Find the principal argument of
Question: Find the principal argument of $(1+i \sqrt{3})^{2}$. Solution: $z=(1+i \sqrt{3})^{2}$ $=1+3 i^{2}+2 \sqrt{3} i$ $=1-3+2 \sqrt{3} i$ $=-2+2 \sqrt{3} i$ Let $\beta$ be an acute angle given by $\tan \beta=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}$. Then, $\tan \beta=\frac{|2 \sqrt{3}|}{|2|}=|\sqrt{3}|$ $\Rightarrow \tan \beta=\left|\tan \frac{\pi}{3}\right|$ $\Rightarrow \beta=\frac{\pi}{3}$ Clearly, $z$ lies in the second quadrant. Therefore, $\arg (z)=\pi-\frac{\pi}{3}=\frac...
Read More →Let f = {(3, 1), (9, 3), (12, 4)} and
Question: Letf= {(3, 1), (9, 3), (12, 4)} andg= {(1, 3), (3, 3) (4, 9) (5, 9)}. Show thatgofandfogare both defined. Also, findfogandgof. Solution: f= {(3, 1), (9, 3), (12, 4)} andg= {(1, 3), (3, 3) (4, 9) (5, 9)}f: {3, 9, 12} {1, 3,4} andg: {1, 3, 4, 5} {3, 9} Co-domain of $f$ is a subset of the domain of $g$. So, gof exists and gof : $\{3,9,12\} \rightarrow\{3,9\}$ $(g \circ f)(3)=g(f(3))=g(1)=3$ $(g \circ f)(9)=g(f(9))=g(3)=3$ $(g \circ f)(12)=g(f(12))=g(4)=9$ $\Rightarrow g o f=\{(3,3),(9,3),...
Read More →Write the least positive integral value of n for which
Question: Write the least positive integral value of $n$ for which $\left(\frac{1+i}{1-i}\right)^{n}$ is real. Solution: $\left(\frac{1+i}{1-i}\right)^{n}$ $=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{n}$ $=\left(\frac{1+i^{2}+2 i}{1-i^{2}}\right)^{n}$ $=\left(\frac{1-1+2 i}{1+1}\right)^{n}$ $\Rightarrow\left(\frac{2 i}{2}\right)^{n}$ $=i^{n}$ For $i^{n}$ to be real, the smallest positive value of $n$ will be 2 . As, $i^{2}=-1$, which is real....
Read More →Factorize:
Question: Factorize: $2 x\left(p^{2}+q^{2}\right)+4 y\left(p^{2}+q^{2}\right)$ Solution: We have: $2 x\left(p^{2}+q^{2}\right)+4 y\left(p^{2}+q^{2}\right)$ $=2\left[x\left(p^{2}+q^{2}\right)+2 y\left(p^{2}+q^{2}\right)\right]$ $=2\left(p^{2}+q^{2}\right)(x+2 y)$...
Read More →Factorize:
Question: Factorize: $2 a(x+y)-3 b(x+y)$ Solution: We have: $2 a(x+y)-3 b(x+y)$ $=(x+y)(2 a-3 b)$...
Read More →Write the argument of −i.
Question: Write the argument of i. Solution: Let $z=-i$ Then, $\operatorname{Re}(z)=0, \operatorname{Im}(z)=-1$ Since, the point $(0,-1)$ representing $z=0-i$ lies on negative direction of imaginary axis. Therefore, $\arg (z)=\frac{-\pi}{2}$ or $\frac{3 \pi}{2}$...
Read More →Solve the following
Question: Write $-1+i \sqrt{3}$ in polar form Solution: Let $z=-1+\sqrt{3} i$. Then, $r=|z|=\sqrt{[-1]^{2}+[\sqrt{3}]^{2}}=2$ Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ $=\sqrt{3}$ $\Rightarrow \alpha=\frac{\pi}{3}$ Since the point representing $z$ lies in the second quadrant. Therefore, the argument of $z$ is given by $\theta=\pi-\alpha$ $=\pi-\frac{\pi}{3}$ $=\frac{2 \pi}{3}$ So, the polar form is $r(\cos \theta+i \sin \theta)$ $\therefore z=2\left(\cos \f...
Read More →Factorize:
Question: Factorize: $27 a^{3} b^{3}-45 a^{4} b^{2}$ Solution: We have: $27 a^{3} b^{3}-45 a^{4} b^{2}$ $=9 a^{3} b^{2}(3 b-5 a)$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}, x \neq 1,-5$ Solution: We have been given $\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}$ $\frac{6}{x^{2}+4 x-5}=\frac{6}{7}$ $x^{2}+4 x-12=0$ $x^{2}+6 x-2 x-12=0$ $x(x+6)-2(x+6)=0$ $(x-2)(x+6)=0$ Therefore, $x-2=0$ $x=2$ or, $x+6=0$ $x=-6$ Hence, $x=2$ or $x=-6$...
Read More →Write 1 − i in polar form.
Question: Write 1 iin polar form. Solution: $z=1-i$ $r=|z|$ $=\sqrt{1+1}$ $=\sqrt{2}$ Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$ $\therefore \tan \alpha=\left|\frac{-1}{1}\right|$ $=\frac{\pi}{4}$ $\Rightarrow \alpha=\frac{\pi}{4}$ Since point $(1,-1)$ lies in the fourth quadrant, the argument of $z$ is given by $\theta=-\alpha=-\frac{\pi}{4}$ Polar form $=r(\cos \theta+i \sin \theta)$ $=\sqrt{2}\left\{\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi...
Read More →Factorize:
Question: Factorize: $18 x^{2} y-24 x y z$ Solution: We have: $18 x^{2} y-24 x y z=6 x y(3 y-4 z)$...
Read More →Factorize:
Question: Factorize: $9 x^{2}+12 x y$ Solution: We have: $9 x^{2}+12 x y=3 x(3 x+4 y)$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $3 x^{2}-2 \sqrt{6} x+2=0$ Solution: We have been given $3 x^{2}-2 \sqrt{6} x+2=0$ $3 x^{2}-\sqrt{6} x-\sqrt{6} x+2=0$ $\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})=0$ $(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0$ Therefore, $\sqrt{3} x-\sqrt{2}=0$ $\sqrt{3} x=\sqrt{2}$ $x=\sqrt{\frac{2}{3}}$ or, $\sqrt{3} x-\sqrt{2}=0$ $\sqrt{3} x=\sqrt{2}$ $x=\sqrt{\frac{2}{3}}$ Hence, $x=\sqrt{\frac{2}{3}}$ or $x=\sqrt{\frac...
Read More →Write the value of
Question: Write the value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}$. Solution: $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}$ $=\frac{i^{4 \times 148}+i^{4 \times 147+2}+i^{4 \times 147}+i^{146 \times 4+2}+i^{4 \times 146}}{i^{4 \times 145+2}+i^{4 \times 145}+i^{4 \times 144+2}+i^{4 \times 144}+i^{4 \times 143+2}}$ $=\frac{1+i^{2}+1+i^{2}+1}{i^{2}+1+i^{2}+1+i^{2}} \quad\left[\because i^{4}=1\right]$$=\f...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}, x \neq 2,4$ Solution: We have been given $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}$ $3\left(x^{2}-5 x+4+x^{2}-5 x+6\right)=10\left(x^{2}-6 x+8\right)$ $4 x^{2}-30 x+50=0$ $2 x^{2}-15 x+25=0$ $2 x^{2}-10 x-5 x+25=0$ $2 x(x-5)-5(x-5)=0$ $(2 x-5)(x-5)=0$ Therefore, $2 x-5=0$ $2 x=5$ $x=\frac{5}{2}$ or, $x-5=0$ $x=5$ Hence, $x=\frac{5}{2}$ or $x=5$....
Read More →If n is any positive integer, write the value of
Question: If $n$ is any positive integer, write the value of $\frac{i^{4 n+1}-i^{4 n-1}}{2}$. Solution: $\frac{i^{4 n+1}-i^{4 n-1}}{2}$ $=\frac{i-\frac{1}{i}}{2} \quad\left(\because i^{4 n}=1, i^{-1}=\frac{1}{i}\right)$ $=\frac{\frac{i^{2}-1}{i}}{2}$ $=\frac{i^{2}-1}{2 i}$ $=\frac{-1-1}{2 i}$ $=\frac{-2}{-2 i}$ $=\frac{-1}{i}$ $=\frac{-i}{i^{2}} \quad\left(\because i^{2}=-1\right)$ $=\frac{-i}{-1}$ $=i$...
Read More →Find gof and fog when f : R → R and g : R → R are defined by
Question: Find gof and fog when $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by (i) $f(x)=2 x+3 \quad$ and $\quad g(x)=x^{2}+5$ (ii) $f(x)=2 x+x^{2} \quad$ and $\quad g(x)=x^{3}$ (iii) $f(x)=x^{2}+8 \quad$ and $\quad g(x)=3 x^{3}+1$ (iv) $f(x)=x \quad$ and $\quad g(x)=|x|$ (v) $f(x)=x^{2}+2 x-3$ and $g(x)=3 x-4$ (vi) $f(x)=8 x^{3} \quad$ and $\quad g(x)=x^{1 / 3}$ Solution: Given, $f: R \rightarrow R$ and $g: R \rightarrow R$ So, gof: $R \rightarrow R$ and fog: $R \rightarrow R$ (i)...
Read More →Solve the following
Question: If 2 and z = 1 + cos +isin , then write the value of|z|z. Solution: $\pi\theta2 \pi$ $\frac{\pi}{2}\frac{\theta}{2}\pi$ (Dividing by 2 ) $z=1+\cos \theta+$ isin $\theta$ $\Rightarrow|z|=\sqrt{(1+\cos \theta)^{2}+\sin ^{2} \theta}$ $\Rightarrow|z|=\sqrt{1+\cos ^{2} \theta+2 \cos \theta+\sin ^{2} \theta}$ $\Rightarrow|z|=\sqrt{1+1+2 \cos \theta}$ $\Rightarrow|z|=\sqrt{2(1+\cos \theta)}$ $\Rightarrow|z|=\sqrt{2 \times 2 \cos ^{2} \frac{\theta}{2}}$ $\Rightarrow|z|=2 \sqrt{\cos ^{2} \frac{...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $a^{2} b^{2} x^{2}+b^{2} x-a^{2} x-1=0$ Solution: We have been given $a^{2} b^{2} x^{2}+b^{2} x-a^{2} x-1=0$ $b^{2} x\left(a^{2} x+1\right)-1\left(a^{2} x+1\right)=0$ $\left(b^{2} x-1\right)\left(a^{2} x+1\right)=0$ Therefore, $b^{2} x-1=0$ $b^{2} x=1$ $x=\frac{1}{b^{2}}$ or, $a^{2} x+1=0$ $a^{2} x=-1$ $x=-\frac{1}{a^{2}}$ Hence, $x=\frac{1}{b^{2}}$ or $x=-\frac{1}{a^{2}}$....
Read More →The zeroes of the polynomial p(x)
Question: The zeroes of the polynomial $p(x)=3 x^{2}-1$ are (a) $\frac{1}{3}$ and 3 (b) $\frac{1}{\sqrt{3}}$ and $\sqrt{3}$ (c) $\frac{-1}{\sqrt{3}}$ and $\sqrt{3}$ (d) $\frac{1}{\sqrt{3}}$ and $\frac{-1}{\sqrt{3}}$ Solution: (d) $\frac{1}{\sqrt{3}}$ and $\frac{-1}{\sqrt{3}}$ Let: $p(x)=3 x^{2}-1$ To find the zeroes of $p(x)$, we have : $p(x)=0 \Rightarrow 3 x^{2}-1=0$ $\Rightarrow 3 x^{2}=1$ $\Rightarrow x^{2}=\frac{1}{3}$ $\Rightarrow x=\pm \frac{1}{\sqrt{3}}$ $\Rightarrow x=\frac{1}{\sqrt{3}}...
Read More →Solve the following
Question: If $x+i y=\sqrt{\frac{a+i b}{c+i d}}$, then write the value of $\left(x^{2}+y^{2}\right)^{2}$ Solution: $x+i y=\sqrt{\frac{a+i b}{c+i d}}$ Taking modulus on both the sides, $|x+i y|=\left|\sqrt{\frac{a+i b}{c+i d}}\right|$ $\Rightarrow|x+i y|=\sqrt{\frac{|a+i b|}{|c+i d|}}$ $\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{c^{2}+d^{2}}}}$ $\left[\because|x+i y|=\sqrt{x^{2}+y^{2}}\right]$ Squaring both the sides, $x^{2}+y^{2}=\sqrt{\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}$ S...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $a b x^{2}+\left(b^{2}-a c\right) x-b c=0$ Solution: We have been given $a b x^{2}+\left(b^{2}-a c\right) x-b c=0$ $a b x^{2}+b^{2} x-a c x-b c=0$ $b x(a x+b)+-c(a x+b)=0$ $(a x+b)(b x-c)=0$ Therefore, $a x+b=0$ $a x=-b$ $x=-\frac{b}{a}$ or, $b x-c=0$ $b x=c$ $x=\frac{c}{b}$ Hence, $x=-\frac{b}{a}$ or $x=\frac{c}{b}$....
Read More →Write the values of the square root of −i.
Question: Write the values of the square root of i. Solution: Let $\sqrt{-i}=x+i y$ Squaring both the sides $-i=x^{2}+y^{2} i^{2}+2 i x y$ $\Rightarrow 2 x y=-1 \quad \ldots$ (i) and $x^{2}-y^{2}=0 \quad \ldots$ (ii) Equation (ii) shows that $x$ and $y$ are of opposite sign. From (ii), $x=\pm y$ From (i) $2(x)(-x)=\frac{-1}{2}$ $\Rightarrow x^{2}=\frac{1}{2}$ $\Rightarrow x=\pm \frac{1}{\sqrt{2}}$[Since $x$ and $y$ have opposite signs, $y=-\frac{1}{\sqrt{2}}$ when $x=\frac{1}{\sqrt{2}}$ and vice...
Read More →The zeroes of the polynomial p(x)
Question: The zeroes of the polynomial $p(x)=x^{2}-3 x$ are (a) 0,0 (b) 0,3 (c) $0,-3$ (d) $3,-3$ Solution: (b) 0, 3 Let: $p(x)=x^{2}-3 x$ Now, we have: $p(x)=0 \Rightarrow x^{2}-3 x=0$ $\Rightarrow x(x-3)=0$ $\Rightarrow x=0$ and $x-3=0$ $\Rightarrow x=0$ and $x=3$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $x^{2}+\left(a+\frac{1}{a}\right) x+1=0$ Solution: We have been given $x^{2}+\left(a+\frac{1}{a}\right) x+1=0$ Therefore, $x^{2}+a x+\frac{1}{a} x+1=0$ $x(x+a)+\frac{1}{a}(x+a)=0$ $\left(x+\frac{1}{a}\right)(x+a)=0$ Therefore, $x+\frac{1}{a}=0$ $x=-\frac{1}{a}$ or, $x+a=0$ $x=-a$ Hence, $x=-\frac{1}{a}$ or $x=-a$....
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