IIT-JEE 2011 · Paper 2 · Mathematics Q45

MathematicsSingle correct+3 / -1Question 45 of 60

If
$\lim_{x\to0}\left[1+x\ln(1+b^2)\right]^{\frac1x}=2b\sin^2\theta,\quad b>0\ \text{and}\ \theta\in(-\pi,\pi],$
then the value of $\theta$ is

\pm\dfrac\pi4

\pm\dfrac\pi3

\pm\dfrac\pi6

\pm\dfrac\pi2

Answer and worked solution

Official answer: D

Limit. $\displaystyle\lim_{x\to0}\left[1+x\ln(1+b^2)\right]^{1/x}=e^{\ln(1+b^2)}=1+b^2$ .

Equation. $\sin^2\theta=\dfrac{1+b^2}{2b}\ge1$ by AM-GM, so equality: $b=1$ and $\sin^2\theta=1$ , giving $\theta=\pm\frac\pi2$ .

(D)

Topics:Limits AM-GM-HM inequalities

IIT-JEE 2011 · Paper 2, Mathematics question 45