[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

Solution

$\sqrt{(\mathrm{x}-2)^{2}+\mathrm{y}^{2}}=2 \sqrt{(\mathrm{x}-3)^{2}+\mathrm{y}^{2}} \quad=\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}+4=4 \mathrm{x}^{2}+4 \mathrm{y}^{2}-24 \mathrm{x}+36$

\[ \begin{aligned} & =3 x^{2}+3 y^{2}-20 x+32=0=x^{2}+y^{2}-\frac{20}{3} x+\frac{32}{3}=0 \\ & =(\alpha, \beta)=\left(\frac{10}{3}, 0\right) \Rightarrow \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \Rightarrow 3(\alpha, \beta, \gamma)=3\left(\frac{10}{3}, \frac{2}{3}\right)=12 \end{aligned} \]

[UPLIFTJEE]

Answer

27

[UPLIFTJEE]

Solution

$\left|\frac{1+\mathrm{ai}}{\mathrm{b}+\mathrm{i}}\right|=1 \Rightarrow|1+\mathrm{ai}|=|\mathrm{b}+\mathrm{i}|$

\[ \begin{aligned} & a^{2}+1=b^{2}+1 \Rightarrow a= \pm b \Rightarrow b=-a \text { as } a b<0 \Rightarrow(a, b) \text { lies on }|z-1|=|2 z| \Rightarrow|a+i b-1|=2|a+i b| \\ & (a-1)^{2}+b^{2}=4\left(a^{2}+b^{2}\right) \Rightarrow(a-1)^{2}=a^{2}=4\left(2 a^{2}\right) \Rightarrow 1-2 a=6 a^{2} \Rightarrow 6 a^{2}+2 a-1=0 \Rightarrow a=-\frac{2 \pm \sqrt{28}}{12}=-\frac{1 \pm \sqrt{7}}{6} \end{aligned} \]

Case $\mathrm{I}: \mathrm{a}=\frac{\sqrt{7}-1}{6} \& \mathrm{~b}=\frac{1-\sqrt{7}}{6} \Rightarrow[\mathrm{a}]=0 \therefore \frac{1+[\mathrm{a}]}{4 \mathrm{~b}}=\frac{6}{4(1-\sqrt{7})}=-\left(\frac{1+\sqrt{7}}{4}\right)$
Case II: $\mathrm{a}=-\frac{1-\sqrt{7}}{6}$ then $[\mathrm{a}]=-1 \Rightarrow \frac{1+[\mathrm{a}]}{4 \mathrm{~b}}=0$ No answer is matching

[UPLIFTJEE]

Answer

Bonus

[UPLIFTJEE]

Solution

$\left|\mathrm{z}-\mathrm{z}_{1}\right|^{2}-\left|\mathrm{z}-\mathrm{z}_{2}\right|^{2}=\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|^{2}$

\[ \begin{aligned} & \Rightarrow\left|(\mathrm{x}-2)^{2}+(\mathrm{y}-3)^{2}\right|-\left|(\mathrm{x}-3)^{2}+(\mathrm{y}-4)^{2}\right|=\left|(2-3)^{2}+(3-4)^{2}\right|=2 \\ & \Rightarrow 2 \mathrm{x}+2 \mathrm{y}=14 \Rightarrow \mathrm{x}+\mathrm{y}=7 \text { (straight line) Sum of intercepts is } 14 \end{aligned} \]

[UPLIFTJEE]

Answer

D

[UPLIFTJEE]

Solution

According to the question, $\operatorname{Re}\left(\frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}\right)=0 \Rightarrow \frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}+\overline{\left(\frac{\mathrm{z}-2 \mathrm{i}}{\mathrm{z}+2 \mathrm{i}}\right)}=0$

\[ \begin{aligned} & \Rightarrow \frac{z-2 i}{z+2 i}+\frac{\bar{z}+2 i}{\bar{z}-2 i}=0 \Rightarrow(z-2 i)(\bar{z}-2 i)+(\bar{z}+2 i)(z+2 i)=0 \\ & \Rightarrow z \bar{z}-2 i \bar{z}-2 i z+4(-1)+z \bar{z}+2 z i+2 \bar{z} i+4(-1)=0 \\ & \Rightarrow 2|z|^{2}=8 \Rightarrow|z|=2 \Rightarrow|z-(6+8 i)|_{\text {maximum }}=10+2=12 \end{aligned} \]

[UPLIFTJEE]

Answer

C

[UPLIFTJEE]

Solution

Since $\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right| \leq\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}\right|+\left|\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right|=\frac{\left|\mathrm{z}_{1}\right|}{\left|\mathrm{z}_{1}\right|}+\frac{\left|\mathrm{z}_{2}\right|}{\left|\mathrm{z}_{2}\right|} \quad \Rightarrow\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2} \mid}{\left|\mathrm{z}_{2}\right|}\right| \leq 2$
$\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right)\left(\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right|\right) \leq 2\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right) \therefore$ statement I is correct
For Statement II : $\because \frac{\mathrm{a}}{|\mathrm{y}-\mathrm{z}|}=\frac{\mathrm{b}}{|\mathrm{z}-\mathrm{x}|}=\frac{\mathrm{c}}{|\mathrm{x}-\mathrm{y}|} \Rightarrow \frac{\mathrm{a}^{2}}{|\mathrm{y}-\mathrm{z}|^{2}}=\frac{\mathrm{b}^{2}}{|\mathrm{z}-\mathrm{x}|^{2}}=\frac{\mathrm{c}^{2}}{|\mathrm{x}-\mathrm{y}|^{2}}=\lambda$
$\mathrm{a}^{2}=\lambda\left(|\mathrm{y}-\mathrm{z}|^{2}\right)=\lambda(\mathrm{y}-\mathrm{z})(\overline{\mathrm{y}}-\overline{\mathrm{z}}) \Rightarrow \mathrm{b}^{2}=\lambda(\mathrm{z}-\mathrm{x})(\overline{\mathrm{z}}-\overline{\mathrm{x}})$ and $\mathrm{c}^{2}=\lambda(\mathrm{x}-\mathrm{y})(\overline{\mathrm{x}}-\overline{\mathrm{y}})$
Now $\frac{\mathrm{a}^{2}}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^{2}}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^{2}}{\mathrm{x}-\mathrm{y}}=\lambda(\overline{\mathrm{y}}-\overline{\mathrm{z}}+\overline{\mathrm{z}}-\overline{\mathrm{x}}+\overline{\mathrm{x}}-\overline{\mathrm{y}})=0$ Statement II is false

[UPLIFTJEE]

Answer

A

[UPLIFTJEE]

Solution

According to the question, $\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 25$

\[ \begin{aligned} & \mathrm{S}_{2}: \operatorname{Im}\left(\frac{\mathrm{z}+(1-\sqrt{3} \mathrm{i})}{(1-\sqrt{3} \mathrm{i})}\right) \geq 0 \Rightarrow \operatorname{Im}\left(\frac{\mathrm{x}+\mathrm{iy}}{1-\sqrt{3} \mathrm{i}}+1\right) \geq 0 \Rightarrow \operatorname{Im}\left(\frac{(\mathrm{x}+\mathrm{iy})(1+\sqrt{3} \mathrm{i})}{4}+1\right) \geq 0 \\ & \Rightarrow \frac{\mathrm{y}+\sqrt{3} \mathrm{x}}{4} \geq 0 \Rightarrow \sqrt{3} \mathrm{x}+\mathrm{y} \geq 0 \quad \ldots(2) \end{aligned} \]

\[ \mathrm{S}_{3}: \mathrm{x} \geq 0 \quad \ldots(3) \text { Shaded area }=\left(\frac{150^{\circ}}{360^{\circ}}\right) \pi(5)^{2}=\frac{5}{12}\left(\pi(5)^{2}\right) \]

[UPLIFTJEE]

Answer

C

[UPLIFTJEE]

Solution

$\left|\mathrm{z}_{1}-2 \mathrm{z}_{2}\right|=2\left|\frac{1}{2}-\mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right|=\left|1-2 \mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right| \Rightarrow\left|\mathrm{z}_{1}-2 \mathrm{z}_{2}\right|^{2}=\left|1-2 \mathrm{z}_{1} \overline{\mathrm{z}}_{2}\right|^{2}\left(\because|\mathrm{z}|^{2}=\mathrm{z} \overline{\mathrm{z}}\right)$

\[ \begin{aligned} & \Rightarrow\left(z_{1}-2 z_{2}\right)\left(\bar{z}_{1}-2 \bar{z}_{2}\right)=\left(1-2 z_{1} \bar{z}_{2}\right)\left(1-2 \bar{z}_{1} z_{2}\right) \Rightarrow z_{1} \bar{z}_{1}-2 z_{2} \bar{z}_{1}-2 z_{1} \bar{z}_{2}+4 z_{2} \bar{z}_{2}=1-2 z_{1} \bar{z}_{2}-2 \bar{z}_{1} z_{2}+4 z_{1} \bar{z}_{1} z_{2} \bar{z}_{2} \\ & \Rightarrow 4\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-\left|z_{1}\right|^{2}-4\left|z_{2}\right|^{2}+1=0 \Rightarrow\left(\left|z_{1}\right|^{2}-1\right)\left(4\left|z_{2}\right|^{2}-1\right)=0 \Rightarrow\left|z_{1}\right|=1 \text { or }\left|z_{2}\right|=\frac{1}{2} \end{aligned} \]

[UPLIFTJEE]

Answer

D

[Author]

Posts