IIT-JEE 2010 · Paper 2, Mathematics question 38

• A. The line from O meets $L_1$ in the plane through O and $L_1$: normal $(2,1,-1)\times(1,-2,1)\parallel(1,3,5)$, plane $x+3y+5z=0$. $L_2$ meets it at $u=\frac13$: $Q=\left(\frac{10}3,-\frac{10}3,\frac43\right)$. Likewise the plane through O and $L_2$ is $3x+y-5z=0$, which meets $L_1$ at $P=(5,-5,2)$. ($P=\frac32Q$, so O, P, Q are collinear.) Then $d^2=\left(\frac53\right)^2+\left(\frac53\right)^2+\left(\frac23\right)^2=6$. → T.
• B. $\tan^{-1}(x+3)-\tan^{-1}(x-3)=\tan^{-1}\dfrac6{x^2-8}$ (for $x^2>8$), and $\sin^{-1}\frac35=\tan^{-1}\frac34$. So $x^2-8=8$, $x=\pm4$. → P, R.
• C. Write $\vec c=\frac14(\vec a-\mu\vec b)$ and let $u=4-\mu$. The two conditions give $|\vec a|^2=u|\vec b|^2$ and $u^2-3u-4=0$, so $u=4$ or $-1$; but $|\vec a|^2>0$ needs $u>0$, so $u=4$ and $\mu=0$. → Q. (The rejected root $u=-1$ gives $\mu=5$, which is why the official key accepts "q, or q and s".)
• D. $\dfrac{\sin(9x/2)}{\sin(x/2)}=1+2\sum_{k=1}^4\cos kx$, whose integral over $[-\pi,\pi]$ is $2\pi$. So the value is $\frac2\pi\cdot2\pi=4$. → R.

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