IIT-JEE 2011 · Paper 2, Mathematics question 59

• A $\vec a+\vec b=\vec c$, so placed head to tail they form the triangle. $\cos\angle(\vec a,\vec b)=\frac{-1+3}{2\cdot2}=\frac12$, an angle of $\frac\pi3$ between the vectors. The internal angle is $\pi-\frac\pi3=\frac{2\pi}3$. → Q.
• B If the identity holds for all $a$, $b$, differentiating in $b$ gives $f(b)-3b=-2b$, so $f(x)=x$ and $f\left(\frac\pi6\right)=\frac\pi6$. → P. (If $a$ and $b$ are fixed numbers, $f$ is not determined; hence the key also accepts all of P-T.)
• C $\displaystyle\int\sec\pi x\,dx=\frac1\pi\ln|\sec\pi x+\tan\pi x|$. At $x=\frac56$ this is $\frac1\pi\ln\sqrt3$; at $x=\frac76$, $\frac1\pi\ln\frac1{\sqrt3}$. The integral is $\frac{\ln3}\pi$, and the value is $\frac{\pi^2}{\ln3}\cdot\frac{\ln3}\pi=\pi$. → S.
• D For $z=e^{i\theta}$, $\arg(1-z)=\frac{\theta-\pi}2\in\left(-\frac\pi2,\frac\pi2\right)$, so $|\operatorname{Arg}\frac1{1-z}|$ approaches $\frac\pi2$ (its supremum). → T.

Answer: A → Q; B → P (or P, Q, R, S, T); C → S; D → T