Answer:
Option A
Explanation:
(A) $\therefore$ $|a|=\sqrt{1+3}=2$
$|b|= \sqrt{1+3}=2$
$ |c|=\sqrt{12}=2 \sqrt{3}$
Using Cosine law,
$\cos C= \frac{|a|^{2}+|b|^{2}-|c|^{2}}{2|a||b|}$
= $\frac{4+4-12}{2 \times 2 \times 2}=\frac{-4}{8}=\frac{-1}{2}$
$\Rightarrow$ $\angle C=120^{0}= \frac{2 \pi}{3}$
(B) $\int_{a}^{b} f(x) dx-3\left(\frac{x^{2}}{2}\right)_{a}^{b}=(a^{2}-b^{2})$
$\Rightarrow \int_{a}^{b} f(x) dx-\frac{3}{2}(b^{2}-a^{2})=(a^{2}-b^{2})$
$\Rightarrow \int_{a}^{b} f(x) dx=(a^{2}-b^{2})+\frac{3}{2}(b^{2}-a^{2})$
= $\frac{b^{2}-a^{2}}{2}$
$\Rightarrow$ $\Rightarrow \int_{a}^{b} f(x) dx=\frac{b^{2}-a^{2}}{2}$
$f(x)=x \Rightarrow f(\frac{\pi}{6})=\frac{\pi}{6}$
(C) $\frac{\pi^{2}}{\log_{e}3}\int_{7/6}^{5/6} \sec(\pi x) dx$
$\Rightarrow\frac{\pi^{2}}{\log_{e}3}\left\{\frac{\log|\sec \pi x+\tan \pi x|}{\pi}\right\}_{7/6}^{5/6}$
$\Rightarrow\frac{\pi}{\log 3}\left\{ \log |\sec \frac{5 \pi}{6}+\tan \frac{5 \pi}{6}|-\log | \sec \frac{7 \pi}{6}+\tan \frac{7 \pi}{6}|\right\}$
$\Rightarrow$ $\frac{\pi}{\log 3}\left\{ \log |\sqrt{3}|-\log |\frac{1}{\sqrt{3}}|\right\}$
$\Rightarrow$ $\frac{\pi}{\log 3} { \log 3}= \pi$
(D) $|arg \frac{1}{(1-z)}|, for |z|=1$
$\Rightarrow$ $|arg (1-z)^{-1}|$
$\Rightarrow$ $|- arg (1-z)| \Rightarrow |arg (1-z)|$
from Figure , arg (z-1) is maximum = $\pi$
