JEE MATHEMATICS MCQ: TRANSFORMATIONS

Solution approach

1. Correct Answer + Core Reason: $\frac{1}{2} \le a \le 1$. The equation has a real solution only when $a$ lies between $\frac{1}{2}$ and $1$, inclusive.
   
   Step-by-Step Solution:
   
   Step 1: Given the equation ${\sin ^4}\,\,\theta + {\cos ^4}\,\,\theta = a$.
   
   Step 2: Rewrite the left side of the equation: ${\sin ^4}\,\,\theta + {\cos ^4}\,\,\theta = {({\sin ^2}\,\,\theta + {\cos ^2}\,\,\theta )^2} - 2{\sin ^2}\,\,\theta {\cos ^2}\,\,\theta = 1 - 2{\sin ^2}\,\,\theta {\cos ^2}\,\,\theta$.
   
   Step 3: Use the identity $\sin 2\theta = 2\sin \theta \cos \theta$. Then, ${\sin ^2}2\theta = 4{\sin ^2}\theta {\cos ^2}\theta$, so ${\sin ^2}\theta {\cos ^2}\theta = \frac{1}{4}{\sin ^2}2\theta$.
   
   Step 4: Substitute this back into the equation: $a = 1 - 2\left( {\frac{1}{4}{{\,\sin }^2}2\theta } \right) = 1 - \frac{1}{2}{\sin ^2}2\theta$.
   
   Step 5: Since $0 \le {\sin ^2}2\theta \le 1$, we have $0 \le \frac{1}{2}{\sin ^2}2\theta \le \frac{1}{2}$.
   
   Step 6: Therefore, $1 - \frac{1}{2} \le 1 - \frac{1}{2}{\sin ^2}2\theta \le 1 - 0$, which means $\frac{1}{2} \le a \le 1$.
   
   Why other options are wrong:
   - Option A: This is incorrect because $a$ must be greater than or equal to $\frac{1}{2}$.
   - Option B: This is incorrect because $a$ must be less than or equal to $1$.
   - Option D: This is incorrect because $a$ must be greater than or equal to $\frac{1}{2}$ and less than or equal to $1$.
   
   Key Concepts:
   - Trigonometric Identities
   - Range of Trigonometric Functions
   
   Memory Aids:
   - Remember the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
   - Recall that the range of ${\sin ^2}x$ is $[0, 1]$.
   
   Quick Tips:
   - Rewrite the expression using trigonometric identities.
   - Find the range of the simplified expression.