Let R be a relation on N defined by x + 2y = 8. The domain of R is
Question: LetRbe a relation onNdefined byx+ 2y= 8. The domain ofRis (a) {2, 4, 8}(b) {2, 4, 6, 8}(c) {2, 4, 6}(d) {1, 2, 3, 4} Solution: (c) {2,4,6}The relationRis defined as $R=\{(x, y): x, y \in N$ and $x+2 y=8\}$ $\Rightarrow R=\left\{(x, y): x, y \in N\right.$ and $\left.y=\frac{(8-x)}{2}\right\}$ Domain of $R$ is all values of $x \in N$ satisfying the relation $R$. Also, there are only three values of $x$ that result in $y$, which is a natural number. These are $\{2,6,4\}$....
Read More →If f(t) = 4t2 − 3t + 6, find
Question: If $f(t)=4 t^{2}-3 t+6$, find (i) $f(0)$, (ii) $f(4)$, (iii) $f(-5)$. Solution: (i) $f(t)=4 t^{2}-3 t+6$ $\Rightarrow f(0)=\left(4 \times 0^{2}-3 \times 0+6\right)$ $=(0-0+6)$ $=6$ (ii) $f(t)=4 t^{2}-3 t+6$ $\Rightarrow f(4)=\left(4 \times 4^{2}-3 \times 4+6\right)$ $=(64-12+6)$ $=58$ (iii) $f(t)=4 t^{2}-3 t+6$ $\Rightarrow f(-5)=\left[4 \times(-5)^{2}-3 \times(-5)+6\right]$ $=(100+15+6)$ $=121$...
Read More →Compute the median for each of the following data:
Question: Compute the median for each of the following data: (i) Solution: (i)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 65 and the corresponding class is $70-90$. Therefore, $70-90$ is the median class. Here, $l=70, f=22, F=43$ and $h=20$ We know that Median $=l+\left\{\frac{\frac{N}{2}-F}{f}\right\} \times h$ $=70+\left\{\frac{50-43}{22}\right\} \times 20$ $=70+\frac{7 \times 20}{22}...
Read More →A relation ϕ from C to R is defined by
Question: A relation $\phi$ from $C$ to $R$ is defined by $x \phi y \Leftrightarrow|x|=y$. Which one is correct? (a) $(2+3 i) \phi 13$ (b) $3 \phi(-3)$ (c) $(1+i) \phi 2$ (d) $i \phi 1$ Solution: (d) $i \phi 1$ $\because|2+3 i|=\sqrt{13} \neq 13$ $|3| \neq-3$ $|1+i|=\sqrt{2} \neq 2$ and $|\mathrm{i}|=1$ So, $(i, 1) \in \phi$...
Read More →If a=cosθ+isinθ, find the value of
Question: If $a=\cos \theta+i \sin \theta$, find the value of $\frac{1+a}{1-a}$ Solution: $\frac{1+a}{1-a}=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}$ $=\frac{(1+\cos \theta)+i \sin \theta}{(1-\cos \theta)-i \sin \theta} \times \frac{(1-\cos \theta)+i \sin \theta}{(1-\cos \theta)+i \sin \theta}$ $=\frac{1-\cos \theta+i \sin \theta+\cos \theta-\cos ^{2} \theta+i \cos \theta \sin \theta+i \sin \theta-i \sin \theta \cos \theta+i^{2} \sin ^{2} \theta}{(1-\cos \theta)^{2}-i^{2} \...
Read More →If p(y) = 4 + 3y − y2 + 5y3,
Question: If $p(y)=4+3 y-y^{2}+5 y^{3}$, find (i) $p(0)$, (ii) $p(2)$, (iii) $p(-1)$ Solution: (i) $p(y)=4+3 y-y^{2}+5 y^{3}$ $\Rightarrow p(0)=\left(4+3 \times 0-0^{2}+5 \times 0^{3}\right)$ $=(4+0-0+0)$ $=4$ (ii) $p(y)=4+3 y-y^{2}+5 y^{3}$ $\Rightarrow p(2)=\left(4+3 \times 2-2^{2}+5 \times 2^{3}\right)$ $=(4+6-4+40)$ $=46$ (iii) $p(y)=4+3 y-y^{2}+5 y^{3}$ $\Rightarrow p(-1)=\left[4+3 \times(-1)-(-1)^{2}+5 \times(-1)^{3}\right]$ $=(4-3-1-5)$ $=-5$...
Read More →If
Question: If $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$, find $(a, b)$ Solution: $\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$ $=\frac{(1-i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+i^{2}-2 i}{1+1}$ $\left[\because i^{2}=-1\right]$ $=\frac{1-1-2 i}{2}$ $=\frac{-2 i}{2}$ $=-i$ ....(1) It is given that, $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$ $\Rightarrow(-i)^{100}=a+i b \quad[$ From $(1)]$ $\Rightarrow i^{4 \times 25}=a+i b$ $\Rightarrow 1+0 i=a+i b$ $\left[\because i^{4}=1\right]$ $\Righta...
Read More →If
Question: If $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$, find $(a, b)$ Solution: $\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$ $=\frac{(1-i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+i^{2}-2 i}{1+1}$ $\left[\because i^{2}=-1\right]$ $=\frac{1-1-2 i}{2}$ $=\frac{-2 i}{2}$ $=-i$ ....(1) It is given that, $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$ $\Rightarrow(-i)^{100}=a+i b \quad[$ From $(1)]$ $\Rightarrow i^{4 \times 25}=a+i b$ $\Rightarrow 1+0 i=a+i b$ $\left[\because i^{4}=1\right]$ $\Righta...
Read More →A relation R is defined from
Question: A relation $R$ is defined from $\{2,3,4,5\}$ to $\{3,6,7,10\}$ by : $x R y \Leftrightarrow x$ is relatively prime to $y$. Then, domain of $R$ is (a) {2, 3, 5}(b) {3, 5}(c) {2, 3, 4}(d) {2, 3, 4, 5} Solution: (d) {2, 3, 4, 5} The relationRis defined as $R=\{(x, y): x \in\{2,3,4,5\}, y \in\{3,6,7,10\}: x$ is relatively prime to $y\}$ $\Rightarrow R=\{(2,3),(2,7),(3,7),(3,10),(4,7),(5,3),(5,7)\}$ Hence, the domain ofRincludes all the values ofx, i.e. {2, 3, 4, 5}....
Read More →If p(x) = 5 − 4x + 2x2,
Question: If $p(x)=5-4 x+2 x^{2}$, find (i) $p(0)$, (ii) $p(3)$, (iii) $p(-2)$ Solution: (i) $p(x)=5-4 x+2 x^{2}$ $\Rightarrow p(0)=\left(5-4 \times 0+2 \times 0^{2}\right)$ $=(5-0+0)$ $=5$ (ii) $p(x)=5-4 x+2 x^{2}$ $\Rightarrow p(3)=\left(5-4 \times 3+2 \times 3^{2}\right)$ $=(5-12+18)$ $=11$ (iii) $p(x)=5-4 x+2 x^{2}$ $\Rightarrow p(-2)=\left[5-4 \times(-2)+2 \times(-2)^{2}\right]$ $=(5+8+8)$ $=21$...
Read More →If
Question: If $\frac{(1+i)^{2}}{2-i}=x+i y$, find $x+y$ Solution: $\frac{(1+i)^{2}}{2-i}=\frac{1^{2}+i^{2}+2 i}{2-i}$ $=\frac{1-1+2 i}{2-i} \quad\left[\because i^{2}=-1\right]$ $=\frac{2 i}{2-1} \times \frac{2+i}{2+i}$ $=\frac{2 i(2+i)}{2^{2}-i^{2}}$ $=\frac{4 i+2 i^{2}}{4+1} \quad \because i^{2}=-1 \mid$ $=\frac{4 i-2}{5}$ $=\frac{-2}{5}+\frac{4}{5} i$ ....(1) It is given that, $\frac{(1+i)^{2}}{2-i}=x+i y$ $\Rightarrow-\frac{2}{5}+\frac{4}{5} i=x+i y$ $[\operatorname{From}(1)]$ $\Rightarrow x=-...
Read More →If the median of the following data is 32.5,
Question: If the median of the following data is 32.5, find the missing frequencies. Solution: Given: Median = 32.5 We prepare the cumulative frequency table, as given below. Now, we have $N=40$ $31+f_{1}+f_{2}=40$ $f_{2}=9-f_{1}$.....(1) Also, $\frac{N}{2}=20$ Since median $=32.5$ so the median class is $30-40$. Here, $l=30, f=12, F=14+f_{1}$ and $h=10$ We know that Median $=I+\left\{\frac{\frac{N}{2}-F}{f}\right\} \times h$ $32.5=30+\left\{\frac{20-\left(14+f_{1}\right)}{12}\right\} \times 10$...
Read More →If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from
Question: IfA= {1, 2, 3},B= {1, 4, 6, 9} andRis a relation fromAtoBdefined by 'xis greater thany'. The range ofRis (a) {1, 4, 6, 9}(b) {4, 6, 9}(c) {1}(d) none of these Solution: (c) {1} Here, $R=\{(x, y): x \in A$ and $y \in B: xy\}$ $\Rightarrow R=\{(2,1),(3,1)\}$ Thus,Range ofR= {1}...
Read More →If
Question: If $\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$, find $(x, y)$ Solution: $\left(\frac{1+i}{1-i}\right)=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$ $=\frac{(1+i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+i^{2}+2 i}{1+1} \quad\left[\because i^{2}=-1\right]$ $=\frac{1-1+2 i}{2}$ $=\frac{2 i}{2}$ $=i \quad \ldots(1)$ Also, $\left(\frac{1-i}{1+i}\right)=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$ $=\frac{(1-i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+i^{2}-2 i}{1+1} \quad\left[\because ...
Read More →If
Question: If $\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$, find $(x, y)$ Solution: $\left(\frac{1+i}{1-i}\right)=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$ $=\frac{(1+i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+i^{2}+2 i}{1+1} \quad\left[\because i^{2}=-1\right]$ $=\frac{1-1+2 i}{2}$ $=\frac{2 i}{2}$ $=i \quad \ldots(1)$ Also, $\left(\frac{1-i}{1+i}\right)=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$ $=\frac{(1-i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+i^{2}-2 i}{1+1} \quad\left[\because ...
Read More →Rewrite each of the following polynomials in standard form.
Question: (i) $x-2 x^{2}+8+5 x^{3}$ (ii) $\frac{2}{3}+4 y^{2}-3 y+2 y^{3}$ (iii) $6 x^{3}+2 x-x^{5}-3 x^{2}$ (iv) $2+t-3 t^{3}+t^{4}-t^{2}$ Solution: A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial. (i) $8+x-2 x^{2}+5 x^{3}$ is a polynomial in standard form as the powers of $x$ are in ascending order. (ii) $\frac{2}{3}-3 y+4 y^{2}+2 y^{3}$ is a polynomial in standard form as the powers of $y$ are in ascending order. (iii) $2...
Read More →The relation 'R' in N × N such that
Question: The relation 'R' inNNsuch that $(a, b) R(c, d) \Leftrightarrow a+d=b+c$ is (a) reflexive but not symmetric(b) reflexive and transitive but not symmetric(c) an equivalence relation(d) none of the these Solution: (c) an equivalence relationWe observe the following properties of relationR. Reflexivity: Let $(a, b) \in N \times N$ $\Rightarrow a, b \in N$ $\Rightarrow a+b=b+a$ $\Rightarrow(a, b) \in R$ So, $R$ is reflexive on $N \times N$. Symmetry: Let $(a, b),(c, d) \in \mathrm{N} \times...
Read More →The median of the following data is 525.
Question: The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observation in the data:\/spanbr data-mce-bogus="1"/ppimg src="https://www.esaral.com/qdb/uploads/2021/12/31/image52898.png" alt="" Solution: Given: Median = 525 We prepare the cumulative frequency table, as given below. Now, we have $N=100$ $76+f_{1}+f_{2}=100$ $f_{2}=24-f_{1}$....(1) So, $\frac{N}{2}=50$ Since median $=525$ so the median class is $500-600$. Here, $l=500, f=20, F=36+...
Read More →Find the smallest positive integer value of m for which
Question: Find the smallest positive integer value of $m$ for which $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ is a real number. Solution: $\frac{(1+i)^{m}}{(1-i)^{m-2}}$ $=\frac{(1+i)^{m}}{(1-i)^{m}} \times(1-i)^{2}$ $=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m} \times\left(1+i^{2}-2 i\right)$ $=\left(\frac{1+i^{2}+2 i}{1-i^{2}}\right)^{m} \times(1-1-2 i)$ $=\left(\frac{1-1+2 i}{1+1}\right)^{m} \times(-2 i)$ $=-2 i\left(i^{m}\right)$ $=-2(i)^{m+1}$ For this to be real, the smallest positive va...
Read More →Give an example of a monomial of degree 5.
Question: (i) Give an example of a monomial of degree 5.(ii) Give an example of a binomial of degree 8.(iii) Give an example of a trinomial of degree 4.(iv) Give an example of a monomial of degree 0. Solution: (i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power ofxin the monomial should be 5. An example of a monomial of degree 5 is $2 x^{5}$. (ii) A polynomial having two terms is called a binomial. Since the degree of required b...
Read More →Find the real values of θ for which the complex number
Question: Find the real values of $\theta$ for which the complex number $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely real. Solution: $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ $=\frac{1+i \cos \theta}{1-2 i \cos \theta} \times \frac{1+2 i \cos \theta}{1+2 i \cos \theta}$ $=\frac{1+2 i \cos \theta+i \cos \theta-2 \cos \theta}{1+4 \cos ^{2} \theta}$ $=\frac{1-2 \cos \theta+i 3 \cos \theta}{1+4 \cos ^{2} \theta}$ For it to be purely real, the imaginary part must be zero. $3 \cos \theta=0$...
Read More →Let
Question: LetA= {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is (a) 1(b) 2(c) 3(d) 4 Solution: (a) 1 The required relation is R.R = {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)}Hence, there is only 1 such relation that is reflexive and symmetric, but not transitive....
Read More →If the median of the following frequency distribution is 28.5 find the missing frequencies:
Question: If the median of the following frequency distribution is 28.5 find the missing frequencies: Solution: Given: Median = 28.5 We prepare the cumulative frequency table, as given below. Now, we have $N=60$ $45+f_{1}+f_{2}=60$ $f_{2}=15-f_{1}$.....(1) $\mathrm{Also}, \frac{N}{2}=30$ Since the median $=28.5$ so the median class is $20-30$. Here, $l=20, f=20, F=5+f_{1}$ and $h=10$ We know that Median $=l+\left\{\frac{\frac{N}{2}-F}{f}\right\} \times h$ $28.5=20+\left\{\frac{30-\left(5+f_{1}\r...
Read More →Find the least positive integral value of n for which
Question: Find the least positive integral value of $n$ for which $\left(\frac{1+i}{1-i}\right)^{n}$ is real. Solution: $\left(\frac{1+i}{1-i}\right)^{n}$ $=\left[\frac{1+i}{1-i} \times\left(\frac{1+i}{1+i}\right)\right]^{n}$ $=\left(\frac{1+i^{2}+2 i}{1-i^{2}}\right)^{n}$ $=\left(\frac{1-1+2 i}{1+1}\right)^{n}$ $=\left(\frac{2 i}{2}\right)^{n}$ $=i^{n}$ For $i^{n}$ to be real, the least positive value of $n$ will be 2 . As $i^{2}=-1$...
Read More →Determine the degree of each of the following polynomials.
Question: Determine the degree of each of the following polynomials. (i) $\frac{4 x-5 x^{2}+6 x^{3}}{2 x}$ (ii) $y^{2}\left(y-y^{3}\right)$ (iii) $(3 x-2)\left(2 x^{3}+3 x^{2}\right)$ (iv) $-\frac{1}{2} x+3$ (v) $-8$ (vi) $x^{-2}\left(x^{4}+x^{2}\right)$ Solution: (i) $\frac{4 x-5 x^{2}+6 x^{3}}{2 x}=\frac{4 x}{2 x}-\frac{5 x^{2}}{2 x}+\frac{6 x^{3}}{2 x}=2-\frac{5}{2} x+3 x^{2}$ Here, the highest power of $x$ is 2 . So, the degree of the polynomial is 2 . (ii) $y^{2}\left(y-y^{3}\right)=y^{3}-y...
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