NIMCET 2025 Paper

We want the range of an expression involving $\sin\; y$ and $\cos y$.A standard trick is to set $t = \sin^2 y$, so $t \in [0,1]$, and rewrite everything in terms of $t$. The result becomes a simple quadratic in $t$, whose minimum and maximum on $[0,1]$ we can find easily. Let$t = \sin^2 y \;\; \Rightarrow \;\; t \in [0,1]$Then$\cos^2 y = 1 - t \;\; \Rightarrow \;\; \cos^4 y = (1-t)^2$So$B = \sin^2 y + \cos^4 y = t + (1-t)^2 .$Expand:$B = t + 1 - 2t + t^2 = t^2 - t + 1$Now find the minimum of $B(t)=t^2-t+1$ on $[0,1]$.Vertex of the parabola:$t = \frac{1}{2}, \;\; B\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} .$At the endpoints:$B(0)=0^2-0+1=1, \;\; B(1)=1^2-1+1=1 .$So the minimum is $\frac{3}{4}$ and the maximum is 1 .$\frac{3}{4} \leq B \leq 1$