Factorize:
Question: Factorize: $x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3}$ Solution: We have: $x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3}=\left(x^{3}-2 x^{2} y\right)+\left(3 x y^{2}-6 y^{3}\right)$ $=x^{2}(x-2 y)+3 y^{2}(x-2 y)$ $=(x-2 y)\left(x^{2}+3 y^{2}\right)$...
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Question: If $n \in \mathbb{N}$, then find the value of $i^{n}+i^{n+1}+i^{n+2}+i^{n+3}$. Solution: $i^{n}+i^{n+1}+i^{n+2}+i^{n+3}$ $=i^{n}+i^{n} \cdot i+i^{n} \cdot i^{2}+i^{n} \cdot i^{3}$ $=i^{n}+i^{n} \cdot i+i^{n} \cdot(-1)+i^{n} \cdot(-i)$ $=i^{n}+i^{n} \cdot i-i^{n}-i^{n} \cdot i$ $=0$ Thus, the value of $i^{n}+i^{n+1}+i^{n+2}+i^{n+3}$ is 0 ....
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Question: Factorize: $8-4 a-2 a^{3}+a^{4}$ Solution: We have: $8-4 a-2 a^{3}+a^{4}=(8-4 a)-\left(2 a^{3}-a^{4}\right)$ $=4(2-a)-a^{3}(2-a)$ $=(2-a)\left(4-a^{3}\right)$...
Read More →Write the conjugate of
Question: Write the conjugate of $\frac{2-i}{(1-2 i)^{2}}$. Solution: $\frac{2-i}{(1-2 i)^{2}}=\frac{2-i}{1+4 i^{2}-4 i}$ $=\frac{2-i}{1-4-4 i}$ $=\frac{2-i}{-3-4 i}$ $=\frac{-2+i}{3+4 i}$ $=\frac{i-2}{3+4 i} \times \frac{3-4 i}{3-4 i}$ $=\frac{3 i-4 i^{2}-6+8 i}{3^{2}-4^{2} i^{2}}$ $=\frac{11 i+4-6}{9+16}$ $=\frac{-2}{25}+\frac{11}{25} i$ $\therefore$ Conjugate of $\frac{2-i}{(1-2 i)^{2}}=\left(-\frac{2}{25}+\frac{11}{25} i\right)=-\frac{2}{25}-\frac{11}{25} i$ Hence, Conjugate of $\frac{2-i}{(...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x-\frac{1}{x}=3, x \neq 0$ Solution: We have been given $x-\frac{1}{x}=3$ $x^{2}-1=3 x$ $x^{2}-3 x-1=0$ $x^{2}-\left(\frac{3-\sqrt{13}}{2}\right) x-\left(\frac{3+\sqrt{13}}{2}\right) x+\left(\frac{3+\sqrt{13}}{2}\right)\left(\frac{3-\sqrt{13}}{2}\right)=0$ $x\left(x-\left(\frac{3-\sqrt{13}}{2}\right)\right)-\left(\frac{3+\sqrt{13}}{2}\right)\left(x-\left(\frac{3-\sqrt{13}}{2}\right)\right)=0$ $\left[x-\left(\frac{3-\sqrt{13}}{2...
Read More →Factorize:
Question: Factorize: $a^{3} b-a^{2} b+5 a b-5 b$ Solution: We have: $a^{3} b-a^{2} b+5 a b-5 b=b\left(a^{3}-a^{2}+5 a-5\right)$ $=b\left[\left(a^{3}-a^{2}\right)+(5 a-5)\right]$ $=b\left[a^{2}(a-1)+5(a-1)\right]$ $=b(a-1)\left(a^{2}+5\right)$...
Read More →For any two complex numbers
Question: For any two complex numbers $z_{1}$ and $z_{2}$ and any two real numbers $a, b$, find the value of $\left|a z_{1}-b z_{2}\right|^{2}+\left|a z_{2}+b z_{1}\right|^{2}$. Solution: $\left|a z_{1}-b z_{2}\right|^{2}+\left|a z_{2}+b z_{1}\right|^{2}=\left(a z_{1}-b z_{2}\right)\left(\overline{a z_{1}-b z_{2}}\right)+\left(a z_{2}+b z_{1}\right)\left(\overline{a z_{2}+b z_{1}}\right)$ $=\left(a z_{1}-b z_{2}\right)\left(a \overline{z_{1}}-b \overline{z_{2}}\right)+\left(a z_{2}+b z_{1}\right...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{1}{x}-\frac{1}{x-2}=3, x \neq 0,2$ Solution: We have been given $\frac{1}{x}-\frac{1}{x-2}=3$ $-2=3 x^{2}-6 x$ $3 x^{2}-6 x+2=0$ $3 x^{2}-(3+\sqrt{3}) x-(3-\sqrt{3}) x+3-\sqrt{3}+\sqrt{3}-1=0$ $x(3 x-3-\sqrt{3})+\left(\frac{-3+\sqrt{3}}{3}\right)(3 x-3-\sqrt{3})=0$ $\left(\frac{3 x-3+\sqrt{3}}{3}\right)(3 x-3-\sqrt{3})=0$ $(\sqrt{3} x-\sqrt{3}+1)(\sqrt{3} x-\sqrt{3}-1)=0$ Therefore, $\sqrt{3} x-\sqrt{3}+1=0$ $\sqrt{3} x=\...
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Question: If $|z+4| \leq 3$, then find the greatest and least values of $|z+1|$. Solution: $|z+1|=|z+4-3|$ $\leq|z+4|+|-3|$ $\leq 3+3$ $=6$ Also, $|z+1| \geq 0$ Thus, $0 \leq|z+1| \leq 6$. Hence, the greatest and least values of $|z+1|$ is 6 and 0 ....
Read More →Factorize:
Question: Factorize: $x^{2}+x y-2 x z-2 y z$ Solution: We have: $x^{2}+x y-2 x z-2 y z=\left(x^{2}+x y\right)-(2 x z+2 y z)$ $=x(x+y)-2 z(x+y)$ $=(x+y)(x-2 z)$...
Read More →Write the value of
Question: Write the value of $\arg (z)+\arg (\bar{z})$ Solution: Letzbe a complex number with argument. Then, $z=r e^{i \theta}$ $\Rightarrow \bar{z}=\overline{r e^{i \theta}}=r e^{-i \theta}$ $\Rightarrow$ argument of $\bar{z}$ is $-\theta$ Thus, $\arg (z)+\arg (\bar{z})=0$....
Read More →Write the value of
Question: Write the value of $\arg (z)+\arg (\bar{z})$ Solution: Letzbe a complex number with argument. Then, $z=r e^{i \theta}$ $\Rightarrow \bar{z}=\overline{r e^{i \theta}}=r e^{-i \theta}$ $\Rightarrow$ argument of $\bar{z}$ is $-\theta$ Thus, $\arg (z)+\arg (\bar{z})=0$....
Read More →Factorize:
Question: Factorize: $x^{3}+2 x^{2}+5 x+10$ Solution: We have: $x^{3}+2 x^{2}+5 x+10=\left(x^{3}+2 x^{2}\right)+(5 x+10)$ $=x^{2}(x+2)+5(x+2)$ $=(x+2)\left(x^{2}+5\right)$...
Read More →Write the sum of the series
Question: Write the sum of the series $i+i^{2}+i^{3}+\ldots$ upto 1000 terms. Solution: We know that, $i+i^{2}+i^{3}+i^{4}=i-1-i+1=0$ $\therefore i+i^{2}+i^{3}+\ldots+i^{1000}$ $=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{5}+i^{6}+i^{7}+i^{8}\right)+\ldots+\left(i^{997}+i^{998}+i^{999}+i^{1000}\right)$ $=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{4} i+i^{4} i^{2}+i^{4} i^{3}+i^{4} i^{4}\right)+\ldots+\left[\left(i^{4}\right)^{249} i+\left(i^{4}\right)^{249} i^{2}+\left(i^{4}\right)^{249} i^{3}+\le...
Read More →Write the sum of the series
Question: Write the sum of the series $i+i^{2}+i^{3}+\ldots$ upto 1000 terms. Solution: We know that, $i+i^{2}+i^{3}+i^{4}=i-1-i+1=0$ $\therefore i+i^{2}+i^{3}+\ldots+i^{1000}$ $=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{5}+i^{6}+i^{7}+i^{8}\right)+\ldots+\left(i^{997}+i^{998}+i^{999}+i^{1000}\right)$ $=\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i^{4} i+i^{4} i^{2}+i^{4} i^{3}+i^{4} i^{4}\right)+\ldots+\left[\left(i^{4}\right)^{249} i+\left(i^{4}\right)^{249} i^{2}+\left(i^{4}\right)^{249} i^{3}+\le...
Read More →Factorize:
Question: Factorize: $x(x+y)^{3}-3 x^{2} y(x+y)$ Solution: We have: $x(x+y)^{3}-3 x^{2} y(x+y)=x(x+y)\left[(x+y)^{2}-3 x y\right]$ $=x(x+y)\left[x^{2}+y^{2}+2 x y-3 x y\right]$ $=x(x+y)\left(x^{2}+y^{2}-x y\right)$...
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Question: Write the value of $\sqrt{-25} \times \sqrt{-9}$. Solution: $\sqrt{-25} \times \sqrt{-9}=5 \sqrt{-1} \times 3 \sqrt{-1}$ $=5 i \times 3 i$ $=15 i^{2}$ $=-15$ Hence, $\sqrt{-25} \times \sqrt{-9}=-15$....
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Question: If $\frac{\left(a^{2}+1\right)^{2}}{2 a-i}=x+i y$, find the value of $x^{2}+y^{2}$ Solution: $\frac{\left(a^{2}+1\right)^{2}}{2 a-i}=x+i y \quad \ldots(1)$ $\Rightarrow\left[\frac{\left(a^{2}+1\right)^{2}}{2 a-i}\right]=\overline{x+i y}$ $\Rightarrow \frac{\left(a^{2}+1\right)^{2}}{2 a+i}=x-i y \quad \ldots(2)$ On multiplying $(1)$ and $(2)$, we get $\frac{\left(a^{2}+1\right)^{2}}{2 a-i} \times \frac{\left(a^{2}+1\right)^{2}}{2 a+i}=(x+i y)(x-i y)$ $\Rightarrow \frac{\left(a^{2}+1\rig...
Read More →Let f = {(1, −1), (4, −2), (9, −3), (16, 4)}
Question: Letf= {(1, 1), (4, 2), (9, 3), (16, 4)} andg= {(1, 2), (2, 4), (3, 6), (4, 8)}. Show thatgofis defined whilefogis not defined. Also, findgof. Solution: f= {(1, 1), (4, 2), (9, 3), (16, 4)} andg= {(1, 2), (2, 4), (3, 6), (4, 8)} $f:\{1,4,9,16\} \rightarrow\{-1,-2,-3,4\}$ and $g:\{-1,-2,-3,4\} \rightarrow\{-2,-4,-6,8\}$ Co-domain off= domain ofg So, gof exists and gof: $\{1,4,9,16\} \rightarrow\{-2,-4,-6,8\}$ $(g o f)(1)=g(f(1))=g(-1)=-2$ $(g o f)(4)=g(f(4))=g(-2)=-4$ $(g o f)(9)=g(f(9))...
Read More →Factorize:
Question: Factorize: $8(3 a-2 b)^{2}-10(3 a-2 b)$ Solution: We have: $8(3 a-2 b)^{2}-10(3 a-2 b)=2(3 a-2 b)[4(3 a-2 b)-5]$ $=2(3 a-2 b)(12 a-8 b-5)$...
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Question: If $|z-5 i|=|z+5 i|$, then find the locus of $z$. Solution: $|z-5 i|=|z+5 i|$ $\Rightarrow|z-5 i|^{2}=|z+5 i|^{2}$ $\Rightarrow(z-5 i)(\overline{z-5 i})=(z+5 i)(\overline{z+5 i}) \quad\left[\because z \bar{z}=|z|^{2}\right]$ $\Rightarrow(z-5 i)(\bar{z}+5 i)=(z+5 i)(\bar{z}-5 i)$ $\Rightarrow z \bar{z}+5 z i-5 \bar{z} i-25 i^{2}=z \bar{z}-5 z i+5 \bar{z} i-25 i^{2}$ $\Rightarrow 5 z i+5 z i=5 \bar{z} i+5 \bar{z} i$ $\Rightarrow 10 z i=10 \bar{z} i$ $\Rightarrow z=\bar{z}$ $\Rightarrow z...
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Question: Factorize: $4(a+b)-6(a+b)^{2}$ Solution: We have: $4(a+b)-6(a+b)^{2}=2(a+b)[2-3(a+b)]$ $=2(a+b)(2-3 a-3 b)$...
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Question: Find $z$, if $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$. Solution: We know that, $z=|z|\{\cos [\arg (z)]+i \sin [\arg (z)]\}$ $\Rightarrow z=4\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)$ $=4\left(-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$ $=4\left(-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)$ $=-2 \sqrt{3}+2 i$ Thus, $z=-2 \sqrt{3}+2 i$....
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Question: Find $z$, if $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$. Solution: We know that, $z=|z|\{\cos [\arg (z)]+i \sin [\arg (z)]\}$ $\Rightarrow z=4\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)$ $=4\left(-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$ $=4\left(-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)$ $=-2 \sqrt{3}+2 i$ Thus, $z=-2 \sqrt{3}+2 i$....
Read More →Factorize:
Question: Factorize: $x(a-5)+y(5-a)$ Solution: We have: $x(a-5)+y(5-a)=x(a-5)-y(a-5)$ $=(a-5)(x-y)$...
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