The equation $\sin ^4 x+\cos ^4 x=a$ has real solutions, then

The equation $\sin ^4 x+\cos ^4 x=a$ has real solutions, then
  1. $1 < a < \frac{5}{2}$
  2. $\frac{1}{2} \leq a \leq 1$
  3. $a \leq \frac{1}{2}$
  4. $0 < $ a $ < 1$

Solution

$\begin{aligned} & \text { } \sin ^4 x+\cos ^4 x=\alpha \text { has real solution. } \\ & \text { Since, } \begin{aligned} & \alpha=\sin ^4 x+\cos ^4 x \\ &=\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cdot \cos ^2 x \\ &=(1)^2-\frac{4}{2} \sin ^2 x \cos ^2 x=1-\frac{(2 \sin x \cos x)^2}{2} \\ &=1-\frac{\left(\sin ^2 2 x\right)}{2} \quad[\because \sin 2 A=2 \sin A \cos A] \\ & \because-1 \leq \sin 2 x \leq 1 \\ & 0 \leq \sin ^2 2 x \leq 1\end{aligned}\end{aligned}$ $\begin{aligned} 0 & \leq \frac{\sin ^2 2 x}{2} \leq \frac{1}{2} \\ -\frac{1}{2} & \leq-\frac{\sin ^2 2 x}{2} \leq 0 \\ 1-\frac{1}{2} & \leq 1-\frac{\sin ^2 2 x}{2} \leq 1 \\ \frac{1}{2} & \leq \alpha \leq 1\end{aligned}$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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