The principal value of the argument of the complex number
Question: The principal value of the argument of the complex number 1 iis ____________. Solution: Since $z=1-i=1+i(-1)$ here $\theta=\tan ^{-1}\left|\frac{-1}{1}\right|$ $\theta=\tan ^{-1} 1=\frac{\pi}{y}$ Sincezlies in 4thquadrant. $\therefore$ argument is given by $\frac{-\pi}{y}$...
Read More →The area of the triangle on the complex plane formed by the complex numbers z, –iz and z + iz is
Question: The area of the triangle on the complex plane formed by the complex numbersz, izandz+izis (a) $|z|^{2}$ (b) $|\bar{z}|^{2}$ (c) $\frac{1}{2}|z|^{2}$ (d) none of these Solution: For any complex number $z,-i z$ represents complex number obtained by rotating $z$ clockwise by $\frac{\pi}{2}$ angle. Hence, $z,-i z$ and $z+i z$ represents a right angled triangle with sides $z,-i z$ and hypotenus $z+i z$ $\therefore$ Area of triangle formed is $=\frac{1}{2}|z \times(-i z)|$ $=\frac{1}{2}|i||z...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $a^{2} x^{2}-3 a b x+2 b^{2}=0$ Solution: We have been given $a^{2} x^{2}-3 a b x+2 b^{2}=0$ $a^{2} x^{2}-2 a b x-a b x+2 b^{2}=0$ $a x(a x-2 b)-b(a x-2 b)=0$ $(a x-b)(a x-2 b)=0$ Therefore, $a x-b=0$ $a x=b$ $x=\frac{b}{a}$ or, $a x-2 b=0$ $a x=2 b$ $x=\frac{2 b}{a}$ Hence, $x=\frac{b}{a}$ or $x=\frac{2 b}{a}$...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$ Solution: We have been given $\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$ $\sqrt{2} x^{2}-4 x+x-2 \sqrt{2}=0$ $\sqrt{2} x(x-2 \sqrt{2})+1(x-2 \sqrt{2})=0$ $(x-2 \sqrt{2})(\sqrt{2} x+1)=0$ Therefore, $x-2 \sqrt{2}=0$ $x=2 \sqrt{2}$ or, $\sqrt{2} x+1=0$ $\sqrt{2} x=-1$ $x=\frac{-1}{\sqrt{2}}$ Hence, $x=2 \sqrt{2}$ or $x=\frac{-1}{\sqrt{2}}$....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$ Solution: We have been given $4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$ $4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3}=0$ $4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0$ $(\sqrt{3} x+2)(4 x-\sqrt{3})=0$ Therefore, $\sqrt{3} x+2=0$ $\sqrt{3} x=-2$ $x=\frac{-2}{\sqrt{3}}$ or, $4 x-\sqrt{3}=0$ $4 x=\sqrt{3}$ $x=\frac{\sqrt{3}}{4}$ Hence, $x=\frac{-2}{\sqrt{3}}$ or $x=\frac{\sqrt{3}}{4}$....
Read More →Solve the following
Question: $\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$ is possible if (a) $z_{2}=\bar{z}_{1}$ (b) $z_{2}=\frac{1}{z_{1}}$ (c) $\arg \left(z_{1}\right)=\arg \left(z_{2}\right)$ (d) $\left|z_{1}\right|=\left|z_{2}\right|$ Solution: Since given $\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$ 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}+z_{2}\right|^{2}=\left(\left|z_{1}\right|+\left|z_{2}\right|\right...
Read More →The real value of θ for which the expression
Question: The real value of $\theta$ for which the expression $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number, is (a) $n \pi+\frac{\pi}{4}$ (b) $n \pi+(-1)^{n} \frac{\pi}{4}$ (c) $2 n \pi \pm \frac{\pi}{2}$ (d) none of theses Solution: Given $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is a real number i. e $\frac{1+i \cos \theta}{1-2 i \cos \theta} \times \frac{1+2 i \cos \theta}{1+2 i \cos \theta}$ $=\frac{(1+i+\cos \theta)(1+2 i \cos \theta)}{1-4 i^{2} \cos ^{2} \theta}$ $=\frac{1...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $\frac{2}{x^{2}}-\frac{5}{x}+2=0$ Solution: We have been given $\frac{2}{x^{2}}-\frac{5}{x}+2=0$ Now we solve the above quadratic equation using factorization method. Therefore, $\frac{2-5 x+2 x^{2}}{x^{2}}=0$ $2 x^{2}-5 x+2=0$ $2 x^{2}-4 x-x+2=0$ $2 x(x-2)-1(x-2)=0$ $(2 x-1)(x-2)=0$ Now, one of the products must be equal to zero for the whole product to be zero. Hence we equate both the products to zero in order to find the val...
Read More →If f : R → R be the function defined by
Question: If $f: R \rightarrow R$ be the function defined by $f(x)=4 x^{3}+7$, show that $f$ is a bijection. Solution: Injectivity:Letxandybe any two elements in the domain (R), such thatf(x) = f(y) $\Rightarrow 4 x^{3}+7=4 y^{3}+7$ $\Rightarrow 4 x^{3}=4 y^{3}$ $\Rightarrow x^{3}=y^{3}$ $\Rightarrow x=y$ So,fis one-one. Surjectivity:Letybe any element in the co-domain(R),such thatf(x) = yfor some elementxinR(domain). $f(x)=y$ $\Rightarrow 4 x^{3}+7=y$ $\Rightarrow 4 x^{3}=y-7$ $\Rightarrow x^{3...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $10 x-\frac{1}{x}=3$ Solution: We have been given $10 x-\frac{1}{x}=3$ Now we solve the above quadratic equation using factorization method. Therefore, $10 x^{2}-3 x-1=0$ $10 x^{2}-5 x+2 x-1=0$ $5 x(2 x-1)+1(2 x-1)=0$ $(5 x+1)(2 x-1)=0$ Now, one of the products must be equal to zero for the whole product to be zero. Hence we equate both the products to zero in order to find the value ofx. Therefore, $5 x+1=0$ $5 x=-1$ $x=\frac{-...
Read More →The vector represented by the complex number 2 – i is rotated about the origin through an angle
Question: The vector represented by the complex number $2-i$ is rotated about the origin through an angle $\frac{\pi}{2}$ in the clockwise direction, the new position of point is (a) 1 + 2i (b) 1 2i (c) 2 + i (d) 1 + 2i Solution: Given $2-\mathrm{i}$ is rotated by $\frac{\pi}{2}$ angle in the clockwise direction about the origin i. e $\theta=-\frac{\pi}{2}$ Let $Z^{\prime}$ denote the new position and $Z$ denote the previous $p$ Hence $Z^{\prime}=Z e^{i \theta}=Z e^{-\frac{i \pi}{2}}$ i.e Z' $=(...
Read More →Let A = {1, 2, 3}. Write all one-one from A to itself.
Question: LetA= {1, 2, 3}. Write all one-one fromAto itself. Solution: A ={1, 2, 3}Number of elements inA= 3 Number of one-one functions = number of ways of arranging 3 elements $=3 !=6$ (i) {(1, 1), (2, 2), (3, 3)}(ii) {(1, 1), (2, 3), (3, 2)}(iii) {(1, 2 ), (2, 2), (3, 3 )}(iv) {(1, 2), (2, 1), (3, 3)}(v) {(1, 3), (2, 2), (3, 1)}(vi) {(1, 3), (2, 1), (3,2 )}...
Read More →Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
Question: Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective: (i) $\{(x, y): x$ is a person, $y$ is the mother of $x\}$ (ii) $\{(a, b): a$ is a person, $b$ is an ancestor of $a\}$ [NCERT EXEMPLAR] Solution: (i)f= {(x,y) :xis a person,yis the mother ofx}As, for each elementxin domain set, there is a unique related elementyin co-domain set.So,fis the function.Injection test:As,ycan be mother of two or more personsSo,fis not injective.Sur...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $25 x(x+1)=-4$ Solution: We have been given $25 x(x+1)=-4$ $25 x^{2}+25 x+4=0$ $25 x^{2}+20 x+5 x+4=0$ $5 x(5 x+4)+1(5 x+4)=0$ $(5 x+1)(5 x+4)=0$ Therefore, $5 x+1=0$ $5 x=-1$ $x=\frac{-1}{5}$ or, $5 x+4=0$ $5 x=-4$ $x=\frac{-4}{5}$ Hence, $x=\frac{-1}{5}$ or $x=\frac{-4}{5}$....
Read More →Solve the following
Question: If $\left(\frac{1+i}{1-i}\right)^{n}=1$, then $n=$ (a) 2m+ 1 (b) 4m (c) 2m (d) 4m+ 1 wheremN Solution: Given :- $\left(\frac{1+i}{1-i}\right)^{n}=1$ Since, $\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$ $=\frac{(1+i)^{2}}{1-i^{2}}$ $=\frac{1+i^{2}+2 i}{1+1}$ i. e $\frac{1+i}{1-i}=\frac{1-1+2 i}{2}=\frac{2 i}{2}=2 i$ $\Rightarrow\left(\frac{1+i}{1-i}\right)^{n}=1$ $\Rightarrow(i)^{n}=1$ $\Rightarrow n=4 \mathrm{~m} \quad$ where $m \in N$ Hence, the correct answer is option B....
Read More →The value of
Question: The value of $(z+3)(\bar{z}+3)$ is equivalent to (a) $|z+3|^{2}$ (b) $|z-3|$ (c) $z^{2}+3$ (d) none of these Solution: Since $|z+3|^{2}$ $=(z+3)(\overline{z+3})$ $=(z+3)(\bar{z}+3)$ $(\because \overline{3}=3$ real $)$ $=(z+3)(\bar{z}+3)$ Hence, the correct answer is option A....
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $3 x^{2}=-11 x-10$ Solution: We have been given $3 x^{2}=-11 x-10$ $3 x^{2}+11 x+10=0$ $3 x^{2}+6 x+5 x+10=0$ $3 x(x+2)+5(x+2)=0$ $(3 x+5)(x+2)=0$ Therefore, $3 x+5=0$ $3 x=-5$ $x=\frac{-5}{3}$ or, $x+2=0$ $x=-2$ Hence, $x=\frac{-5}{3}$ or $x=-2$....
Read More →Solve this
Question: If $p(x)=x^{2}-2 \sqrt{2} x+1$, then $p(2 \sqrt{2})=$ ? (a) 0 (b) 1 (c) $4 \sqrt{2}$ (d) $-1$ Solution: (b) 1 $p(x)=x^{2}-2 \sqrt{2} x+1$ $\therefore p(2 \sqrt{2})=(2 \sqrt{2})^{2}-2 \sqrt{2} \times(2 \sqrt{2})+1$ $=8-8+1$ $=1$...
Read More →Let A = [−1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
Question: LetA= [-1, 1]. Then, discuss whether the following functions fromAto itself are one-one, onto or bijective: (i) $f(x)=\frac{x}{2}$ (ii) $g(x)=|x|$ (iii) $h(x)=x^{2}$ [NCERT EXEMPLAR] Solution: (i) $f: A \rightarrow A$, given by $f(x)=\frac{x}{2}$\ Injection test:Letxandybe any two elements in the domain (A), such thatf(x) =f(y). $f(x)=f(y)$ $\frac{x}{2}=\frac{y}{2}$ $x=y$ So,fis one-one.Surjection test:Letybe any element in the co-domain (A), such thatf(x) =yfor some elementxinA(domain...
Read More →The real value of α for which the expression
Question: The real value of $\alpha$ for which the expression $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real, is (a) $(n+1) \frac{\pi}{2}$ (b) $(2 n+1) \frac{\pi}{2}$ (c)n (d) none of these wherenN. Solution: Given $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real i. e $\frac{1-i \sin \alpha}{1+2 i \sin \alpha} \times\left(\frac{1-2 i \sin \alpha}{1-2 i \sin \alpha}\right)$ $=\frac{1-i \sin \alpha-2 i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}$ $=\frac...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $48 x^{2}-13 x-1=0$ Solution: We have been given $48 x^{2}-13 x-1=0$ $48 x^{2}-16 x+3 x-1=0$ $16 x(3 x-1)+1(3 x-1)=0$ $(16 x+1)(3 x-1)=0$ Therefore, $16 x+1=0$ $16 x=-1$ $x=\frac{-1}{16}$ or, $3 x-1=0$ $3 x=1$ $x=\frac{1}{3}$ Hence, $x=\frac{-1}{16}$ or $x=\frac{1}{3}$...
Read More →The real value of α for which the expression
Question: The real value of $\alpha$ for which the expression $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real, is (a) $(n+1) \frac{\pi}{2}$ (b) $(2 n+1) \frac{\pi}{2}$ (c)n (d) none of these wherenN. Solution: Given $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real i. e $\frac{1-i \sin \alpha}{1+2 i \sin \alpha} \times\left(\frac{1-2 i \sin \alpha}{1-2 i \sin \alpha}\right)$ $=\frac{1-i \sin \alpha-2 i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}$ $=\frac...
Read More →Solve this
Question: If $p(x)=x=4$, then $p(x)+p(-x)=?$ (a) 0 (b) 4 (c) $2 x$ (d) 8 Solution: (d) 8 Let: $p(x)=(x+4)$ $\therefore p(-x)=(-x)+4$ $=-x+4$ Thus, we have $p(x)+p(-x)=\{(x+4)+(-x+4)\}$ = 4 + 4 =8...
Read More →Solve the following quadratic equations by factorization:
Question: Solve the following quadratic equations by factorization: $5 x^{2}-3 x-2=0$ Solution: We have been given $5 x^{2}-3 x-2=0$ $5 x^{2}-5 x+2 x-2=0$ $5 x(x-1)+2(x-1)=0$ $(5 x+2)(x-1)=0$ Therefore, $5 x+2=0$ $5 x=-2$ $x=\frac{-2}{5}$ or, $x-1=0$ $x=1$ Hence, $x=\frac{-2}{5}$ or $x=1$....
Read More →sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for
Question: sinx+icos 2xand cosxisin 2xare conjugate to each other for (a)x=n (b) $x=\left(n+\frac{1}{2}\right) \frac{\pi}{2}$ (c)x= 0 (d) No value ofx Solution: Given sinx+icos 2xand cosxisin 2xare conjugate to each other i.e sinx+icos 2x= cosxisin 2x i.e sinxicos 2x= cosxisin 2x on comparing real and imaginary part, sinx= cosxand cos 2x= sin 2x i.e. sinx= cosxand 2cos2x 1 = 2 sinxcosx i.e 2cos2x 1 = 2 cosxcosx ( sinx= cosx) i.e 2cos2x 1 = 2cos2x i.e 1 = 0 which is a false statement. Hence no val...
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