Aspire Faculty ID #11970 · Topic: NIMCET 2025 · Just now
NIMCET 2025

If $B=sin^2 y+cos^4 y$, then for all real y

Solution

Let $B = \sin^{2}y + \cos^{4}y$. 
Put $t = \cos^{2}y$, so $0 \le t \le 1$. 
Then $\sin^{2}y = 1 - t$. 
So, $B = (1 - t) + t^{2} = t^{2} - t + 1$. 
Minimum of the quadratic $t^{2} - t + 1$ occurs at $t = \dfrac{1}{2}$: 
$B_{\min} = \dfrac{1}{4} - \dfrac{1}{2} + 1 = \dfrac{3}{4}$. 
Maximum occurs at $t = 0$ or $t = 1$: 
$B_{\max} = 1$. 
Hence, $B \in \left[\dfrac{3}{4},, 1\right]$.
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