Question

Let S be the set of all $\alpha \in R$ such that the equation, $\cos 2x+\alpha \sin x=2\alpha -7$ has a solution. Then S is equal to:

& A)\left[ 2,6 \right] \\\ & B)\left[ 3,7 \right] \\\ & C)R \\\ & D)\left[ 1,4 \right] \\\ \end{aligned}$$

Answer 1

Let us consider the equation $\cos 2x+\alpha \sin x=2\alpha -7$ as equation (1). We know that $\cos 2x=1-2{{\sin }^{2}}x$. Now let us consider this as equation (2). Now let us substitute equation (2) in equation (1). Let us assume this equation as equation (3). We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are equal to $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. By using this formula, we should find the roots of equation (3). We know that $-1\le \sin x\le 1$. Now by using this condition, we should find the elements of set S.

Complete step-by-step answer :
From the question, it is given that the equation $\cos 2x+\alpha \sin x=2\alpha -7$.
Let us consider
$\cos 2x+\alpha \sin x=2\alpha -7....(1)$
We know that $\cos 2x=1-2{{\sin }^{2}}x$.
$\cos 2x=1-2{{\sin }^{2}}x...(2)$
Now let us substitute equation (1) in equation (2), then we get

& \Rightarrow 1-2{{\sin }^{2}}x+\alpha \sin x=2\alpha -7 \\\ & \Rightarrow 2{{\sin }^{2}}x-\alpha \sin x+2\alpha -7-1=0 \\\ & \Rightarrow 2{{\sin }^{2}}x-\alpha \sin x+2\alpha -8=0....(3) \\\ \end{aligned}$$ We know that the roots of the quadratic equation $$a{{x}^{2}}+bx+c=0$$ are equal to $$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$. Now we should find the roots of equation (3). $$\begin{aligned} & \Rightarrow \sin x=\dfrac{\alpha \pm \sqrt{{{\alpha }^{2}}-4\left( 2 \right)\left( 2\alpha -8 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow \sin x=\dfrac{\alpha \pm \sqrt{{{\alpha }^{2}}-8\left( 2\alpha -8 \right)}}{4} \\\ & \Rightarrow \sin x=\dfrac{\alpha \pm \sqrt{{{\alpha }^{2}}-16\alpha +64}}{4} \\\ & \Rightarrow \sin x=\dfrac{\alpha \pm \sqrt{{{\left( \alpha -8 \right)}^{2}}}}{4} \\\ & \Rightarrow \sin x=\dfrac{\alpha \pm \left( \alpha -8 \right)}{4} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \sin x=\dfrac{\alpha +\alpha -8}{4},\dfrac{\alpha -\left( \alpha -8 \right)}{4} \\\ & \Rightarrow \sin x=\dfrac{2\alpha -8}{4},\dfrac{\alpha -\alpha +8}{4} \\\ & \Rightarrow \sin x=\dfrac{\alpha -4}{2},\dfrac{8}{4} \\\ & \Rightarrow \sin x=\dfrac{\alpha -4}{2},2 \\\ \end{aligned}$$ We know that $$-1\le \sin x\le 1$$. So, it is clear that we cannot take the value of sinx as equal to 2. So, we can take only $$\dfrac{\alpha -4}{2}$$as the value of sinx. $$\Rightarrow -1\le \dfrac{\alpha -4}{2}\le 1$$ Now by using cross multiplication, then we get $$\begin{aligned} & \Rightarrow -2\le \alpha -4\le 2 \\\ & \Rightarrow 4-2\le \alpha \le 4+2 \\\ & \Rightarrow 2\le \alpha \le 6....(4) \\\ \end{aligned}$$ From equation (4), it is clear that $$\alpha \in \left[ 2,6 \right]$$. **Hence, option A is correct.** **Note** : Students may have a misconception that $$0\le \sin x\le 1$$. If this misconception is followed, then we get So, we can take only $$\dfrac{\alpha -4}{2}$$as the value of sinx. $$\Rightarrow 0\le \dfrac{\alpha -4}{2}\le 1$$ Now by using cross multiplication, then we get $$\begin{aligned} & \Rightarrow 0\le \alpha -4\le 2 \\\ & \Rightarrow 4\le \alpha \le 4+2 \\\ & \Rightarrow 4\le \alpha \le 6....(1) \\\ \end{aligned}$$ From equation (1), it is clear that $$\alpha \in \left[ 4,6 \right]$$. But we know this is wrong.