Question

In a triangle $ABC$, angles $A,B,C$ are in A.P.Then
${lim}_{x\to c}\frac{\sqrt{(3-4\sin A\sin C)}}{|A-C|}$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\because A,B,C$ are in A.P
$\Rightarrow 2B=A+C$ and $A+B+C={180}^0$
$\Rightarrow (A+C)+B={180}^0$
$\Rightarrow 2B+B=3B={180}^0$ or $B={60}^0$
Using cosine rule,
$\therefore \cos B=\cos {60}^0\Rightarrow \frac{1}{2}=\frac{a^2+c^2-b^2}{2ac}$
$\Rightarrow a^2+c^2-b^2=ac$
$\Rightarrow a^2+c^2=b^2+ac$
$\Rightarrow {(a-c)}^2=b^2-ac$
$\Rightarrow |a-c|=\sqrt{(b^2-ac)}$
Using sine Rule,
$\Rightarrow |\sin A-\sin C|=\sqrt{{\sin }^{2}B-\sin A\sin C}$
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\left|\sin \left(\frac{A-C}{2}\right)\right|=\sqrt{\frac{3}{4}-\sin A\sin C}$
$\therefore {lim}_{x\to c}\frac{\sqrt{\frac{3}{4}-\sin A\sin C}}{|A-C|}={lim}_{x\to c}\frac{2\left|\sin \left(\frac{A-C}{2}\right)\right|}{\frac{|A-C|}{2}\times 2}=\left|1\right|=1$