Question

Match the statements given in Column I with the values given in Column II

Column - I	Column - II
(A) If $\overrightarrow{a}=\hat{j}+\sqrt{3}\hat{k},\overrightarrow{b}=-\hat{j}+\sqrt{3}\hat{k}\text{and}\overrightarrow{c}=2\sqrt{3}\hat{k}$ form a triangle, then internal angle of the triangle between $\overrightarrow{a}\text{and}\overrightarrow{b}$ is	(p) $\frac{\pi }{6}$
(B) If ${\int }_a^b\left(f\right(x)-3x)dx=a^2-b^2,$ then the value of $f\left(\frac{\pi }{6}\right)$ is	(q) $\frac{2\pi }{3}$
(C) The value of $\frac{{\pi }^2}{\ln 3}{\int }_{\frac{7}{6}}^{\frac{5}{6}}\sec \left(\pi x\right)dx$ is	(r) $\frac{\pi }{3}$
(D) the maximum value of $\left|Arg\left(\frac{1}{1-z}\right)\right||z|=1,z\ne 1$ is given by	(s) $\pi$
	(t) $\frac{\pi }{2}$

• A.
• B. A(r), B(p), C(s), D(t*)

Answer 1

The correct option is B A(r), B(p), C(s), D(t*)

(A) We have,
$\overrightarrow{a}=\hat{j}+\sqrt{3}\hat{k}$
$\overrightarrow{b}=\hat{j}+\sqrt{3}\hat{k}$
$\overrightarrow{c}=2\sqrt{3}\hat{k}$

We observe that
$\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}.\overrightarrow{b}.=|\overrightarrow{c}{|}^2$
$\Rightarrow 4+4+8\cos \theta =12$
$\cos \theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$

(B) We have,
${\int }_a^b\left(f\right(x)-3x)dx=a^2-b^2$
Keeping a constant and differentiating both sides w.r.t. b, we get
$f\left(b\right)-3b=-2b$
$\Rightarrow f\left(b\right)=b$
$\Rightarrow f\left(\frac{\pi }{6}\right)=\frac{\pi }{6}$

(C) Using $\int \sec \pi xdx=\frac{1}{\pi }\ln |\sec \pi x+\tan \pi x|,$ we get
$\frac{{\pi }^2}{\ln 3}{\int }_{\frac{7}{6}}^{\frac{5}{6}}\sec \pi xdx=\pi$

(D) $\left|\arg \left(\frac{1}{1-z}\right)\right|=|\arg \left(1\right)-\arg (1-z)|$
$=|\arg (1-z)|$
But z lies on $|z|=1$

Hence it is obvious from geometry that $\arg (1–z)$ can be very-very close to $\frac{\pi }{2}$, if the chosen complex number is very close to 1.
Let us suppose that $z=\cos \theta +i\sin \theta$
$\frac{1}{1-z}=\frac{1}{2}(1+i\cot \frac{\theta }{2})$
Now $-\pi <\theta \le \pi ,\theta \ne 0$
$\arg \left(\frac{1}{1-z}\right)=\frac{\pi }{2}-\frac{\theta }{2},0<\theta \le \frac{\pi }{2}$
and, $\arg \left(\frac{1}{1-z}\right)=\pi -(\frac{\pi }{2}-\frac{\theta }{2}),-\frac{\pi }{2}<\theta <\theta$
$=\frac{\pi }{2}+\frac{\theta }{2},-\frac{\pi }{2}<\theta <0$

So in both the cases we can observe that $\arg \left(\frac{1}{1-z}\right)$ can take values which are infinitesimally closer to $\frac{\pi }{2}$, but given the statement of the problem, this value can’t be achieved.

It can be seen that if one takes z to be a complex number which is very-very close to $(1+0i)$, the argument of $\left(\frac{1}{1-z}\right)$ will be very close to $\frac{\pi }{2}$.