$$ ext { Match the statements given in Column I with the values given in Column II. } $$

Question

$$ 	ext { Match the statements given in Column I with the values given in Column II. } $$

Aicharya · Accepted Answer

(A) $	herefore|\mathbf{a}|=\sqrt{1+3}=2$ $|\mathbf{b}|=\sqrt{1+3}=2$ $|\mathbf{c}|=\sqrt{12}=2 \sqrt{3}$ Using cosine law, $ \begin{aligned} \cos C & =\frac{|\mathbf{a}|^2+|\mathbf{b}|^2-|\mathbf{c}|^2}{2|\mathbf{a}||\mathbf{b}|} \ & =\frac{4+4-12}{2 	imes 2 	imes 2}=\frac{-4}{8}=\frac{-1}{2} \ \Rightarrow \quad \angle C & =120^{\circ}=\frac{2 \pi}{3} \end{aligned} $ (B) $ \begin{aligned} & \int_a^b f(x) d x-3\left(\frac{x^2}{2}ight)_a^b=\left(a^2-b^2ight) \ & \Rightarrow \int_a^b f(x) d x-\frac{3}{2}\left(b^2-a^2ight)=\left(a^2-b^2ight) \ & \Rightarrow \int_a^b f(x) d x=\left(a^2-b^2ight)+\frac{3}{2}\left(b^2-a^2ight) \ & =\frac{b^2-a^2}{2} \ & \Rightarrow \int_a^b f(x) d x=\frac{b^2-a^2}{2} \ & \end{aligned} $ $ f(x)=x \Rightarrow f\left(\frac{\pi}{6}ight)=\frac{\pi}{6} $ (C) $ \begin{aligned} & \frac{\pi^2}{\log _e 3} \int_{7 / 6}^{5 / 6} \sec (\pi x) d x \ & \Rightarrow \frac{\pi^2}{\log _e 3}\left\{\frac{\log |\sec \pi x+	an \pi x|}{\pi}ight\}_{7 / 6}^{5 / 6} \ & \Rightarrow \frac{\pi}{\log 3}\left\{\log \left|\sec \frac{5 \pi}{6}+	an \frac{5 \pi}{6}ight|ight. \ & \left.\quad-\log \left|\sec \frac{7 \pi}{6}+	an \frac{7 \pi}{6}ight|ight\} \ & \Rightarrow \frac{\pi}{\log 3}\left\{\log |\sqrt{3}|-\log \left|\frac{1}{\sqrt{3}}ight|ight\} \ & \Rightarrow \frac{\pi}{\log 3}\{\log 3\}=\pi \end{aligned} $ (D) $ \begin{aligned} & \left|\arg \frac{1}{(1-z)}ight|, 	ext { for }|z|=1 \ & \Rightarrow\left|\arg (1-z)^{-1}ight| \ & \Rightarrow|-\arg (1-z)| \Rightarrow|\arg (1-z)| \end{aligned} $ From figure, $\arg (z-1)$ is maximum $=\pi$

$$ \text { Match the statements given in Column I with the values given in Column II. } $$

Solution