If z1, z2 and z3, z4 are two pairs of conjugate complex numbers,
Question: If z1, z2and z3, z4are two pairs of conjugate complex numbers, then find arg(z1/z4) + arg(z2/z3). Solution: According to the question, We have, z1and z2are conjugate complex numbers. The negative side of the real axis = r1(cos 1 i sin 1) = r1[cos (-1) + I sin (-1)] Similarly, z3= r2(cos 2 i sin 2) ⇒z4= r2[cos (-2) + I sin (-2)] $\Rightarrow \arg \left(\frac{\mathrm{z}_{1}}{\mathrm{z}_{4}}\right)+\arg \left(\frac{\mathrm{z}_{2}}{\mathrm{z}_{3}}\right)=\arg \left(\mathrm{z}_{1}\right)-\a...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: In the interval $(1,2)$, function $f(x)=2|x-1|+3|x-2|$ is A. monotonically increasing B. monotonically decreasing C. not monotonic D. constant Solution: Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly decreading on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ Given:- $f(x)=2(x-1)+3(2-x)$ $f(x)=-x+4$ $\frac{\mathrm{d}(\mathrm{f}(\mathrm{x}))}{\mathrm{dx}}=-1=\mathrm...
Read More →If |z1| = 1 (z1 ≠ –1) and z2 = (z1 – 1) / (z + 1),
Question: If |z1| = 1 (z1 1) and z2= (z1 1) / (z + 1),then show that the real part of z2 is zero. Solution: According to the question, Let z1 = x + iy $\Rightarrow\left|z_{1}\right|=\sqrt{x^{2}+y^{2}}=1$ $\mathrm{Z}_{2}=\frac{\mathrm{z}_{1}-1}{\mathrm{z}_{1}+1}=\frac{\mathrm{x}+\mathrm{iy}-1}{\mathrm{x}+\mathrm{iy}+1}$ $=\frac{[(x-1)+i y][(x+1)-i y]}{[(x+1)+i y][(x+1)-i y]}$ $=\frac{\left(x^{2}-1\right)+y^{2}+i[(x+1) y-(x-1) y]}{(x+1)^{2}+y^{2}}$ $=\frac{x^{2}+y^{2}-1+2 i y}{(x+1)^{2}+y^{2}}$ Si...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: Function $f(x)=\cos x-2 \lambda x$ is monotonic decreasing when A. $\lambda\frac{1}{2}$ B. $\lambda\frac{1}{2}$ C. $\lambda2$ D. $\lambda2$ Solution: Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly decreasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ Given:- $f(x)=\cos x-2 \lambda x$ $\frac{d(f(x))}{d x}=-\sin x-2 \lambda=f^{\prime}(x)$ for decreasing functio...
Read More →z1 and z2 are two complex numbers such
Question: z1and z2are two complex numbers such that |z1| = |z2| and arg (z1) + arg (z2) = , then show thatz1= z̅2. Solution: According to the question, Let z1= |z1| (cos 1+ I sin 1) and z2= |z2| (cos 2+ I sin 2) Given that |z1| = |z2| And arg (z1) + arg (z2) = ⇒1+ 2= ⇒1= 2 Now, z1= |z2| (cos ( 2) + I sin ( 2)) ⇒z1= |z2| (-cos 2+ I sin 2) ⇒z1= -|z2| (cos 2 I sin 2) ⇒z1= [|z2| (cos 2 I sin 2)] Hence, z1= -z̅2 Hence proved....
Read More →If (z – 1)/(z + 1) is a purely imaginary number (z ≠ –1),
Question: If (z 1)/(z + 1) is a purely imaginary number (z 1), then find the value of |z|. Solution: According to the question, Let z = x + iy Now, $\frac{z-1}{z+1}=\frac{x+i y-1}{x+i y+1}$ $=\frac{[(x-1)+i y][(x+1)-i y]}{[(x+1)+i y][(x+1)-i y]}$ $=\frac{\left(x^{2}-1\right)+y^{2}+i[(x+1) y-(x-1) y]}{(x+1)^{2}+y^{2}}$ According to the question, we have, $\frac{z-1}{z+1}$ is purely imaginary. $\Rightarrow \frac{\left(x^{2}-1\right)+y^{2}}{(x+1)^{2}+y^{2}}=0$ ⇒x2 1 + y2= 0 ⇒x2+ y2= 1 ⇒(x2+ y2)= 1 ...
Read More →Find the equation of the line drawn through the point of intersection of the
Question: Find the equation of the line drawn through the point of intersection of the lines x y = 7 and 2x + y = 2 and passing through the origin. Solution: Suppose the given two lines intersect at a point $P\left(x_{1}, y_{1}\right)$. Then, $\left(x_{1}, y_{1}\right)$ satisfies each of the given equations. $x-y=7 \ldots$ (i) $2 x+y=2 \ldots$ (ii) Now, we find the point of intersection of eq. (i) and (ii) Multiply the eq. (i) by 2, we get $2 x-2 y=14 \ldots$ (iii) On subtracting eq. (iii) from ...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: The function $f(x)=x^{2} e^{-x}$ is monotonic increasing when A. $x \in R-[0,2]$ B. $0x2$ C. $2x\infty$ D. $x0$ Solution: $f(x)=x^{2} e^{-x}$ $\frac{\mathrm{d}(\mathrm{f}(\mathrm{x}))}{\mathrm{dx}}=x \mathrm{e}^{-\mathrm{x}}(2-\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})$ for $f^{\prime}(x)=0$ $\Rightarrow x^{2} e^{-x}=0$ $\Rightarrow x(2-x)=0$ $x=2, x=0$ $f(x)$ is increasing in $(0,2)$...
Read More →Show that |(z – 2) / (z – 3)| = 2 represents a circle.
Question: Show that |(z 2) / (z 3)| = 2 represents a circle. Find its centre and radius. Solution: According to the question, We have, |(z 2) / (z 3)| = 2 Substituting z = x + iy, we get ⇒ |(x + iy 2) / (x + iy 3)| = 2 ⇒|x 2 + iy| = 2 |x 3 + iy| ⇒ ((x 2)2+ y2) = 2((x 3)2+ y2) ⇒x2 4x + 4 + y2= 4 (x2 6x + 9 + y2) ⇒3x2+ 3y2 20x + 32 = 0 $\Rightarrow x^{2}+y^{2}-\frac{20}{3} x+\frac{32}{3}=0$ $\Rightarrow\left(x-\frac{10}{3}\right)^{2}+y^{2}+\frac{32}{3}-\frac{100}{9}=0$ $\Rightarrow\left(x-\frac{10...
Read More →If arg (z – 1) = arg (z + 3i),
Question: If arg (z 1) = arg (z + 3i), then find x 1 : y. where z = x + iy Solution: According to the question, Let z = x + iy Given that, arg (z 1) = arg (z + 3i) ⇒arg (x + iy 1) = arg (x + iy + 3i) ⇒ arg (x 1 + iy) = arg (x + I (y) = /4 $\Rightarrow \tan ^{-1} \frac{y}{x-1}=\tan ^{-1} \frac{y+3}{x}$ $\Rightarrow \frac{y}{x-1}=\frac{y+3}{x}$ ⇒xy = xy y + 3x 3 ⇒3x 3 = y ⇒(x 1)/y = 1/3 Hence, (x 1): y = 1: 3...
Read More →If |z + 1| = z + 2 (1 + i),
Question: If |z + 1| = z + 2 (1 + i), then find z. Solution: According to the question, We have, |z + 1| = z + 2 (1 + i) Substituting z = x + iy, we get, ⇒|x + iy + 1| = x + iy + 2 (1 + i) We know, |z| = (x2+ y2) ((x + 1)2+ y2) = (x + 2) + i(y + 1) Comparing real and imaginary parts, ⇒ ((x + 1)2+ y2) = x + 2 And 0 = y + 2 ⇒y = -2 Substituting the value of y in((x + 1)2+ y2) = x + 2, ⇒(x + 1)2+ (-2)2= (x + 2)2 ⇒x2+ 2x + 1 + 4 = x2+ 4x + 4 ⇒2x = 1 Hence, x = Hence, z = x + iy = 2i...
Read More →Solve that equation |z| = z + 1 + 2i.
Question: Solve that equation |z| = z + 1 + 2i. Solution: According to the question, We have, |z| = z + 1 + 2i Substituting z = x + iy, we get, ⇒|x + iy| = x + iy + 1 + 2i We know that, |z| = (x2+ y2) (x2+ y2) = (x + 1) + i(y + 2) Comparing real and imaginary parts, We get, (x2+ y2) = (x + 1) And 0 = y + 2 ⇒y = -2 Substituting the value of y in(x2+ y2) = (x + 1), We get, ⇒x2+ (-2)2= (x + 1)2 ⇒x2+ 4 = x2+ 2x + 1 Hence, x = 3/2 Hence, z = x + iy = 3/2 2i...
Read More →Show that the complex number z,
Question: Show that the complex number z, satisfying the condition arg ((z-1)/(z+1)) = /4 lies on a circle. Solution: According to the question, Let z = x + iy arg ((z-1)/(z+1)) = /4 ⇒arg (z 1) arg (z + 1) = /4 ⇒arg (x + iy 1) arg (x + iy + 1) = /4 ⇒ arg (x 1 + iy) arg (x + 1 + iy) = /4 $\Rightarrow \tan ^{-1} \frac{y}{x-1}+\tan ^{-1} \frac{y}{x+1}=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}\left[\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left(\frac{y}{x-1}\right)\left(\frac{y}{x+1}\right)}\right]=\frac{\...
Read More →If the real part of ( z̅ + 2)/ ( z̅ – 1) is 4,
Question: If the real part of ( z̅ + 2)/ ( z̅ 1) is 4, then show that the locus of the point representing z in the complex plane is a circle. Solution: According to the question, Let z = x + iy Now, $\frac{\bar{z}+2}{\bar{z}-1}=\frac{x-i y+2}{x-i y-1}$ $=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}$ $=\frac{(x-1)(x+2)+y^{2}+i[(x+2) y-(x-1) y]}{(x-1)^{2}+y^{2}}$ According to the question, we have, real part $=4$. $\Rightarrow \frac{(x-1)(x+2)+y^{2}}{(x-1)^{2}+y^{2}}=4$ ⇒x2+ x 2 + y2= 4 (...
Read More →If z = x + iy, then show that zz̅ + 2(z + z̅) + b = 0 where bϵR,
Question: If z = x + iy, then show that zz̅ + 2(z + z̅) + b = 0 where bϵR, representing z in the complex plane is a circle. Solution: According to the question, We have, z = x + iy ⇒z̅= x iy Now, we also have, z z̅+ 2 (z + z̅) + b = 0 ⇒(x + iy) (x iy) + 2 (x + iy + x iy) + b = 0 ⇒x2+ y2+ 4x + b = 0 The equation obtained represents the equation of a circle....
Read More →If (1 + i)z = (1 – i) z̅,
Question: If (1 + i)z = (1 i)z̅,then show thatz = iz̅. Solution: According to the question, We have, $(1+i) z=(1-i) \bar{z}$ $\Rightarrow \mathrm{z}=\frac{1-\mathrm{i}}{1+\mathrm{i}} \overline{\mathrm{z}}$ Rationalizing the denominator, We get, $=\frac{(1-i)(1-i)}{(1+i)(1-i)} \bar{z}$ $=\frac{(1-i)^{2}}{\left(1-i^{2}\right)} \bar{z}$ $=\frac{1-2 \mathrm{i}+\mathrm{i}^{2}}{1+1} \overline{\mathrm{z}}$ $=\frac{1-2 \mathrm{i}-1}{2} \overline{\mathrm{z}}$ = -iz̅ Hence proved....
Read More →If a = cos θ + i sin θ,
Question: If a = cos + i sin , find the value of $\frac{1+a}{1-a}$ Solution: According to the question, We have, a = cos + i sin $\Rightarrow \frac{1+a}{1-a}=\frac{(1+\cos \theta)+i \sin \theta}{(1-\cos \theta)-i \sin \theta}$ $=\frac{2 \cos ^{2} \frac{\theta}{2}+12 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}-12 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$ $=\frac{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)}{2 \sin \frac{\theta...
Read More →Prove the following
Question: if $\left(\frac{1-i}{1+i}\right)^{100}=a+i b ;$ then find $(a, b)$ Solution: According to the question, We have, $a+i b=\left(\frac{1-i}{1+i}\right)^{100}$ $=\left[\frac{1-i}{1+i} \cdot \frac{1-i}{1-i}\right]^{100}$ $=\left[\frac{(1-i)^{2}}{1-i^{2}}\right]^{100}$ $=\left(\frac{1-2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\right)^{100}$ $=\left(-\frac{2 \mathrm{i}}{2}\right)^{100}$ $=\left(i^{4}\right)^{25}$ = 1 Hence, (a, b) = (1, 0)...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: Let $f(x)=x^{3}-6 x^{2}+15 x+3 .$ Then, A. $f(x)0$ for all $x \in R$ B. $f(x)f(x+1)$ for all $x \in R$ C. $f(x)$ in invertible D. $f(x)0$ for all $x \in R$ Solution: Formula:- (i) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ (ii) If $f(x)$ is strictly increasing function on interval $[a, b]$, then $f^{-1}$ exist an...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: Let $f(x)=\tan ^{-1}(g(x))$, where $g(x)$ is monotonically increasing for $0x\frac{\pi}{2} .$ Then, $f(x)$ is A. increasing on $\left(0, \frac{\pi}{2}\right)$ B. decreasing on $\left(0, \frac{\pi}{2}\right)$ C. increasing on $\left(0, \frac{\pi}{4}\right)$ and decreasing on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ D. none of these Solution: Formula:- (i) The necessary and sufficient condition for differentiable function defined on $(a, ...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: If the function $f(x)=2 \tan x+(2 a+1) \log _{e}|\sec x|+(a-2) x$ is increasing on $R$, then A. $a \in\left(\frac{1}{2}, \infty\right)$ B. $a \in\left(-\frac{1}{2}, \frac{1}{2}\right)$ C. $a=\frac{1}{2}$ D. $a \in R$ Solution: Formula:- (i) $a x^{2}+b x+c0$ for all $x \Rightarrow a0$ and $b^{2}-4 a c0$ (ii) $a x^{2}+b x+c0$ for all $x \Rightarrow a0$ and $b^{2}-4 a c0$ (iii) The necessary and sufficient condition for differentiable functio...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: The function $f(x)=\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right)$ is of the following types: A. even and increasing B. odd and increasing C. even and decreasing D. odd and decreasing Solution: Formula:- (i) if $f(-x)=f(x)$ then function is even (ii) if $f(-x)=-f(x)$ then function is odd (iii) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ fo...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: Let $f(x)=x^{3}+a x^{2}+b x+5 \sin ^{2} x$ be an increasing function on the set $R$. Then, $a$ and $b$ satisfy. A. $a^{2}-3 b-150$ B. $a^{2}-3 b+150$ C. $a^{2}-3 b+150$ D. $a0$ and $b0$ Solution: Formula:- (i) $a x^{2}+b x+c0$ for all $x \Rightarrow a0$ and $b^{2}-4 a c0$ (ii) $a x^{2}+b x+c0$ for all $x \Rightarrow a0$ and $b^{2}-4 a c0$ (iii) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be str...
Read More →Find the equation of the line drawn through the point of intersection of the
Question: Find the equation of the line drawn through the point of intersection of the lines x 2y + 3 = 0 and 2x 3y + 4 = 0 and passing through the point (4, -5). Solution: Suppose the given two lines intersect at a point $P\left(x_{1}, y_{1}\right) .$ Then, $\left(x_{1}, y_{1}\right)$ satisfies each of the given equations. $x-2 y+3=0 \ldots(\mathrm{i})$ $2 x-3 y+4=0 \ldots(\mathrm{ii})$ Now, we find the point of intersection of eq. (i) and (ii) Multiply the eq. (i) by 2 , we get $2 x-4 y+6=0 \l...
Read More →Mark the correct alternative in the following:
Question: Mark the correct alternative in the following: If the function $f(x)=2 x^{2}-k x+5$ is increasing on $[1,2]$, then $k$ lies in the interval. A. $(-\infty, 4)$ B. $(4, \infty)$ C. $(-\infty, 8)$ D. $(8, \infty)$ Solution: Formula:- The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)0$ for all $x \in(a, b)$ $f(x)=2 x^{2}-k x+5$ $\mathrm{d}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{dx}}\right)...
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