Mathematics - Definite Integration Question with Solution | TestHub

(A) $\therefore |\mathbf{a}|=\sqrt{1+3}=2$ $|\mathbf{b}|=\sqrt{1+3}=2$ $|\mathbf{c}|=\sqrt{12}=2\sqrt{3}$
Using cosine law, $$\begin{array}{rl}\cos C& =\frac{|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-|\mathbf{c}{|}^2}{2|\mathbf{a}||\mathbf{b}|}\\ & =\frac{4+4-12}{2\times 2\times 2}=\frac{-4}{8}=\frac{-1}{2}\\ \Rightarrow \angle C& =120^{\circ }=\frac{2\pi }{3}\end{array}$$ (B) $$\begin{array}{rl}& {\int }_a^{b}f(x)dx-3{\left(\frac{x^2}{2}\right)}_a^b=\left(a^2-b^2\right)\\ & \Rightarrow {\int }_a^{b}f(x)dx-\frac{3}{2}\left(b^2-a^2\right)=\left(a^2-b^2\right)\\ & \Rightarrow {\int }_a^{b}f(x)dx=\left(a^2-b^2\right)+\frac{3}{2}\left(b^2-a^2\right)\\ & =\frac{b^2-a^2}{2}\\ & \Rightarrow {\int }_a^{b}f(x)dx=\frac{b^2-a^2}{2}\\ & \end{array}$$ $$f(x)=x\Rightarrow f\left(\frac{\pi }{6}\right)=\frac{\pi }{6}$$ (C) $$\begin{array}{rl}& \frac{{\pi }^2}{{\log }_{e}3}{\int }_{7/6}^{5/6}\sec (\pi x)dx\\ & \Rightarrow \frac{{\pi }^2}{{\log }_{e}3}{\left\{\frac{\log |\sec \pi x+\tan \pi x|}{\pi }\right\}}_{7/6}^{5/6}\\ & \Rightarrow \frac{\pi }{\log 3}\{\log \left|\sec \frac{5\pi }{6}+\tan \frac{5\pi }{6}\right|\\ & -\log \left|\sec \frac{7\pi }{6}+\tan \frac{7\pi }{6}\right|\}\\ & \Rightarrow \frac{\pi }{\log 3}\left\{\log |\sqrt{3}|-\log \left|\frac{1}{\sqrt{3}}\right|\right\}\\ & \Rightarrow \frac{\pi }{\log 3}\{\log 3\}=\pi \end{array}$$ (D) $$\begin{array}{rl}& \left|\arg \frac{1}{(1-z)}\right|,\text{ for }|z|=1\\ & \Rightarrow \left|\arg (1-z{)}^{-1}\right|\\ & \Rightarrow |-\arg (1-z)|\Rightarrow |\arg (1-z)|\end{array}$$
From figure, $\arg (z-1)$ is maximum $=\pi$.