Verify that
Question: Verify that (i) 1 and 2 are the zeros of the polynomial $p(x)=x^{2}-3 x+2$. (ii) 2 and $-3$ are the zeros of the polynomial $q(x)=x^{2}+x-6$. (iii) 0 and 3 are the zeros of the polynomial $r(x)=x^{2}-3 x$. Solution: (i) $p(x)=x^{2}-3 x+2=(x-1)(x-2)$ $\Rightarrow p(1)=(1-1) \times(1-2)$ $=0 \times(-1)$ $=0$ Also, $p(2)=(2-1)(2-2)$ $=(-1) \times 0$ $=0$ Hence, 1 and 2 are the zeroes of the given polynomial. (ii) $p(x)=x^{2}+x-6$ $\Rightarrow p(2)=2^{2}+2-6$ $=4-4$ $=0$ Also, $p(-3)=(-3)^...
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Question: IfA= {1, 2, 3}, then a relationR= {(2, 3)} onAis(a) symmetric and transitive only(b) symmetric only(c) transitive only(d) none of these Solution: (c) transitive only The relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.R is transitive by default because there is only one element in it....
Read More →Solve the following
Question: If $\frac{z-1}{z+1}$ is purely imaginary number $(z \neq-1)$, find the value of $|z|$. Solution: Let $z=x+i y$. Then, $\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$ $=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$ $=\frac{x^{2}+x-i x y-x-1+i y+i x y+i y-i^{2} y^{2}}{(x+1)^{2}-i^{2} y^{2}}$ $=\frac{x^{2}+y^{2}-1+2 i y}{x^{2}+1+2 x+y^{2}} \quad\left[\because i^{2}=-1\right]$ If $\frac{z-1}{z+1}$ is purely imaginary number, then $\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0...
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Question: IfA= {a,b,c,d}, then a relationR= {(a,b), (b,a), (a,a)} onAis (a) symmetric and transitive only(b) reflexive and transitive only(c) symmetric only(d) transitive only Solution: (a) symmetric and transitive only Reflexivity: Since $(b, b) \notin R, R$ is not reflexive on $A$. Symmetry : Since $(a, b) \in R$ and $(b, a) \in R, R$ is symmetric on $A$. Transitivity : Since $(a, b) \in R, \quad(b, a) \in R$ and $(a, a) \in R, R$ is transitive on $A$....
Read More →Solve the system of equations Re
Question: Solve the system of equations $\operatorname{Re}\left(z^{2}\right)=0,|z|=2$. Solution: Let $z=x+i y$. Then. $z^{2}=(x+i y)^{2}$ $=x^{2}+i^{2} y^{2}+2 i x y$ $=x^{2}-y^{2}+2 i x y$ $\left[\because i^{2}=-1\right]$ and $|z|=\sqrt{x^{2}+y^{2}}$ According to the question, $\operatorname{Re}\left(z^{2}\right)=0$ and $|z|=2$ $\Rightarrow x^{2}-y^{2}=0$ and $\sqrt{x^{2}+y^{2}}=2$ $\Rightarrow x^{2}-y^{2}=0$ and $x^{2}+y^{2}=4$ On Adding both the equations, we get $2 x^{2}=4$ $\Rightarrow x^{2...
Read More →The shirt sizes worn by a group of 200 persons,
Question: The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows: Find the modal shirt size worn by the group. Solution: Here, shirt size 40 has the maximum number of persons.Hence, the mode shirt size is 40....
Read More →If R is a relation on the set
Question: IfRis a relation on the setA= {1, 2, 3} given byR= {(1, 1), (2, 2), (3, 3)}, thenRis(a) reflexive(b) symmetric(c) transitive(d) all the three options Solution: (d) all the three options $R=\{(a, b): a=b$ and $a, b \in A\}$ Reflexivity: Let $a \in A$. Then, $a=a$ $\Rightarrow(a, a) \in R$ 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 $a \in A$ ...
Read More →Solve the system of equations Re
Question: Solve the system of equations $\operatorname{Re}\left(z^{2}\right)=0,|z|=2$. Solution: Let $z=x+i y$. Then. $z^{2}=(x+i y)^{2}$ $=x^{2}+i^{2} y^{2}+2 i x y$ $=x^{2}-y^{2}+2 i x y$ $\left[\because i^{2}=-1\right]$ and $|z|=\sqrt{x^{2}+y^{2}}$ According to the question, $\operatorname{Re}\left(z^{2}\right)=0$ and $|z|=2$ $\Rightarrow x^{2}-y^{2}=0$ and $\sqrt{x^{2}+y^{2}}=2$ $\Rightarrow x^{2}-y^{2}=0$ and $x^{2}+y^{2}=4$ On Adding both the equations, we get $2 x^{2}=4$ $\Rightarrow x^{2...
Read More →Find the mode of the following data:
Question: Find the mode of the following data: (i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7,4(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15 Solution: (i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 The frequency table for the given data We observe that the value 5 has the maximum frequency. Hence, the mode of data is 5. (ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 The frequency table for the given data W...
Read More →Verify that:
Question: Verify that: (i) 4 is a zero of the polynomial $p(x)=x-4$. (ii) $-3$ is a zero of the polynomial $q(x)=x+3$. (iii) $\frac{2}{5}$ is a zero of the polynomial, $f(x)=2-5 x$. (iv) $\frac{-1}{2}$ is a zero of the polynomial $g(y)=2 y+1$. Solution: (i) $p(x)=x-4$ $\Rightarrow p(4)=4-4$ = 0Hence, 4 is the zero of the given polynomial. (ii) $p(x)=(-3)+3$ $\Rightarrow p(3)=0$ Hence, 3 is the zero of the given polynomial. (iii) $p(x)=2-5 x$ $\Rightarrow p\left(\frac{2}{5}\right)=2-5 \times\left...
Read More →If R is a relation on the set
Question: If $R$ is a relation on the set $A=\{1,2,3,4,5,6,7,8,9\}$ given by $x R y \Leftrightarrow y=3 x$, then $R=$ Solution: (d) none of theseThe relationRis defined as $R=\{(x, y): x, y \in A: y=3 x\}$ $\Rightarrow R=\{(1,3),(2,6),(3,9)\}$...
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Question: If $(1+i) z=(1-i) \bar{z}$, then show that $z=-i \bar{z}$. Solution: $(1+i) z=(1-i) \bar{z}$ $\Rightarrow \frac{z}{\bar{z}}=\frac{1-i}{1+i}$ $\Rightarrow \frac{z}{\bar{z}}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$ $\Rightarrow \frac{z}{\bar{z}}=\frac{1+i^{2}-2 i}{1-i^{2}}$ $\Rightarrow \frac{z}{\bar{z}}=\frac{1-1-2 i}{1+1}$ $\left[\because i^{2}=-1\right]$ $\Rightarrow \frac{z}{\bar{z}}=\frac{-2 i}{2}$ $\Rightarrow \frac{z}{\bar{z}}=-i$ $\Rightarrow z=-i \bar{z}$ Hence, $z=-i \bar{z}$....
Read More →The following table gives the distribution of the life time of 400 neon lamps:
Question: The following table gives the distribution of the life time of 400 neon lamps: Find the median life. Solution: We prepare the cumulative frequency table, as given below. We have, $N=400$ So, $\frac{N}{2}=200$ Now, the cumulative frequency just greater than 200 is 216 and the corresponding class is $3000-3500$. Therefore, $3000-3500$ is the median class. Here, $I=3000, f=86, F=130$ and $h=500$ We know that Median $=l+\left\{\frac{\frac{N}{2}-F}{f}\right\} \times h$ $=3000+\left\{\frac{2...
Read More →For a positive integer n, find the value of
Question: For a positive integer $n$, find the value of $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}$. Solution: $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}=(1-i)^{n}\left(1-\frac{i^{4}}{i}\right)^{n}$ $\left[\because i^{4}=1\right]$ $=(1-i)^{n}\left(1-i^{3}\right)^{n}$ $=(1-i)^{n}(1+i)^{n}$ $\left[\because i^{3}=-i\right]$ $=[(1-i)(1+i)]^{n}$ $=\left(1-i^{2}\right)^{n}$ $=2^{n} \quad\left[\because i^{2}=-1\right]$ Thus, the value of $(1-i)^{n}\left(1-\frac{1}{i}\right)^{n}$ is $2^{n}$....
Read More →If R is the largest equivalence relation on a set A and S is any relation on A, then
Question: IfRis the largest equivalence relation on a setAandSis any relation onA, then (a) $R \subset S$ (b) $S \subset R$ (c) $R=S$ (d) none of these Solution: (b) $S \subset R$ Since $R$ is the largest equivalence relation on set $A$, $R \subseteq A \times A$ Since $S$ is any relation on $A$, $S \subset A \times A$ So, SR...
Read More →The length of 40 leaves of a plant are measured correct to the nearest millimeter,
Question: The length of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table: Find the mean length of leaf. Solution: Calculation for mean. Mean length of the leaf $=\frac{1}{N} \sum f_{i} x_{i}=\frac{1}{40} \times 5880=147$ Calculation for median. The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have Now, we have $N=40$ So, $\frac{N}{2}=20$ Now, the ...
Read More →If p(x) = x3 + x2 – 9x – 9,
Question: If $p(x)=x^{3}+x^{2}-9 x-9$, find $p(0), p(3), p(-3)$ and $p(-1)$. What do you conclude about the zero of $p(x) ?$ Is 0 a zero of $p(x)$ ? Solution: $p(x)=x^{3}+x^{2}-9 x-9$ Putting $x=0$ in $(1)$, we get $p(0)=0^{3}+0^{2}-9 \times 0-9=0+0-0-9=-9 \neq 0$ Thus,x= 0 isnota zero ofp(x).Puttingx = 3 in (1), we get $p(3)=3^{3}+3^{2}-9 \times 3-9=27+9-27-9=0$ Thus,x= 3 is a zero ofp(x).Puttingx = 3 in (1), we get $p(-3)=(-3)^{3}+(-3)^{2}-9 \times(-3)-9=-27+9+27-9=0$ Thus,x= 3 is a zero ofp(x...
Read More →Evaluate the following:
Question: Evaluate the following: (i) $2 x^{3}+2 x^{2}-7 x+72$, when $x=\frac{3-5 i}{2}$ (ii) $x^{4}-4 x^{3}+4 x^{2}+8 x+44$, when $x=3+2 i$ (iii) $x^{4}+4 x^{3}+6 x^{2}+4 x+9$, when $x=-1+i \sqrt{2}$ (iv) $x^{6}+x^{4}+x^{2}+1$, when $x=\frac{1+i}{\sqrt{2}}$ (v) $2 x^{4}+5 x^{3}+7 x^{2}-x+41$, when $x=-2-\sqrt{3} i$ Solution: (i) $x=\frac{3-5 i}{2}$ $\Rightarrow x^{2}=\left(\frac{3-5 i}{2}\right)^{2}$ $=\frac{9+25 i^{2}-30 i}{4}$ $=\frac{-16-30 i}{4}$ $\Rightarrow x^{3}=\frac{-16-30 i}{4} \times...
Read More →Evaluate the following:
Question: Evaluate the following: (i) $2 x^{3}+2 x^{2}-7 x+72$, when $x=\frac{3-5 i}{2}$ (ii) $x^{4}-4 x^{3}+4 x^{2}+8 x+44$, when $x=3+2 i$ (iii) $x^{4}+4 x^{3}+6 x^{2}+4 x+9$, when $x=-1+i \sqrt{2}$ (iv) $x^{6}+x^{4}+x^{2}+1$, when $x=\frac{1+i}{\sqrt{2}}$ (v) $2 x^{4}+5 x^{3}+7 x^{2}-x+41$, when $x=-2-\sqrt{3} i$ Solution: (i) $x=\frac{3-5 i}{2}$ $\Rightarrow x^{2}=\left(\frac{3-5 i}{2}\right)^{2}$ $=\frac{9+25 i^{2}-30 i}{4}$ $=\frac{-16-30 i}{4}$ $\Rightarrow x^{3}=\frac{-16-30 i}{4} \times...
Read More →Let
Question: LetA= {1, 2, 3} andB= {(1, 2), (2, 3), (1, 3)} be a relation onA. Then,Ris(a) neither reflexive nor transitive(b) neither symmetric nor transitive(c) transitive(d) none of these Solution: (c) transitive Reflexivity : Since $(1,1) \notin B, B$ is not reflexive on $A$. Symmetry : Since $(1,2) \in B$ but $(2,1) \notin B, B$ is not symmetric on $A$. Transitivity : Since $(1,2) \in B,(2,3) \in B$ and $(1,3) \in B, B$ is transitive on $A$....
Read More →A life insurance agent found the following data for distribution of ages of 100 policy holders.
Question: A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years. Solution: We prepare the cumulative frequency, table as given below. Now, we have $N=100$ So, $\frac{N}{2}=50$ Now, the cumulative frequency just greater than 50 is 78 and the corresponding class is $35-40$. Therefore, $35-40$ is the median class. Here, $l=35, f=33, F=45$ and $h...
Read More →If p(x)=x3−3x2+2x,
Question: If $p(x)=x^{3}-3 x^{2}+2 x$, find $p(0), p(1), p(2)$. What do you conclude? Solution: $p(x)=x^{3}-3 x^{2}+2 x$ Putting $x=0$ in $(1)$, we get $p(0)=0^{3}-3 \times 0^{2}+2 \times 0=0$ Thus, $x=0$ is a zero of $p(x)$. Putting $x=1$ in (1), we get $p(1)=1^{3}-3 \times 1^{2}+2 \times 1=1-3+2=0$ Thus, $x=1$ is a zero of $p(x)$. Putting $x=2$ in (1), we get $p(2)=2^{3}-3 \times 2^{2}+2 \times 2=8-3 \times 4+4=8-12+4=0$ Thus, $x=2$ is a zero of $p(x)$....
Read More →Let
Question: LetR= {(a,a), (b,b), (c,c), (a,b)} be a relation on setA=a,b,c. Then,Ris(a) identify relation(b) reflexive(c) symmetric(d) antisymmetric Solution: (b) reflexive Reflexivity : Since $(a, a) \in R \forall a \in A, R$ is reflexive on $A$. Symmetry: Since $(a, b) \in R$ but $(b, a) \notin R, R$ is not symmetric on $A$. $\Rightarrow R$ is not antisymmetric on $A$. Also, $R$ is not an identity relation on $A$....
Read More →R is a relation from {11, 12, 13} to {8, 10, 12} defined by
Question: $R$ is a relation from $\{11,12,13\}$ to $\{8,10,12\}$ defined by $y=x-3$. Then, $R^{-1}$ is (a) {(8, 11), (10, 13)}(b) {(11, 8), (13, 10)}(c) {(10, 13), (8, 11)}(d) none of these Solution: (a) {(8, 11), (10, 13)}The relationRis defined by $R=\{(x, y): x \in\{11,12,13\}, y \in\{8,10,12\}: y=x-3\}$ $\Rightarrow R=\{(11,8),(13,10)\}$ So, $R^{-1}=\{(8,11),(10,13)\}$...
Read More →A survey regarding the height (in cm) of 51 girls
Question: A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained: Find the median height. Solution: We prepare the cumulative frequency, table as given below. Now, we have $N=51$ So, $\frac{N}{2}=25.5$ Now, the cumulative frequency just greater than $25.5$ is 40 and the corresponding class is $150-155$. Therefore, $150-155$ is the median class. $l=150, f=11, F=29$ and $h=5$ We know that Median $=l+\left\{\frac{\frac{N}{2}-F}{f}\...
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