By actual division, find the quotient and the remainder when
Question: By actual division, find the quotient and the remainder when $\left(x^{4}+1\right)$ is divided by $(x-1)$. Verify that remainder $=f(1)$. Solution: Let $f(x)=x^{4}+1$ and $g(x)=x-1$ Quotient $=x^{3}+x^{2}+x+1$ Remainder $=2$ Verification: Putting $x=1$ in $f(x)$, we get $f(1)=1^{4}+1=1+1=2=$ Remainder, when $f(x)=x^{4}+1$ is divided by $g(x)=x-1$...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: The maximum number of equivalence relations on the setA= {1, 2, 3} is (a) 1 (b) 2 (c) 3 (d) 5 Solution: Consider the relation $R_{1}=\{(1,1)\}$ It is clearly reflexive, symmetric and transitive Similarly, $R_{2}=\{(2,2)\}$ and $R_{3}=\{(3,3)\}$ are reflexive, symmetric and transitive Also, $R_{4}=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$ It is reflexive as $(a, a) \in R_{4}$ for all $a \in\{1,2,3\}$ It is symmetric as $(a, b) \in R_{4} \...
Read More →The following is the distribution of height of students of a certain class in a certain city:
Question: The following is the distribution of height of students of a certain class in a certain city: Find the average height of maximum number of students. Solution: The given data is an inclusive series. So, firstly convert it into an exclusive series as given below. Here, the maximum frequency is 142 so the modal class is 165.5168.5. Therefore, $l=165.5$ $h=3$ $f=142$ $f_{1}=118$ $f_{2}=127$ Now, Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=165.5+\frac{142-118}{284-118-127} \times 3...
Read More →If 2 and 0 are the zeros of the polynomial
Question: If 2 and 0 are the zeros of the polynomial $f(x)=2 x^{3}-5 x^{2}+a x+b$ then find the values of $a$ and $b$. Hint $f(2)=0$ and $f(0)=0$ Solution: It is given that 2 and 0 are the zeroes of the polynomial $f(x)=2 x^{3}-5 x^{2}+a x+b$. $\therefore f(2)=0$ $\Rightarrow 2 \times 2^{3}-5 \times 2^{2}+a \times 2+b=0$ $\Rightarrow 16-20+2 a+b=0$ $\Rightarrow-4+2 a+b=0$ $\Rightarrow 2 a+b=4$ .......(1) Also, $f(0)=0$ $\Rightarrow 2 \times 0^{3}-5 \times 0^{2}+a \times 0+b=0$ $\Rightarrow 0-0+0...
Read More →Show that
Question: Show that 1 +i10+i20+i30is a real number. Solution: $1+i^{10}+i^{20}+i^{30}$ $=1+i^{4 \times 2+2}+i^{4 \times 5}+i^{4 \times 7+2}$ $=1+\left[\left(i^{4}\right)^{2} \times i^{2}\right]+\left(i^{4}\right)^{5}+\left[\left(i^{4}\right)^{7} \times i^{2}\right]$ $=1+i^{2}+1+i^{2}$ $\left(\because i^{4}=1\right)$ $=1-1+1-1$ $\left(\because i^{2}=-1\right)$ = 0 This is a real number. Hence proved....
Read More →The following data gives the distribution of total monthly household expenditure of 200 families of a villages.
Question: The following data gives the distribution of total monthly household expenditure of 200 families of a villages. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure: Solution: Here, the maximum frequency is 40 so the modal class is 15002000. Therefore, $l=1500$ $h=500$ $f=40$ $f_{1}=24$ $f_{2}=33$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=1500+\frac{40-24}{80-24-33} \times 500$] $=1500+\frac{16}{23} \times 500$ $=1500+347....
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: The relation $S$ defined on the set $\mathbf{R}$ of all real number by the rule $a S b$ iff $a \geq b$ is (a) an equivalence relation(b) reflexive, transitive but not symmetric(c) symmetric, transitive but not reflexive(d) neither transitive nor reflexive but symmetric Solution: We have, $S=\{(a, b): a \geq b ; a, b \in \mathbf{R}\}$ As, $a=a \forall a \in \mathbf{R}$ $\Rightarrow(a, a) \in S$ So, $S$ is reflexive relation Let $(a...
Read More →Find the square root of the following complex numbers:
Question: Find the square root of the following complex numbers: (i) $-5+12 i$ (ii) $-7-24 i$ (iii) $1-\mathrm{i}$ (iv) $-8-6 i$ (v) $8-15 i$ (vi) $-11-60 \sqrt{-1}$ (vii) $1+4 \sqrt{-3}$ (viii) $4 i$ (ix) -i Solution: $\sqrt{z}=\pm\left[\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}+i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right] \quad$, if $\operatorname{Im}(z)0$ $\sqrt{z}=\pm\left[\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}-i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right]$, if $\operatorname{Im...
Read More →Find the square root of the following complex numbers:
Question: Find the square root of the following complex numbers: (i) $-5+12 i$ (ii) $-7-24 i$ (iii) $1-\mathrm{i}$ (iv) $-8-6 i$ (v) $8-15 i$ (vi) $-11-60 \sqrt{-1}$ (vii) $1+4 \sqrt{-3}$ (viii) $4 i$ (ix) -i Solution: $\sqrt{z}=\pm\left[\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}+i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right] \quad$, if $\operatorname{Im}(z)0$ $\sqrt{z}=\pm\left[\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}-i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right]$, if $\operatorname{Im...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Let $A=\{1,2,3\}$ and consider the relation $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$. Then, $R$ is (a) reflexive but not symmetric (b) reflexive but not transitive(c) symmetric and transitive (d) neither symmetric nor transitive Solution: We have, $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ As, $(a, a) \in R \forall a \in A$ So, $R$ is reflexive relation Also, $(1,2) \in R$ but $(2,1) \notin R$ So, $R$ is not symmetric relation...
Read More →In the set Z of all integers,
Question: In the setZof all integers, which of the following relationRis not an equivalence relation? (a) $x R y$ : if $x \leq y$ (b) $x R y$ : if $x=y$ (c) $x R y:$ if $x-y$ is an even integer (d) $x R y:$ if $x \equiv y(\bmod 3)$ Solution: (a) $x R y:$ if $x \leq y$ Clearly, $R$ is not symmetric because $xy$ does not imply $yx$. Hence, (a) is not an equivalence relation....
Read More →The marks in science of 80 students of class X are given below:
Question: The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science. Solution: Here, the maximum frequency is 20 so the modal class is 5060. Therefore, $l=50$ $h=10$ $f=20$ $f_{1}=13$ $f_{2}=5$ Now, Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=50+\frac{20-13}{40-13-5} \times 10$ $=50+\frac{7}{22} \times 10$ $=50+\frac{70}{22}$ $=50+3.17$ Thus, the mode of the marks obtained by the students in science is 53.17....
Read More →Find the number of solutions of
Question: Find the number of solutions of $z^{2}+|z|^{2}=0$ Solution: Let $z=x+i y$. Then, $|z|=\sqrt{x^{2}+y^{2}}$ $\therefore z^{2}+|z|^{2}=0$ $\Rightarrow(x+i y)^{2}+\left(\sqrt{x^{2}+y^{2}}\right)^{2}=0$ $\Rightarrow x^{2}+i^{2} y^{2}+2 i x y+x^{2}+y^{2}=0$ $\Rightarrow x^{2}-y^{2}+2 i x y+x^{2}+y^{2}=0$ $\Rightarrow 2 x^{2}+2 i x y=0$ $\Rightarrow 2 x(x+i y)=0$ $\Rightarrow x=0$ or $x+i y=0$ $\Rightarrow x=0$ or $z=0$ For $x=0, z=0+i y$ Thus, there are infinitely many solutions of the form ...
Read More →S is a relation over the set R of all real numbers and it is given by
Question: Sis a relation over the setRof all real numbers and it is given by $(a, b) \in S \Leftrightarrow a b \geq 0$. Then, $S$ is (a) symmetric and transitive only(b) reflexive and symmetric only(c) antisymmetric relation(d) an equivalence relation Solution: (d) an equivalence relation Reflexivity: Let $a \in R$ Then $a a=a^{2}0$ $\Rightarrow(a, a) \in R \forall a \in R$ So,Sis reflexive onR. Symmetry: Let $(a, b) \in S$ Then, $(a, b) \in S$ $\Rightarrow a b \geq 0$ $\Rightarrow b a \geq 0$ $...
Read More →Find the zero of the polynomial:
Question: Find the zero of the polynomial: (i) $p(x)=x-5$ (ii) $q(x)=x+4$ (iii) $r(x)=2 x+5$ (iv) $f(x)=3 x+1$ (v) $g(x)=5-4 x$ (vi) $h(x)=6 x-2$ (vii) $p(x)=a x, a \neq 0$ (viii) $q(x)=4 x$ Solution: (i) $p(x)=0 \Rightarrow x-5=0$ $\Rightarrow x=5$ Hence, 5 is the zero of the polynomial $p(x)$. (ii) $q(x)=0 \Rightarrow x+4=0$ $\Rightarrow x=-4$ Hence, $-4$ is the zero of the polynomial $q(x)$. (iii) $r(x)=0 \Rightarrow 2 x+5=0$ $\Rightarrow t=\frac{-5}{2}$ Hence, $\frac{-5}{2}$ is the zero of t...
Read More →Solve the following
Question: If $z_{1}, z_{2}, z_{3}$ are complex numbers such that $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1$, then find the value of $\left|z_{1}+z_{2}+z_{3}\right|$. Solution: $\left|z_{1}+z_{2}+z_{3}\right|=\left|\frac{z_{1} \overline{z_{1}}}{\overline{z_{1}}}+\frac{z_{2} \overline{z_{2}}}{\overline{z_{2}}}+\frac{z_{3} \overline{z_{3}}}{\overline{z_{3}}}\right|$ $=\left|\frac{\left|z_{1}\right|^{2}}{\overline{z_{1}}}...
Read More →Compare the modal ages of two groups of students appearing for an entrance test:
Question: Compare the modal ages of two groups of students appearing for an entrance test: Solution: Forgroup A The maximum frequency is 78 so the modal class is 1820. Therefore, $l=18$ $h=2$ $f=78$ $f_{1}=50$ $f_{2}=46$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h$ $=18+\frac{78-50}{156-50-46} \times 2$ $=18+\frac{14}{15}$ $=18+0.93$ Mode $=18.93$ Forgroup B The maximum frequency 89 so modal class 1820. Therefore, $l=18$ $h=2$ $f=89$ $f_{1}=54$ $f_{2}=40$ $\Rightarrow$ Mode $...
Read More →What is the smallest positive integer n for which
Question: What is the smallest positive integer $n$ for which $(1+i)^{2 n}=(1-i)^{2 n} ?$ Solution: $(1+i)^{2 n}=(1-i)^{2 n}$ $\Rightarrow\left[(1+i)^{2}\right]^{n}=\left[(1-i)^{2}\right]^{n}$ $\Rightarrow\left(1^{2}+i^{2}+2 i\right)^{n}=\left(1^{2}+i^{2}-2 i\right)^{n}$ $\Rightarrow(1-1+2 i)^{n}=(1-1-2 i)^{n} \quad\left[\because i^{2}=-1\right]$ $\Rightarrow(2 i)^{n}=(-2 i)^{n}$ $\Rightarrow(2 i)^{n}=(-1)^{n}(2 i)^{n}$ $\Rightarrow(-1)^{n}=1$ $\Rightarrow n$ is a multiple of 2 Thus, the smalles...
Read More →The relation
Question: The relationR= {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is(a) symmetric only(b) reflexive only(c) an equivalence relation(d) transitive only Solution: (c) an equivalence relation $R=\{(a, b): a=b$ and $a, b \in A\}$ Reflexivity: Let $a \in A$ Here, $a=a$ $\Rightarrow(a, a) \in R$ for all $a \in A$ So, $R$ is reflexive on $A$. Symmetry: Let $a, b \in A$ such that $(a, b) \in R$. Then, $(a, b) \in R$ $\Rightarrow a=b$ $\Rightarrow b=a$ $\Rightarrow(b, a) \in R$ for all $a \in A$ So,...
Read More →Find the mode of the following distribution.
Question: Find the mode of the following distribution. Solution: (i) Here, maximum frequency is 28 so the modal class is 4050. Therefore, $l=40$ $h=10$ $f=28$ $f_{1}=12$ $f_{2}=20$ $\Rightarrow$ Mode $=l+\frac{f-f_{1}}{2 f+f_{1}-f_{2}} \times h$ $=40+\frac{28-12}{2 \times 28-12-20} \times 10$ $=40+\frac{16}{24} \times 10$ $=40+\frac{80}{12}$ $=40+6.67$ Mode $=46.67$ (ii) Here, maximum frequency is 75 so the modal class is 2025. Therefore, $l=20$, $h=5$ $f=75$ $f_{1}=45$ $f_{2}=35$ $\Rightarrow$ ...
Read More →Solve the equation
Question: Solve the equation $|z|=z+1+2 i$ Solution: Let $z=x+i y$. Then, $|z|=\sqrt{x^{2}+y^{2}}$ $\therefore|z|=z+1+2 i$ $\Rightarrow \sqrt{x^{2}+y^{2}}=(x+i y)+1+2 i$ $\Rightarrow \sqrt{x^{2}+y^{2}}=(x+1)+i(y+2)$ $\Rightarrow \sqrt{x^{2}+y^{2}}=(x+1)$ and $y+2=0$ $\Rightarrow x^{2}+y^{2}=(x+1)^{2}$ and $y=-2$ $\Rightarrow x^{2}+y^{2}=x^{2}+1+2 x$ and $y=-2$ $\Rightarrow y^{2}=2 x+1$ and $y=-2$ $\Rightarrow 4=2 x+1$ and $y=-2$ $\Rightarrow 2 x=3$ and $y=-2$ $\Rightarrow x=\frac{3}{2}$ and $y=-...
Read More →If
Question: If $|z+1|=z+2(1+i)$, find $z$ Solution: Let $z=x+i y$. Then, $z+1=(x+1)+i y$ $\Rightarrow|z+1|=\sqrt{(x+1)^{2}+y^{2}}$ $\therefore|z+1|=z+2(1+i)$ $\Rightarrow \sqrt{x^{2}+2 x+1+y^{2}}=(x+i y)+2+2 i$ $\Rightarrow \sqrt{x^{2}+2 x+1+y^{2}}=(x+2)+i(y+2)$ $\Rightarrow \sqrt{x^{2}+2 x+1+y^{2}}=(x+2)$ and $y+2=0$ $\Rightarrow x^{2}+2 x+1+y^{2}=(x+2)^{2}$ and $y=-2$ $\Rightarrow x^{2}+2 x+1+y^{2}=x^{2}+4 x+4$ and $y=-2$ $\Rightarrow y^{2}=2 x+3$ and $y=-2$ $\Rightarrow 4=2 x+3$ and $y=-2$ $\Ri...
Read More →Let A = {1, 2, 3}. Then,
Question: LetA= {1, 2, 3}. Then, the number of equivalence relations containing (1, 2) is(a) 1(b) 2(c) 3(d) 4 Solution: (b) 2 There are 2 equivalence relations containing {1, 2}.R= {(1, 2)}S= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}...
Read More →Let R be the relation on the set A = {1, 2, 3, 4} given by
Question: LetRbe the relation on the setA= {1, 2, 3, 4} given by R= {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,(a)Ris reflexive and symmetric but not transitive(b)Ris reflexive and transitive but not symmetric(c)Ris symmetric and transitive but not reflexive(d)Ris an equivalence relation Solution: (b) R is reflexive and transitive but not symmetric. Reflexivity : Clearly, $(a, a) \in R \forall a \in A$ So, $R$ is reflexive on $A$. Symmetry : Since $(1,2) \in R$, but $(2,1) \n...
Read More →Solve the following
Question: If $z_{1}$ is a complex number other than $-1$ such that $\left|z_{1}\right|=1$ and $z_{2}=\frac{z_{1}-1}{z_{1}+1}$, then show that the real parts of $z_{2}$ is zero. Solution: Let $z=x+i y$. Then, $z_{2}=\frac{z_{1}-1}{z_{1}+1}$ $=\frac{x+i y-1}{x+i y+1}$ $=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$ $=\frac{x^{2}+x-i x y-x-1+i y+i x y+i y-i^{2} y^{2}}{(x+1)^{2}-i^{2} y^{2}}$ $=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+1+2 x+y^{2}} \quad\left[\because i^{2}=-1\right]$ Now, ...
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