Question

If $\cos \frac{x}{2}+\cos \frac{x}{2^2}+\cos \frac{x}{2^3}+\dots +\cos \frac{x}{2^n}=f\left(x\right)$ , then $\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\dots +\frac{1}{2^n}\tan \frac{x}{2^n}=$

• A. $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$
• B. $\frac{f\left(x\right)}{f^{\prime }\left(x\right)}$
• C. $\frac{-f^{\prime }\left(x\right)}{f\left(x\right)}$
• D. $0$

Answer 1

The correct option is C $\frac{-f^{\prime }\left(x\right)}{f\left(x\right)}$

$\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}....\cos \frac{x}{2^n}=f\left(x\right)$

LEt $\frac{1}{2^{}}\tan \frac{x}{2^{}}+\frac{1}{2^2}\tan \frac{x}{2^2}+....\frac{1}{2^n}\tan \frac{x}{2^n}==g\left(x\right)$

$\cos \frac{x}{2}=f_1$; $\cos \frac{x}{2^2}=f_2$; $\cos \frac{x}{2^3}=f_3$;.......$\cos \frac{x}{2^n}=f_n$

Differentiating all $f_1$ till $f_n$

$\frac{d}{dx}\left(\cos \frac{x}{2}\right)=-\frac{1}{2}\sin \frac{x}{2}$

.........................................

$\frac{d}{dx}\left(\cos \frac{x}{2^n}\right)=-\frac{1}{2^n}\sin \frac{x}{2^n}=f_n^{\prime }$

$\Rightarrow$ $-\frac{f_1^{\prime }}{f_1}=\frac{-(-\frac{1}{2}\sin \frac{x}{2^{}})}{\cos \left(\frac{x}{2}\right)}=\frac{1}{2}\tan \frac{x}{2^{}}=g_1$

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$-\frac{f_n^{\prime }}{f_n}=\frac{-(-\frac{1}{2^n}\sin \frac{x}{2^n})}{\cos \left(\frac{x}{2^n}\right)}=\frac{1}{2^n}\tan \frac{x}{2^n}=g_n$

$g_1^{\prime }.....g_n=-\frac{f_1^{\prime }}{f_1}+....\frac{f_n^{\prime }}{f_n}$

$g_n=\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$