Write the values of the square root of i.
Question: Write the values of the square root ofi.\ Solution: Let the square root of $i$ be $x+i y$. $\Rightarrow \sqrt{i}=x+i y$ $\Rightarrow i=x^{2}+y^{2} i^{2}+2 i x y$ $\Rightarrow i=x^{2}-y^{2}+2 i x y$ (Squaring both the sides) Comparing both the sides: $x^{2}-y^{2}=0$ ...(i) and $2 x y=1$ ...(ii) By equation (ii), we find that $x$ and $y$ are of the same sign.From equation (i), $x=\pm y$ $\therefore x y=\frac{1}{2}, x^{2}=\frac{1}{2}$ $x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{2}}$ $...
Read More →Write the values of the square root of i.
Question: Write the values of the square root ofi.\ Solution: Let the square root of $i$ be $x+i y$. $\Rightarrow \sqrt{i}=x+i y$ $\Rightarrow i=x^{2}+y^{2} i^{2}+2 i x y$ $\Rightarrow i=x^{2}-y^{2}+2 i x y$ (Squaring both the sides) Comparing both the sides: $x^{2}-y^{2}=0$ ...(i) and $2 x y=1$ ...(ii) By equation (ii), we find that $x$ and $y$ are of the same sign.From equation (i), $x=\pm y$ $\therefore x y=\frac{1}{2}, x^{2}=\frac{1}{2}$ $x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{2}}$ $...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x^{2}-x-a(a+1)=0$ Solution: We have been given $x^{2}-x-a(a+1)=0$ $x^{2}+a x-(a+1) x-a(a+1)=0$ $x(x+a)-(a+1)(x+a)=0$ $(x-(a+1))(x+a)=0$ Therefore, $x-(a+1)=0$ $x=(a+1)$ or, $x+a=0$ $x=-a$ Hence, $x=a+1$ or $x=-a$....
Read More →For what value of k is the polynomial p(x)
Question: For what value of $k$ is the polynomial $p(x)=2 x^{3}-k x+3 x+10$ exactly divisible by $(x+2) ?$ (a) $-\frac{1}{3}$ (b) $\frac{1}{3}$ (c) 3 (d) $-3$ Solution: (d) 3 Let: $p(x)=2 x^{3}-k x^{2}+3 x+10$ Now, $x+2=0 \Rightarrow x=-2$ $p(x)$ is completely divisible by $(x+2)$. $\therefore p(-2)=0$ $\Rightarrow 2 \times(-2)^{3}-k \times(-2)^{2}+3 \times(-2)+10=0$ $\Rightarrow-16-4 k-6+10=0$ $\Rightarrow-12-4 k=0$ $\Rightarrow 4 k=-12$ $\Rightarrow k=\frac{-12}{4}$ $\Rightarrow k=-3$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $a\left(x^{2}+1\right)-x\left(a^{2}+1\right)=0$ Solution: We have been given $a\left(x^{2}+1\right)-x\left(a^{2}+1\right)=0$ $a x^{2}-\left(a^{2}+1\right) x+a=0$ $a x^{2}-a^{2} x-x+a=0$ $a x(x-a)-1(x-a)=0$ $(a x-1)(x-a)=0$ Therefore, $a x-1=0$ $a x=1$ $x=\frac{1}{a}$ or, $x-a=0$ $x=a$ Hence, $x=\frac{1}{a}$ or $x=a$....
Read More →Solve the following
Question: If $\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg \left(\frac{z_{1}}{z_{2}}\right)=\pi$, then $z_{1}+z_{2}=$ ______________________ Solution: Let |z1| = |z2| =r Let argz1=1and argz2=2 $\Rightarrow z_{1}=r e^{i \theta_{1}}$ and $z_{2}=r e^{i \theta_{2}}$ Since $\arg \left(\frac{z_{1}}{z_{2}}\right)=\pi$ i. e $\arg z_{1}-\arg z_{2}=\pi$ i. $e \theta_{1}-\theta_{2}=\pi$ i. e $\theta_{1}=\theta_{2}+\pi$ $e^{i \theta_{1}}=e^{i\left(\theta_{2}+\pi\right)}$ $=e^{i \pi} e^{i \theta_{2}}$ $=...
Read More →Let $R=\left\{\left(a, a^{3}\right): a\right.$ is a prime number less than 5$\}$ be a relation.
[question] Question. Let $R=\left\{\left(a, a^{3}\right): a\right.$ is a prime number less than 5$\}$ be a relation. Find the range of $R$. [CBSE 2014] [/question] [solution] Solution: We have, $R=\left\{\left(a, a^{3}\right): a\right.$ is a prime number less than 5$\}$ Or, $R=\{(2,8),(3,27)\}$ So, the range of $R$ is $\{8,27\}$. [/solution]...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $(a+b)^{2} x^{2}-4 a b x-(a-b)^{2}=0$ Solution: We have been given $(a+b)^{2} x^{2}-4 a b x-(a-b)^{2}=0$ $(a+b)^{2} x^{2}-(a+b)^{2} x+(a-b)^{2} x-(a-b)^{2}=0$ $(a+b)^{2} x(x-1)+(a-b)^{2}(x-1)=0$ $\left((a+b)^{2} x+(a-b)^{2}\right)(x-1)=0$ Therefore, $(a+b)^{2} x+(a-b)^{2}=0$ $(a+b)^{2} x=-(a-b)^{2}$ $x=-\left(\frac{a-b}{a+b}\right)^{2}$ or, $x-1=0$ $x=1$ Hence, $x=-\left(\frac{a-b}{a+b}\right)^{2}$ or $x=1$....
Read More →Solve the following
Question: Letz1andz2be two complex numbers such that |z1+z2| = |z1| + |z2|, then arg (z1) arg (z2) = ____________. Solution: Given for complex number $z_{1}$ and $z_{2}$ $\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$ i. e $\left|z_{1}+z_{2}\right|^{2}=\left(\left|z_{1}\right|+\left|z_{2}\right|\right)^{2}$ i. e $\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+2\left|z_{1}\right|\left|z_{2}\right| \cos \theta$ $=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+2\left|z_{1}\right|\lef...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x^{2}+2 a b=(2 a+b) x$ Solution: We have been given $x^{2}+2 a b=(2 a+b) x$ $x^{2}-(2 a+b) x+2 a b=0$ $x^{2}-2 a x-b x+2 a b=0$ $x(x-2 a)-b(x-2 a)=0$ $(x-2 a)(x-b)=0$ Therefore, $x-2 a=0$ $x=2 a$ or, $x-b=0$ $x=b$ Hence, $x=2 a$ or $x=b$....
Read More →For any two complex numbers
Question: For any two complex numbers $z_{1}, z_{2}$ and any real numbers $a, b,\left|a z_{1}-b z_{2}\right|^{2}+\left|b z_{1}+a z_{2}\right|^{2}=$ _________________ Solution: For complexz1andz2and real numbersaandb $\left|a z_{1}-b z_{2}\right|^{2}=\left(a z_{1}-b z_{2}\right)\left(\overline{a z_{1}-b z_{2}}\right)$ $=\left(a z_{1}-b z_{2}\right)\left(a \bar{z}_{1}-b \bar{z}_{2}\right)$ $=a^{2} z_{1} \bar{z}_{1}-a b z_{1} \bar{z}_{2}-a b z_{2} \bar{z}_{1}+b^{2} z_{2} \bar{z}_{2}$ $\left|a z_{1}...
Read More →For any two complex numbers
Question: For any two complex numbers $z_{1}, z_{2}$ and any real numbers $a, b,\left|a z_{1}-b z_{2}\right|^{2}+\left|b z_{1}+a z_{2}\right|^{2}=$ _________________ Solution: For complexz1andz2and real numbersaandb $\left|a z_{1}-b z_{2}\right|^{2}=\left(a z_{1}-b z_{2}\right)\left(\overline{a z_{1}-b z_{2}}\right)$ $=\left(a z_{1}-b z_{2}\right)\left(a \bar{z}_{1}-b \bar{z}_{2}\right)$ $=a^{2} z_{1} \bar{z}_{1}-a b z_{1} \bar{z}_{2}-a b z_{2} \bar{z}_{1}+b^{2} z_{2} \bar{z}_{2}$ $\left|a z_{1}...
Read More →Find the value
Question: If $\left(x^{100}+2 x^{99}+k\right)$ is divisible by $(x+1)$, then the value of $k$ is (a) 1(b) 2(c) 2(d) 3 Solution: (a) 1 Let: $p(x)=x^{100}+2 x^{99}+k$ Now, $x+1=0 \Rightarrow x=-1$ $(x+1)$ is divisible by $p(x)$ $\therefore p(-1)=0$ $\Rightarrow(-1)^{100}+2 \times(-1)^{99}+k=0$ $\Rightarrow 1-2+k=0$ $\Rightarrow-1+k=0$ $\Rightarrow k=1$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{a}{x-a}+\frac{b}{x-b}=\frac{2 c}{x-c}$ Solution: We have been given, $\frac{a}{(x-a)}+\frac{b}{(x-b)}=\frac{2 c}{(x-c)}$ $a(x-b)(x-c)+b(x-a)(x-c)=2 c(x-a)(x-b)$ $a\left(x^{2}-(b+c) x+b c\right)+b\left(x^{2}-(a+c) x+a c\right)=2 c\left(x^{2}-(a+b) x+a b\right)$ $(a+b-2 c) x^{2}-(2 a b-a c-b c) x=0$ $x[(a+b-2 c) x-(2 a b-a c-b c)]=0$ Therefore, $x=0$ or, $(a+b-2 c) x-(2 a b-a c-b c)=0$ $(a+b-2 c) x=(2 a b-a c-b c)$ $x=\frac...
Read More →If 0 < arg (z) < π,
Question: If 0 arg (z) , then arg (z) arg (z) = ____________. Solution: For 0 argz Letz = r(cos,isin) i.e argz = Then z = r(cos+isin) $=-r(+\cos \theta+i(+\sin \theta))$ $=(-1) r e^{i \theta}$ $=e^{i \pi} r e^{i \theta}$ $=r e^{i(\theta+\pi)}$ i.e arg (z) = + ⇒ argz arg(z) = =...
Read More →If 0 < arg (z) < π,
Question: If 0 arg (z) , then arg (z) arg (z) = ____________. Solution: For 0 argz Letz = r(cos,isin) i.e argz = Then z = r(cos+isin) $=-r(+\cos \theta+i(+\sin \theta))$ $=(-1) r e^{i \theta}$ $=e^{i \pi} r e^{i \theta}$ $=r e^{i(\theta+\pi)}$ i.e arg (z) = + ⇒ argz arg(z) = =...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $7 x+\frac{3}{x}=35 \frac{3}{5}$ Solution: We have been given, $7 x+\frac{3}{x}=35 \frac{3}{5}$ $7 x^{2}+3=\left(35+\frac{3}{5}\right) x$ $7 x^{2}-\left(35+\frac{3}{5}\right) x+3=0$ Therefore, $7 x^{2}-35 x-\frac{3}{5} x+3=0$ $7 x(x-5)-\frac{3}{5}(x-5)=0$ $\left(7 x-\frac{3}{5}\right)(x-5)=0$ Therefore, $7 x-\frac{3}{5}=0$ $7 x=\frac{3}{5}$ $x=\frac{3}{35}$ or, $x-5=0$ $x=5$ Hence, $x=\frac{3}{35}$ or $x=5$....
Read More →Solve this
Question: If $(x+2)$ and $(x-1)$ are factors of $\left(x^{3}+10 x^{2}+m x+n\right)$, then (a) $m=5, n=-3$ (b) $m=7, n=-18$ (c) $m=17, n=-8$ (d) $m=23, n=-19$ Solution: (b) $m=7, n=-18$ Let: $p(x)=x^{3}+10 x^{2}+m x+n$ Now, $x+2=0 \Rightarrow x=-2$ $(x+2)$ is a factor of $p(x)$. So, we have $p(-2)=0$ $\Rightarrow(-2)^{3}+10 \times(-2)^{2}+m \times(-2)+n=0$ $\Rightarrow-8+40-2 m+n=0$ $\Rightarrow 32-2 m+n=0$ $\Rightarrow 2 m-n=32$ ........(i) Now, $x-1=0 \Rightarrow x=1$ Also, $(x-1)$ is a factor ...
Read More →Solve the following
Question: If $z_{1}=\sqrt{3}+i \sqrt{3}$ and $z_{2}=\sqrt{3}+i$, then the point representing $\frac{z_{1}}{z_{2}}$ lies in _____________________ Solution: Let $z_{1}=\sqrt{3}(1+i)$ and $z_{2}=\sqrt{3}+i$ then $\frac{z_{1}}{z_{2}}=\frac{\sqrt{3}(1+i)}{\sqrt{3}+i} \times \frac{\sqrt{3}-i}{\sqrt{3}-i}$ $=\frac{\sqrt{3}(1+i)(\sqrt{3}-i)}{3-i^{2}}$ $=\frac{\sqrt{3}\left(\sqrt{3}-i+\sqrt{3} i-i^{2}\right)}{3+1}$ $=\frac{\sqrt{3}}{4}(\sqrt{3}+1+i(\sqrt{3}-1))$ Since $\sqrt{3}+1$ and $\sqrt{3}-10$ $\Rig...
Read More →The value of
Question: The value of $\frac{i^{4 n+1}-i^{4 n-1}}{2}$ is _________________ Solution: Let $z=\frac{i^{4 n+1}-i^{4 n-1}}{2}$ $=\frac{i^{4 n} \cdot i-i^{4 n}-i^{-1}}{2}$ $=\frac{i-i^{-1}}{2}$ $=\frac{i-\frac{1}{i}}{2}$ $=\frac{i^{2}-1}{2 i}$ $=-\frac{1-1}{2 i}$ $=-\frac{2}{2 i}$ $=-\frac{1}{i} \times \frac{i}{i}$ $=-\frac{i}{-1}$ i. e $\frac{i^{4 n+1}-i^{4 n-1}}{2}=i$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $(x-5)(x-6)=\frac{25}{(24)^{2}}$ Solution: We have been given that, $(x-5)(x-6)=\frac{25}{(24)^{2}}$ $x^{2}-11 x+30-\frac{25}{576}=0$ $x^{2}-11 x+\frac{17255}{576}=0$ $x^{2}-\frac{145}{24} x-\frac{119}{24} x+\frac{17255}{576}=0$ $x\left(x-\frac{145}{24}\right)-\frac{119}{24}\left(x-\frac{145}{24}\right)=0$ $\left(x-\frac{119}{24}\right)\left(x-\frac{145}{24}\right)=0$ Therefore, $x-\frac{119}{24}=0$ $x=\frac{119}{24}$ or, $x-\fr...
Read More →Let f : N → N be defined by
Question: Let $f: \mathbf{N} \rightarrow \mathbf{N}$ be defined by $f(n)=\left\{\begin{array}{l}n+1, \text { if } n \text { is odd } \\ n-1, \text { if } n \text { is even }\end{array}\right.$ Show thatfis a bijection. [CBSE 2012, NCERT] Solution: We have, $f(n)= \begin{cases}n+1, \text { if } n \text { is odd } \\ n-1, \text { if } n \text { is even }\end{cases}$ Injection test: Case I: If $n$ is odd, Let $x, y \in \mathbf{N}$ such that $f(x)=f(y)$ As, $f(x)=f(y)$ $\Rightarrow x+1=y+1$ $\Righta...
Read More →Let f : N → N be defined by
Question: Let $f: \mathbf{N} \rightarrow \mathbf{N}$ be defined by $f(n)=\left\{\begin{array}{l}n+1, \text { if } n \text { is odd } \\ n-1, \text { if } n \text { is even }\end{array}\right.$ Show thatfis a bijection. [CBSE 2012, NCERT] Solution: We have, $f(n)= \begin{cases}n+1, \text { if } n \text { is odd } \\ n-1, \text { if } n \text { is even }\end{cases}$ Injection test: Case I: If $n$ is odd, Let $x, y \in \mathbf{N}$ such that $f(x)=f(y)$ As, $f(x)=f(y)$ $\Rightarrow x+1=y+1$ $\Righta...
Read More →The principal argument of
Question: The principal argument ofi 1097is ____________. Solution: Let $z=i^{-1097}$ $z=\frac{1}{(i)^{1096} \cdot i}$ $=\frac{1}{\left(i^{4}\right)^{274}} \times \frac{1}{i}$ $=\frac{1}{1} \times \frac{1}{i} \quad\left(\because i^{4}=1\right)$ $=\frac{1}{i}$ $=\frac{1}{i} \times \frac{i}{i}$ $z=-i=\cos \left(\frac{\pi}{2}\right)-i \sin \frac{\pi}{2}$ Hence, principle argument is $-\frac{\pi}{2}$....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$ Solution: We have been given, $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$ $\frac{(x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2)}{(x-1)(x-4)(x-2)(x-3)}=\frac{1}{6}$ $\frac{3\left(x^{2}-5 x+6\right)}{\left(x^{2}-5 x+4\right)\left(x^{2}-5 x+6\right)}=\frac{1}{6}$ $18=x^{2}-5 x+4$ $x^{2}-5 x-14=0$ $x^{2}-7 x+2 x-14=0$ $x(x-7)+2(x-7)=0$ $(...
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