Question

Area of the region enclosed by the solutions of the inequality $(1.25{)}^{{\log }_3|z-1|}\le (0.64{)}^{{\log }_{1/9}2}\text{and}Re\left(Z\right)\le 0$ is

• A. $(\frac{4\pi }{3}-\sqrt{3}){\text{unit}}^2$
• B. $4\frac{\pi }{3}{\text{unit}}^2$
• C. $\sqrt{3}{\text{unit}}^2$
• D. $(\frac{4\pi }{3}+\sqrt{3}){\text{unit}}^2$

Answer 1

The correct option is A $(\frac{4\pi }{3}-\sqrt{3}){\text{unit}}^2$

${\left(\frac{125}{100}\right)}^{{\log }_3|z-1|}\le {\left(\frac{64}{100}\right)}^{{\log }_{\frac{1}{9}}2}$


or ${\left(\frac{5}{4}\right)}^{{\log }_3|z-1|}\le {\left(\frac{5}{4}\right)}^{-2{\log }_{\frac{1}{9}}2}$

or ${\left(\frac{5}{4}\right)}^{{\log }_3|z-1|}\le {\left(\frac{5}{4}\right)}^{{\log }_{3}2}$

$\Rightarrow {\log }_3|z-1|\le {\log }_{3}2$

$\Rightarrow |z-1|\le 2$

$\Rightarrow$ z lies inside the circle of radius 2 with centre $\equiv (1,0)$

Also, $\because Re\left(z\right)<0$

$\Rightarrow$ Area of the required region is the shaded area which is equal to $=ArCPSQC-Ar\Delta CPQ$

Equation of the circle is $|z–1|=2$

$\Rightarrow (x-1{)}^2+y^2=4$

On putting x = 0 we get,

$y^2=3$

$\therefore y=\pm \sqrt{3}$

$\Rightarrow p\equiv (0,\sqrt{3})\text{and}Q\equiv (0,-\sqrt{3})$

$\Rightarrow \angle PCO=\frac{\pi }{3}$

$\Rightarrow ArCPSQC=\frac{4\pi }{3}\text{sq. unit}$

Now, $Ar\Delta CPQ=\frac{1}{2}\times 1\times 2\sqrt{3}=\sqrt{3}$

$\therefore$ The sought result $=(\frac{4\pi }{3}-\sqrt{3})$ sq. unit