NIMCET 2025 — Mathematics PYQ

1. Substitute ${\cos }^{2}y=1-{\sin }^{2}y$ into the expression for $B$:

$B={\sin }^{2}y+(1-{\sin }^{2}y{)}^2$

2. Let $x={\sin }^{2}y$.
Since $y$ is a real number, $0\le {\sin }^{2}y\le 1$, so $0\le x\le 1$.

$B=x+(1-x{)}^2$
$B=x+1-2x+x^2$
$B=x^2-x+1$

3. To find the range of this quadratic function for $0\le x\le 1$, we can find the vertex and evaluate the function at the endpoints.

The x-coordinate of the vertex is given by $x=-(-1)/(2\cdot 1)=1/2$.
Since $1/2$ is within the interval, the minimum value of $B$ occurs at $x=1/2$.
Minimum value of
$B=(1/2{)}^2-(1/2)+1=1/4-1/2+1=1/4-2/4+4/4=3/4$.

4. Now, evaluate $B$ at the endpoints of the interval for $x$:

At $x=0$ (i.e., ${\sin }^{2}y=0$): $B=0^2-0+1=1$.
At $x=1$ (i.e., ${\sin }^{2}y=1$): $B=1^2-1+1=1$.

5. Comparing the values, the maximum value of $B$ is $1$.

Therefore, the range of $B$ is $3/4\le B\le 1$.

The final answer is (a).