Question

If $g\left(x\right)={(4{\cos }^{4}x-2\cos 2x-\frac{\cos 4x}{2}-x^7)}^{1/7}$ for $x\in R,$ then the value of $g\left(g\left(100\right)\right)$ is equal to

• A. value of $50\left(\frac{{\cos }^2{1}^{\circ }-{\cos }^2{2}^{\circ }}{\sin 3^{\circ }\sin 1^{\circ }}\right)$
• B. value of $\frac{200}{3}\sum _{n=1}^{\infty }\left(\frac{5^n-2^n}{{10}^n}\right)$
• C. value of $\frac{400}{3}\sum _{n=1}^{\infty }\left(\frac{5^n-2^n}{{10}^n}\right)$
• D. value of $50\left({12}^{\left(\frac{1-a-b}{2(1-b)}\right)}\right)$ if ${60}^a=3$ and ${60}^b=5$

Answer 1

Answer: (C), (D)

value of $50\left({12}^{\left(\frac{1-a-b}{2(1-b)}\right)}\right)$ if ${60}^a=3$ and ${60}^b=5$

$g\left(x\right)={({\left(2{\cos }^{2}x\right)}^2-2\cos 2x-\frac{\cos 4x}{2}-x^7)}^{1/7}={({\left(2{\cos }^{2}x\right)}^2-2\cos 2x-\frac{2{\cos }^2\left(2x\right)-1}{2}-x^7)}^{1/7}={({(1+\cos 2x)}^2-2\cos 2x-{\cos }^2(2x)+\frac{1}{2}-x^7)}^{1/7}={({(1+\cos 2x)}^2-{(1+\cos 2x)}^2+\frac{3}{2}-x^7)}^{1/7}$
$\Rightarrow g\left(x\right)={(\frac{3}{2}-x^7)}^{1/7}$

$g\left(g\left(x\right)\right)={[\frac{3}{2}-{\left\{{(\frac{3}{2}-x^7)}^{1/7}\right\}}^7]}^{1/7}=x$
$g\left(g\left(100\right)\right)=100$

Now, $50\left(\frac{{\cos }^2{1}^{\circ }-{\cos }^2{2}^{\circ }}{\sin 3^{\circ }\sin 1^{\circ }}\right)$
$=50\left(\frac{{\sin }^2{2}^{\circ }-{\sin }^2{1}^{\circ }}{\sin 3^{\circ }\sin 1^{\circ }}\right)=50\left(\frac{\sin 3^{\circ }\sin 1^{\circ }}{\sin 3^{\circ }\sin 1^{\circ }}\right)=50$


$\frac{400}{3}\sum _{n=1}^{\infty }({\left(\frac{5}{10}\right)}^n-{\left(\frac{2}{10}\right)}^n)=\frac{400}{3}\left[\sum _{n=1}^{\infty }({\left(\frac{1}{2}\right)}^n-{\left(\frac{1}{5}\right)}^n)\right]=\frac{400}{3}[\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{1}{5}}{1-\frac{1}{5}}]=\frac{400}{3}\times \frac{3}{4}=100$

$a={\log }_{60}3$ and $b={\log }_{60}5$
$\frac{1-a-b}{2(1-b)}$
$=\frac{{\log }_{60}60-{\log }_{60}15}{2{\log }_{60}12}$
$=\frac{{\log }_{60}4}{{\log }_{60}{12}^2}$
$={\log }_{{12}^2}4={\log }_{12}2$

$\therefore 50\left({12}^{\left(\frac{1-a-b}{2(1-b)}\right)}\right)$
$=50\left({12}^{{\log }_{12}2}\right)$
$=50\cdot 2=100$