Question

Assertion :$sin\left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+4$ has only one solution Reason: The smallest positive value of x in degrees, for which $tan(x+{100}^o)=tan(x+{50}^o)tanxtan(x-{50}^o)$ is ${30}^o$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Reason:
$\tan (x+100)=\tan (x+50)\tan x\tan (x-50)$
$\Rightarrow \frac{\tan (x+100)}{\tan x}=\tan (x+50)\tan (x-50)$
$\Rightarrow \frac{\sin (x+100)}{\cos (x+100)}.\frac{\cos x}{\sin x}=\frac{\sin (x+50)\sin (x-50)}{\cos (x+50)\cos (x-50)}$
$\Rightarrow \frac{\sin (2x+100)+\sin 100}{\sin (2x+100)-\sin 100}=\frac{\cos 100-\cos 2x}{\cos 100+\cos 2x}$
$\Rightarrow (\sin (2x+100)+\sin 100)(\cos 100+\cos 2x)=(\cos 100-\cos 2x)(\sin (2x+100)-\sin 100)\Rightarrow \sin (2x+100)\cos 100+\sin (2x+100)\cos 2x+\sin 100\cos 100+\sin 100\cos 2x=\cos 100\sin (2x+100)-\cos 100\sin 100-\cos 2x\sin (2x+100)+\cos 2x\sin 100\Rightarrow 2\sin (2x+100)\cos 2x+2\sin 100\cos 100=0\Rightarrow \sin (4x+100)+\sin 100+\sin 200=0\Rightarrow \sin (4x+100)+2\sin 150\cos 50=0\Rightarrow \sin (4x+100)+\sin (90-50)=0\Rightarrow \sin (4x+100)=-\sin 40$
$\Rightarrow x=\frac{1}{4}(n\pi -{(-1)}^n(-40))$
For $n=0$ $x=30$
Assertion
$\sin \left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+4={(x-\sqrt{3})}^2+1$
Since R.H.S >1
So, the solution exists if and only if $x-\sqrt{3}=0\Rightarrow x=\sqrt{3}$