NIMCET 2025 Paper

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