Question

The least value of the expression $\frac{1}{2bx-(x^2+b^2+{\sin }^{2}x)},xϵ[-1,0],bϵ[2,3]$ is

• A. $-\frac{1}{4}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8+{\sin }^{2}1}$
• D. $-\frac{1}{9}$

Answer 1

The correct option is A $-\frac{1}{4}$

Let $y=\frac{1}{2bx-(x^2+b^2+{\sin }^{2}x)},xϵ[-1,0],bϵ[2,3]$
To minimise y, we maximise the denominator if it is positive and we minimise the magnitude of the denominator if it is negative.
Thus, denominator is $2bx-x^2-b^2-{\sin }^{2}x=-(x-b{)}^2-{\sin }^{2}x$
Now, we need the minimum value of its magnitude since it will be a negative number as both the terms are square numbers.
Thus, we minimise $(x-b{)}^2+{\sin }^{2}x$.
Substituting $x$ as $0$ will make ${\sin }^{2}x$ zero and then, to minimise $(x-b{)}^2$, we take $b$ as $2$.
Thus, the minimum value of $y$ becomes $\frac{-1}{4}$