If three complex numbers
Question: If three complex numbersz1,z2andz3are in A.P., then points representing them lie on ____________. Solution: Sincez1,z2andz3are in A.P Hence $2 z_{2}=z_{1}+z_{3}$ i. e $z_{2}=\frac{z_{1}+z_{3}}{z_{2}}$ i.ez2is the mid-point of line joiningz1andz3 ⇒z1,z2andz3lie on a straight....
Read More →The points representing the complex number z for which
Question: The points representing the complex numberzfor which |z+ 1| |z 1| lie on the left side of ____________. Solution: Given |z+ 1| |z 1| Letz = x + iywherex, yR i. e $|x+i y+1||x+i y-1|$ $\Rightarrow|(x+1)+i y||(x-1)+i y|$ Squaring both sides, we get $|(x+1)+i y|^{2}|(x-1)+i y|^{2}$ i.e $(x+1)^{2}+y^{2}(x-1)^{2}+y^{2}$ i. e $x^{2}+1+2 x+y^{2}x^{2}+1-2 x+y^{2}$ i. e $2 x0$ i. e $x0$ Hence, $|z+1||z-1|$ lies on left side of $y$-axis...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$ Solution: We have been given $\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$ $\frac{x^{2}+a^{2}-2 a x+x^{2}+b^{2}-2 b x}{x^{2}-(a+b) x+a b}=\frac{a^{2}+b^{2}}{a b}$ $2 a b x^{2}-2(a b)(a+b) x+a b\left(a^{2}+b^{2}\right)=\left(a^{2}+b^{2}\right) x^{2}-(a+b)\left(a^{2}+b^{2}\right) x+a b\left(a^{2}+b^{2}\right)$ $(a-b)^{2} x^{2}-(a+b)(a-b)^{2} x=0$ $x(a-b)^{2}(x-(...
Read More →The points representing the complex number z for which
Question: The points representing the complex numberzfor which |z+ 1| |z 1| lie on the left side of ____________. Solution: Given |z+ 1| |z 1| Letz = x + iywherex, yR i. e $|x+i y+1||x+i y-1|$ $\Rightarrow|(x+1)+i y||(x-1)+i y|$ Squaring both sides, we get $|(x+1)+i y|^{2}|(x-1)+i y|^{2}$ i.e $(x+1)^{2}+y^{2}(x-1)^{2}+y^{2}$ i. e $x^{2}+1+2 x+y^{2}x^{2}+1-2 x+y^{2}$ i. e $2 x0$ i. e $x0$ Hence, $|z+1||z-1|$ lies on left side of $y$-axis...
Read More →If a complex number coincides with its conjugate,
Question: If a complex number coincides with its conjugate, then it lies on ____________. Solution: Let $z=x+i y$ and $\bar{z}=\overline{x+i y}$ $\bar{z}=x-i y$ Since $z=\bar{z}$ (given) $\Rightarrow x+i y=x-i y$ $\Rightarrow i y=-i y$ $\Rightarrow 2 i y=0$ i.e $y=0$ Then $z$ lies an $x$-axis....
Read More →Show that f : R → R, given by
Question: Show that $f: R \rightarrow R$, given by $f(x)=x-[x]$, is neither one-one nor onto. Solution: We have, $f(x)=x-[x]$ Injection test: $f(x)=0$ for all $x \in \mathbf{Z}$ So,fis a many-one function.Surjection test: Range $(f)=[0,1) \neq \mathbf{R}$. So,fis an into function.Therefore,fis neither one-one nor onto....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{m}{n} x^{2}+\frac{n}{m}=1-2 x$ Solution: We have been given $\frac{m}{n} x^{2}+\frac{n}{m}=1-2 x$ $\frac{m^{2} x^{2}+n^{2}}{m n}=1-2 x$ $m^{2} x^{2}+2 m n x+\left(n^{2}-m n\right)=0$ $m^{2} x^{2}+m n x+m n x+\left[n^{2}-(\sqrt{m n})^{2}\right]=0$ $m^{2} x^{2}+m n x+m n x+(n+\sqrt{m n})(n-\sqrt{n m})+(m \sqrt{m n} x-m \sqrt{m n} x)=0$ $\left[m^{2} x^{2}+m n x+m \sqrt{m n} x\right]+[m n x-m \sqrt{m n} x+(n+\sqrt{m n})(n-\sq...
Read More →The argument of the complex number
Question: The argument of the complex number $(-1+i \sqrt{3})(1+i)(\cos \theta+i \sin \theta)$ is ____________________ Solution: For $z=(-1+i \sqrt{3})(1+i)(\cos \theta+i \sin \theta)$ $\arg z=\arg (-1+i \sqrt{3})+\arg (1+i)+\arg (\cos \theta+i \sin \theta)$ $\arg (-1+i \sqrt{3}):-z_{1}=-1+i \sqrt{3}$ $\theta_{1}=\tan ^{-1}\left|\left(\frac{\sqrt{3}}{1}\right)\right|$ $\theta=\frac{\pi}{3}$ Sincez1lies in II quadrant $\Rightarrow \arg z_{1}=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$ $\arg (1+i):-z_{2}=1...
Read More →Given A = {2, 3, 4}, B = {2, 5, 6, 7}.
Question: GivenA= {2, 3, 4},B= {2, 5, 6, 7}. Construct an example of each of the following: (i) an injective map fromAtoB(ii) a mapping fromAtoBwhich is not injective(iii) a mapping fromAtoB. Solution: (i) {(2, 7), (3, 6), (4, 5)}(ii) {(2, 2), (3, 2), (4, 5)}(iii) {(2, 5), (3, 6), (4, 7)}Disclaimer: There are many more possibilities of each case....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{4}{x}-3=\frac{5}{2 x+3}, x \neq 0,-\frac{3}{2}$ Solution: $\frac{4}{x}-3=\frac{5}{2 x+3}$ $\Rightarrow \frac{4-3 x}{x}=\frac{5}{2 x+3}$ $\Rightarrow(4-3 x)(2 x+3)=5 x$ $\Rightarrow 8 x+12-6 x^{2}-9 x=5 x$ $\Rightarrow-6 x^{2}-6 x+12=0$ $\Rightarrow x^{2}+x-2=0$ $\Rightarrow x^{2}+2 x-x-2=0$ $\Rightarrow x(x+2)-1(x+2)=0$ $\Rightarrow(x-1)(x+2)=0$ $\Rightarrow x-1=0$ or $x+2=0$ $\Rightarrow x=1$ or $x=-2$ Hence, the factors...
Read More →The complex number cosθ + i sinθ
Question: The complex number cos+ i sin__________ be zero for any. Solution: z= cos+isincan never be zero for anybecausez =0 ⇒ cos =0 and sin= 0 Since no such value ofexists ⇒ cos+isin can never be zero....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$ Solution: $\frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$ $\Rightarrow \frac{1}{2 a+b+2 x}-\frac{1}{2 a}=\frac{1}{b}+\frac{1}{2 x}$ $\Rightarrow \frac{2 a-(2 a+b+2 x)}{(2 a+b+2 x)(2 a)}=\frac{2 x+b}{2 b x}$ $\Rightarrow \frac{-1(2 x+b)}{4 a^{2}+2 a b+4 a x}=\frac{2 x+b}{2 b x}$ $\Rightarrow-2 b x(2 x+b)=\left(4 a^{2}+2 a b+4 a x\right)(2 x+b)$ $\Rightarr...
Read More →The multiplication of a non-zero complex number by i rotates it through
Question: The multiplication of a non-zero complex number byi rotates it through____________ in the anti-clockwise direction. Solution: Letz=x+iytheniz=ix+i2y i.eiz=ixy i.eiz= y+ix $\tan \theta_{1}$ for $z=x+i y, \tan \theta_{1}=\frac{y}{x}$ $\tan \theta_{2}$ for $i z=i x-y, \tan \theta_{2}=\frac{x}{-y}$ Since $\tan \theta_{1} \times \tan \theta_{2}=\frac{y}{x} \times \frac{x}{-y}=-1$ i.e iz is rotated by $90^{\circ}$....
Read More →Suppose f1 and f2 are non-zero one-one functions from R to R.
Question: Suppose $f_{1}$ and $f_{2}$ are non-zero one-one functions from $R$ to $R$. Is $\frac{f_{1}}{f_{2}}$ necessarily one-one? Justify your answer. Here, $\frac{f_{1}}{f_{2}}: R \rightarrow R$ is given by $\left(\frac{f_{1}}{f_{2}}\right)(x)=\frac{f_{1}(x)}{f_{2}(x)}$ for all $x \in R$. Solution: We know that $f_{1}: R \rightarrow R$, given by $f_{1}(x)=x^{3}$ and $f_{2}(x)=x$ are one-one. Injectivity of $f_{1}$ : Letxandybe two elements in the domainR,such that $f_{1}(x)=f_{2}(y)$ $\Righta...
Read More →The multiplication of a non-zero complex number by i rotates it through
Question: The multiplication of a non-zero complex number byi rotates it through____________ in the anti-clockwise direction. Solution: Letz=x+iytheniz=ix+i2y i.eiz=ixy i.eiz= y+ix $\tan \theta_{1}$ for $z=x+i y, \tan \theta_{1}=\frac{y}{x}$ $\tan \theta_{2}$ for $i z=i x-y, \tan \theta_{2}=\frac{x}{-y}$ Since $\tan \theta_{1} \times \tan \theta_{2}=\frac{y}{x} \times \frac{x}{-y}=-1$ i.e $i z$ is rotated by $90^{\circ}$....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: Solution: $\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}$ $\Rightarrow \frac{(x-2)+2(x-3)}{(x-3)(x-2)}=\frac{8}{x}$ $\Rightarrow \frac{x-2+2 x-6}{x^{2}-2 x-3 x+6}=\frac{8}{x}$ $\Rightarrow \frac{3 x-8}{x^{2}-5 x+6}=\frac{8}{x}$ $\Rightarrow x(3 x-8)=8\left(x^{2}-5 x+6\right)$ $\Rightarrow 3 x^{2}-8 x=8 x^{2}-40 x+48$ $\Rightarrow 5 x^{2}-32 x+48=0$ $\Rightarrow 5 x^{2}-20 x-12 x+48=0$ $\Rightarrow 5 x(x-4)-12(x-4)=0$ $\Rightarrow(5 x-...
Read More →If the point representing a complex number lies in the third quadrant,
Question: If the point representing a complex number lies in the third quadrant, then the point representing its conjugate lies in the ____________. Solution: Ifzlies in III quadrant. i.ez= xiy;x, y 0 $\Rightarrow \bar{z}=-\overline{x-i y}=-x+i y$ i.e $\bar{z}$ lies in II quadrant....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $3 x^{2}-14 x-5=0$ Solution: We have been given $3 x^{2}-14 x-5=0$ $3 x^{2}-15 x+x-5=0$ $3 x(x-5)+1(x-5)=0$ $(3 x+1)(x-5)=0$ Therefore, $3 x+1=0$ $3 x=-1$ $x=\frac{-1}{3}$ or, $x-5=0$ $x=5$ Hence, $x=\frac{-1}{3}$ or $x=5$....
Read More →If (2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = x + iy,
Question: If (2 +i) (2 + 2i) (2 + 3i) ...... (2 +ni) =x+iy, then 5.8.13...(4 +n2) = ____________. Solution: Given:- (2 +i) (2 + 2i) (2 + 3i) ...... (2 +ni) =x+iy Taking modulus both sides, we get |(2 +i) (2 + 2i) (2 + 3i) ...... (2 +ni)= |x + iy| By squaring both sides, we get $|2+i|^{2}|2+2 i|^{2}|2+3 i|^{2} \ldots \ldots|2+n i|^{2}=|x+i y|^{2}$ i.e $(4+1)(4+4)(4+9) \ldots \ldots\left(4+n^{2}\right)=x^{2}+y^{2}$ i.e $5.8 .13 \ldots \ldots\left(4+n^{2}\right)=x^{2}+y^{2}$...
Read More →Solve this
Question: If $(x+5)$ is a factor of $p(x)=x^{3}-20 x+5 k$, then $k=?$ (a) $-5$ (b) 5 (c) 3 (d) $-3$ Solution: (b) 5 $(x+5)$ is a factor of $p(x)=x^{3}-20 x+5 k$. $\Rightarrow(-5)^{3}-20 \times(-5)+5 k=0$ $\Rightarrow-125+100+5 k=0$ $\Rightarrow 5 k=25$ $\Rightarrow k=5$...
Read More →Show that if f1 and f2 are one-one maps from R to R, then the product
Question: Show that if $f_{1}$ and $f_{2}$ are one-one maps from $R$ to $R$, then the product $f_{1} \times f_{2}: R \rightarrow R$ defined by $\left(f_{1} \times f_{2}\right)(x)=f_{1}(x) f_{2}(x)$ need not be one-one. Solution: We know that $f_{1}: R \rightarrow R$, given by $f_{1}(x)=x$, and $f_{2}(x)=x$ are one-one. Proving $f_{1}$ is one-one: Letxandybe two elements in the domainR,such that $f_{1}(x)=f_{1}(y)$ $\Rightarrow x=y$ Proving $f_{2}$ is one-one: Letxandybe two elements in the domai...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: Solution: We have been given $\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=\frac{5}{2}$ $2\left(x^{2}+1-2 x+4 x^{2}+1+4 x\right)=5\left(2 x^{2}-x-1\right)$ $10 x^{2}+4 x+4=10 x^{2}-5 x-5$ $9 x+9=0$ Therefore, $9 x=-9$ $x=\frac{-9}{9}$ $x=-1$ Hence, $x=-1$....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: Solution: We have been given $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}$ $6\left(x^{2}+1+2 x-x^{2}-1+2 x\right)=5\left(x^{2}-1\right)$ $5 x^{2}-24 x-5=0$ $5 x^{2}-25 x+x-5=0$ $5 x(x-5)+1(x-5)=0$ $(5 x+1)(x-5)=0$ Therefore, $5 x+1=0$ $5 x=-1$ $x=\frac{-1}{5}$ or, $x-5=0$ $x=5$ Hence, $x=\frac{-1}{5}$ or $x=5$....
Read More →The zeros of the polynomial p(x)
Question: The zeros of the polynomial $p(x)=2 x^{2}+7 x-4$ are (a) $4, \frac{-1}{2}$ (b) $4, \frac{1}{2}$ (c) $-4, \frac{1}{2}$ (d) $-4, \frac{-1}{2}$ Solution: The given polynomial is $p(x)=2 x^{2}+7 x-4$. Putting $x=\frac{1}{2}$ in $p(x)$, we get $p\left(\frac{1}{2}\right)=2 \times\left(\frac{1}{2}\right)^{2}+7 \times \frac{1}{2}-4=\frac{1}{2}+\frac{7}{2}-4=4-4=0$ Therefore, $x=\frac{1}{2}$ is a zero of the polynomial $p(x)$. Puttingx= 4 inp(x), we get $p(-4)=2 \times(-4)^{2}+7 \times(-4)-4=32...
Read More →The conjugate of the complex number
Question: The conjugate of the complex number $\frac{1-i}{1+i}$ is Solution: $\frac{(\overline{1-i})}{(\overline{1+i})}$ i. e conjugate of $\frac{1-i}{1+i}$ $=\frac{(\overline{1-i})}{(\overline{1+i})}$ $=\frac{1+i}{1-i}$ $=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$ $=\frac{(1+i)^{2}}{1-i^{2}}$ $=\frac{(1+i)^{2}}{1+1}=\frac{1}{2}\left(1+i^{2}+2 i\right)$ Conjugate of $\frac{1-i}{1+i}=\frac{1}{2}(2 i)=i$...
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