Question

In the given figure, $AC$ and $BC$ are two tangents to the circle at the points $A$ and $B,$ respectively. $f\left(\theta \right)$ be the area of triangular region $\Delta ABO$. The triangular region $\Delta ABC$ is divided into two regions, one outside the circle whose area is $h\left(\theta \right)$ and another one inside the circle whose area is $g\left(\theta \right)$. $(\theta =\angle AOB)$
$$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I)}\underset{\theta \to 0}{lim}\frac{f\left(\theta \right)}{g\left(\theta \right)}& \text{(P)}-2\\ \text{(II)}\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{h\left(\theta \right)}& \text{(Q)}-1\\ \text{(III)}\underset{\theta \to 0}{lim}\frac{h\left(\theta \right)}{f\left(\theta \right)}& \text{(R)}0\\ \text{(IV)}\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}& \text{(S)}1\\ & \text{(T)}2\\ & \text{(U)}\text{Does not exist}\end{array}$$
Which of the following is the only INCORRECT combination?

• A. $\left(I\right),\left(U\right)$
• B. $\left(II\right),\left(T\right)$
• C. $\left(III\right),\left(P\right)$
• D. $\left(IV\right),\left(R\right)$

Answer 1

The correct option is C $\left(III\right),\left(P\right)$


$\text{Area of}\Delta ABO=\frac{1}{2}\times \text{Base}\times \text{Height}=\frac{1}{2}\left(2r\sin \theta /2\right)\left(r\cos \theta /2\right)=\frac{1}{2}{r}^2\sin \theta$


$f\left(\theta \right)=\frac{1}{2}{r}^2\sin \theta$
$g\left(\theta \right)=\frac{1}{2}{r}^2\theta -\frac{1}{2}{r}^2\sin \theta =\frac{1}{2}{r}^2(\theta -\sin \theta )$

$h\left(\theta \right)=\text{Area of OACB - Area of circular arc OAB}=\frac{1}{2}\times r\times R-\frac{1}{2}\times r^2\times \theta =\frac{1}{2}\times r\times r\tan \frac{\theta }{2}-\frac{1}{2}\times r^2\times \theta (\because R=r\tan \frac{\theta }{2})$
$=\frac{1}{2}{r}^2[2\tan \theta /2-\theta ]$

$\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}=\underset{\theta \to 0}{lim}\frac{\theta -\sin \theta }{\sin \theta }$
$\text{Applying L'Hospital's Rule}$
$=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{\cos \theta }=\frac{1-1}{1}=0$

$\underset{\theta \to 0}{lim}\frac{f\left(\theta \right)}{g\left(\theta \right)}=\frac{1}{\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{f\left(\theta \right)}}=\frac{1}{0}\Rightarrow \text{Doesn't exist}$

$\underset{\theta \to 0}{lim}\frac{g\left(\theta \right)}{h\left(\theta \right)}=\underset{\theta \to 0}{lim}\frac{\theta -\sin \theta }{2\tan \theta /2-\theta }$
$\text{Applying L'Hospital's Rule}$
$=\underset{\theta \to 0}{lim}\frac{1-\cos \theta }{{\sec }^2\theta /2-1}=\underset{\theta \to 0}{lim}\frac{\sin \theta }{{\sec }^2\theta /2.\tan \theta /2}=\underset{\theta \to 0}{lim}\frac{2}{{\sec }^2\theta /2}=2$

$\underset{\theta \to 0}{lim}\frac{h\left(\theta \right)}{f\left(\theta \right)}=\underset{\theta \to 0}{lim}\frac{2\tan \theta /2-\theta }{\sin \theta }$
$\text{Applying L'Hospital's Rule}$
$=\underset{\theta \to 0}{lim}\frac{{\sec }^2\theta /2-1}{\cos \theta }=\frac{1-1}{1}=0$