JEE Advance - Mathematics (2016 - Paper 1 Offline - No. 1)
Let $$S = \left\{ {x \in \left( { - \pi ,\pi } \right):x \ne 0, \pm {\pi \over 2}} \right\}.$$ The sum of all distinct solutions of the equation $$\sqrt 3 \,\sec x + \cos ec\,x + 2\left( {\tan x - \cot x} \right) = 0$$ in the set S is equal to
$$ - {{7\pi } \over 9}$$
$$ - {{2\pi } \over 9}$$
0
$${{5\pi } \over 9}$$
Explanation
Let us consider
$$ S=\left\{x \in(-\pi, \pi), x \neq 0, \pm \frac{\pi}{2}\right\} $$
The given equation is
$$ \begin{aligned} & \sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0 \\\\ & \Rightarrow \frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0 \\\\ & \Rightarrow \sqrt{3} \sin x+\cos x+2\left(\sin ^2 x-\cos ^2 x\right)=0 \\\\ & \Rightarrow \sqrt{3} \sin x+\cos x=2 \cos 2 x \\\\ & \Rightarrow \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos 2 x \\\\ & \Rightarrow \cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}=\cos 2 x \\\\ & \Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right) \\\\ & \Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right) \quad\quad(x \in I) \end{aligned} $$
Case 1: When
$$ 2 x=2 n \pi+x-\frac{\pi}{3}, $$
we have $x=2 n \pi-\frac{\pi}{3}$.
If $n=0$, we get $x=-\frac{\pi}{3}$.
If $n=1$, we get $x=2 \pi-\frac{\pi}{3}$.
If $n=-1, x=-2 \pi-\frac{\pi}{3}$.
Case 2: When
$2 x=2 n \pi-x+\frac{\pi}{3}$, we get $x=\frac{2 n \pi}{3}+\frac{\pi}{9}$.
If $n=0$, we get $x=\frac{\pi}{9}$.
If $n=1$, we get $x=\frac{2 \pi}{3}+\frac{\pi}{9}$.
If $n=2$, we get $n=2 x=\frac{4 \pi}{3}+\frac{\pi}{9}$.
If $n=-1$, we get $x=\frac{-2 \pi}{3}+\frac{\pi}{9}$.
Therefore, the sum of all distinct solutions of the given equation is
$$ \frac{-\pi}{3}+\frac{\pi}{9}+\frac{2 \pi}{3}+\frac{\pi}{9}-\frac{2 \pi}{3}+\frac{\pi}{9}=0 $$
$$ S=\left\{x \in(-\pi, \pi), x \neq 0, \pm \frac{\pi}{2}\right\} $$
The given equation is
$$ \begin{aligned} & \sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0 \\\\ & \Rightarrow \frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0 \\\\ & \Rightarrow \sqrt{3} \sin x+\cos x+2\left(\sin ^2 x-\cos ^2 x\right)=0 \\\\ & \Rightarrow \sqrt{3} \sin x+\cos x=2 \cos 2 x \\\\ & \Rightarrow \frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos 2 x \\\\ & \Rightarrow \cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}=\cos 2 x \\\\ & \Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right) \\\\ & \Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right) \quad\quad(x \in I) \end{aligned} $$
Case 1: When
$$ 2 x=2 n \pi+x-\frac{\pi}{3}, $$
we have $x=2 n \pi-\frac{\pi}{3}$.
If $n=0$, we get $x=-\frac{\pi}{3}$.
If $n=1$, we get $x=2 \pi-\frac{\pi}{3}$.
If $n=-1, x=-2 \pi-\frac{\pi}{3}$.
Case 2: When
$2 x=2 n \pi-x+\frac{\pi}{3}$, we get $x=\frac{2 n \pi}{3}+\frac{\pi}{9}$.
If $n=0$, we get $x=\frac{\pi}{9}$.
If $n=1$, we get $x=\frac{2 \pi}{3}+\frac{\pi}{9}$.
If $n=2$, we get $n=2 x=\frac{4 \pi}{3}+\frac{\pi}{9}$.
If $n=-1$, we get $x=\frac{-2 \pi}{3}+\frac{\pi}{9}$.
Therefore, the sum of all distinct solutions of the given equation is
$$ \frac{-\pi}{3}+\frac{\pi}{9}+\frac{2 \pi}{3}+\frac{\pi}{9}-\frac{2 \pi}{3}+\frac{\pi}{9}=0 $$
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