot on a road for
Question: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised in the table given below. Find the mode of the data:/spanbr data-mce-bogus="1"/ppimg src="https://www.esaral.com/qdb/uploads/2021/12/31/image14651.png" alt="" Solution: The given data is shown below. Here, the maximum frequency is 20 so the modal class is 4050. Therefore, $l=40$ $h=10$ $f=20$ $f_{1}=12$ $f_{2}=11$ $\therefore$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \ti...
Read More →Using the remainder theorem, find the remainder,
Question: Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=2 x^{3}-9 x^{2}+x+15, g(x)=2 x-3 .$ Solution: $p(x)=2 x^{3}-9 x^{2}+x+15$ $g(x)=2 x-3=2\left(x-\frac{3}{2}\right)$ By remainder theorem, when $p(x)$ is divided by $(2 x-3)$, then the remainder $=p\left(\frac{3}{2}\right)$. Putting $x=\frac{3}{2}$ in $p(x)$, we get $p\left(\frac{3}{2}\right)=2 \times\left(\frac{3}{2}\right)^{3}-9 \times\left(\frac{3}{2}\right)^{2}+\frac{3}{2}+15=\frac{27}{4}-\...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Let $L$ denote the set of all straight lines in a plane. Let a relation $R$ be defined by $I R m$ iff $/$ is perpendicular to $m$ for all $I, m \in L$. Then, $R$ is (a) reflexive (b) symmetric (c) transitive (d) none of these [NCERT EXEMPLAR] Solution: We have, $R=\{(l, m): l$ is perpendicular to $m ; l, m \in L\}$ As, $l$ is not perpencular to $l$ $\Rightarrow(l, l) \notin R$ So, $R$ is not reflexive relation Let $(l, m) \in R$ $...
Read More →Using the remainder theorem, find the remainder,
Question: Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=3 x^{4}-6 x^{2}-8 x-2, g(x)=x-2$. Solution: $p(x)=3 x^{4}-6 x^{2}-8 x-2$ $g(x)=x-2$ By remainder theorem, whenp(x) is divided by (x2), then the remainder =p(2).Puttingx= 2 inp(x), we get $p(2)=3 \times 2^{4}-6 \times 2^{2}-8 \times 2-2=48-24-16-2=6$ Remainder = 6Thus, the remainder whenp(x) is divided byg(x) is 6....
Read More →The given distribution shows the number of runs scored by
Question: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Solution: The given data is shown below. Here, the maximum frequency is 18 so the modal class is 4000-5000. Therefore, $l=4000$ $h=1000$ $f=18$ $f_{1}=4$ $f_{2}=9$ $\therefore$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=4000+\frac{18-4}{2 \times 18-4-9} \times 1000$ $=4000+\frac{14}{23} \times 1000$ $=4000+608.7$ $=4608.7$ Thus, the mode of the dat...
Read More →Using the remainder theorem, find the remainder,
Question: Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=2 x^{3}-7 x^{2}+9 x-13, g(x)=x-3$. Solution: $p(x)=2 x^{3}-7 x^{2}+9 x-13$ $g(x)=x-3$ By remainder theorem, when $p(x)$ is divided by $(x-3)$, then the remainder $=p(3)$. Putting $x=3$ in $p(x)$, we get $p(3)=2 \times 3^{3}-7 \times 3^{2}+9 \times 3-13=54-63+27-13=5$ Remainder = 5Thus, the remainder whenp(x) is divided byg(x) is 5....
Read More →The following distribution gives the state-wise teacher-student
Question: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures: Solution: Here, the maximum frequency is 10 so the modal class is 3035. Therefore, $l=30$ $h=5$ $f=10$ $f_{1}=9$ $f_{2}=3$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=30+\frac{10-9}{20-9-3} \times 5$ $=30+\frac{1}{8} \times 5$ $=30+0.625$ Mode $=30.6$ Thus, the mode of the data is 30.6. Now, ...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: LetRbe a relation on the setNof natural numbers defined bynRmiffndividesm. Then,Ris (a) Reflexive and symmetric (b) Transitive and symmetric (c) Equivalence (d) Reflexive, transitive but not symmetric [NCERT EXEMPLAR] Solution: We have, $R=\{(m, n): n$ divides $m ; m, n \in \mathbf{N}\}$ As, $m$ divides $m$ $\Rightarrow(m, m) \in R \forall m \in \mathbf{N}$ So, $R$ is reflexive Since, $(2,1) \in R$ i. e. 1 divides 2 but 2 cannot d...
Read More →Verify the division algorithm for the polynomials
Question: Verify the division algorithm for the polynomials $p(x)=2 x^{4}-6 x^{3}+2 x^{2}-x+2$ and $g(x)=x+2$. Solution: $p(x)=2 x^{4}-6 x^{3}+2 x^{2}-x+2$ and $g(x)=x+2$ Quotient $=2 x^{3}-10 x^{2}+22 x-45$ Remainder = 92Verification: Divisor $\times$ Quotient + Remainder $=(x+2) \times\left(2 x^{3}-10 x^{2}+22 x-45\right)+92$ $=x\left(2 x^{3}-10 x^{2}+22 x-45\right)+2\left(2 x^{3}-10 x^{2}+22 x-45\right)+92$ $=2 x^{4}-10 x^{3}+22 x^{2}-45 x+4 x^{3}-20 x^{2}+44 x-90+92$ $=2 x^{4}-6 x^{3}+2 x^{2...
Read More →The following data is gives the information on the observed lifetimes (in hours) of 225 electrical components:
Question: The following data is gives the information on the observed lifetimes (in hours) of 225 electrical components: Determine the modal lifetimes of the components. Solution: Here, the maximum frequency of electrical components is 61 so the modal class is 6080. Therefore, $l=60$ $h=20$ $f=61$ $f_{1}=52$ $f_{2}=38$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=60+\frac{61-52}{122-52-38} \times 20$ $=60+\frac{9}{32} \times 20$ Mode $=65.625 \mathrm{hrs}$ Thus, the modal l...
Read More →The following table show the ages of the patients admitted in a hospital during a year:
Question: The following table show the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Solution: Here, the maximum frequency is 23 so the modal class is 3545. Therefore, $l=35$ $h=10$ $f=23$ $f_{1}=21$ $f_{2}=14$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=35+\frac{2}{46-35} \times 10$ $=35+\frac{2}{11} \times 10$ $=35+\frac{20}{11}$ $=35+1.80$ Mode $=...
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