MCQ
Match the statements given in Column $I$ with the values given in Column $II$
Column $I$ Column $II$
$(A)$ If $\vec{a}=\hat{j}+\sqrt{3} \hat{k}, \vec{b}=-\hat{j}+\sqrt{3} \hat{k}$ and $\vec{c}=2 \sqrt{3} \hat{k}$ form a triangle, then the internal angle of the triangle between $\vec{a}$ and $\vec{b}$ is $(p)$ $\frac{\pi}{6}$
$(B)$ If $\int_a^b(f(x)-3 x) d x=a^2-b^2$, then the value of $f\left(\frac{\pi}{6}\right)$ is $(q)$ $\frac{2 \pi}{3}$
$(C)$ The value of $\frac{\pi^2}{\ln 3} \int_{1 / 6}^{5 / 6} \sec (\pi x) d x$ is $(r)$ $\frac{\pi}{3}$
$(D)$ The maximum value of $\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right|$ for $|z|=1, \quad z \neq 1$ is given by $(s)$ $\pi$
  $(t)$ $\frac{\pi}{2}$
  • $(A) \rightarrow q (B) \rightarrow p\  or \ p, q, r, s \ and \ t (C) \rightarrow  s(D) \rightarrow t$
  • B
    $(A) \rightarrow p (B) \rightarrow p\  or \ p, q, s \ and \ t (C) \rightarrow  t(D) \rightarrow r$
  • C
    $(A) \rightarrow s (B) \rightarrow q\  or \ p, q, r, s \ and \ t (C) \rightarrow  q(D) \rightarrow p$
  • D
    $(A) \rightarrow s (B) \rightarrow r\  or \ p, q, r, s \ and \ t (C) \rightarrow  q(D) \rightarrow s$

Answer

Correct option: A.
$(A) \rightarrow q (B) \rightarrow p\  or \ p, q, r, s \ and \ t (C) \rightarrow  s(D) \rightarrow t$
a
$(A)$ $|\vec{a}|=2,|\vec{b}|=2,|\vec{c}|=2 \sqrt{3}$

Angle between $\vec{a}$ and $b$ is

$\cos \theta=\frac{\left|\hat{a}^2+\right| \vec{b}^2-\vec{c}^2}{2|\vec{a}||\vec{b}|}=\frac{4+4-12}{2 \times 2 \times 2}=-\frac{1}{2}$

$\Rightarrow \theta=\frac{2 \pi}{3}$

$(B)$ $\int_a^b(f(x)-3 x) d x=a^2-b^2$

$\Rightarrow \int_a^b f(x) d x-\int_a^b 3 x d x=a^2-b^2$

$\Rightarrow \int_a^b f(x) d x=\frac{b^2-a^2}{2}$

$\Rightarrow f(x)=x$

$\Rightarrow f\left(\frac{\pi}{6}\right)=\frac{\pi}{6}$

$(C)$ $I=\frac{\pi^2}{\ln 3} \int_{7 / 6}^{5 / 6} \sec (\pi x) d x$

$=\frac{\pi^2}{\ln 3}[\ln |(\sec (\pi x)+\tan (\pi x))|]_{7 / 6}^{5 / 6}=\frac{\pi^2}{\ln 3} \times \frac{1}{\pi} \ln 3$

$=\pi$

$(D)$ Let $u=\frac{1}{1-z}$

$\Rightarrow z=1-\frac{1}{u}$

Given that $|z|=1 \Rightarrow\left|1-\frac{1}{u}\right|=1$

$\Rightarrow|u-1|=|u|$

$\therefore$ locus of $u$ is perpendicular bisector of line segment joining $(1,0)$ and $(0,0)$ in complex plane.

$\Rightarrow$ maximum argument approaches $\frac{\pi}{2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free