NIMCET 2021 — Mathematics PYQ

1. Area under $y=\tan x$ from $-\frac{\pi }{3}$ to $\frac{\pi }{3}$:

Since $\tan x$ is an odd function and symmetric about the origin, the area above the X-axis for $x\in [0,\pi /3]$ is equal to the magnitude of the area below the X-axis for $x\in [-\pi /3,0]$.

$$A_1={\int }_{-\pi /3}^{\pi /3}|\tan x|dx=2{\int }_0^{\pi /3}\tan xdx$$

$$A_1=2[\ln |\sec x|{]}_0^{\pi /3}=2(\ln 2-\ln 1)=2\ln 2$$

2. Area under $y=\cot x$ from $\frac{\pi }{6}$ to $\frac{3\pi }{2}$:

The area is ${\int }_{\pi /6}^{3\pi /2}|\cot x|dx$. Note that $\cot x$ changes sign at $x=\pi$:

• For $x\in [\pi /6,\pi ]$, \cot x > 0.

• For $x\in [\pi ,3\pi /2]$, \cot x < 0.

  $$A_2={\int }_{\pi /6}^{\pi }\cot xdx+{\int }_{\pi }^{3\pi /2}(-\cot x)dx$$

  Evaluating these integrals (using limits to handle the vertical asymptotes):

  $$A_2=[\ln |\sin x|{]}_{\pi /6}^{\pi }-[\ln |\sin x|{]}_{\pi }^{3\pi /2}$$

  Both integrals diverge to infinity as they approach the asymptotes ($x=\pi$ for the first, $x=\pi$ for the second). Therefore, the area is not finite.

Final Answer

Since the integral of the area is divergent, the area is not represented by any of the finite values given in options A, B, or C.

The correct option is D (None of these).