Quadratic EquationHard
Question
The greatest value of the function $f(x) = \frac{1}{2bx^{2} - x^{4} - 3b^{2}}$ on the interval $\lbrack - 2,1\rbrack$ depending on the parameter 'b' is/are
Options
A.$- \frac{1}{3b^{2}}$ if $b \in \lbrack 0,2\rbrack$
B.$\frac{1}{4b - 4 - 3b^{2}}$ if $b \in \lbrack 0,4\rbrack$
C.$\frac{1}{8b - 16 - 3b^{2}}$ if $b \leq 2$
D.$- \frac{1}{3b^{2}}$ if $b \geq 2$
Solution
$f(x) = \frac{1}{2bx^{2} - x^{4} - 3b^{2}}\ x \in \lbrack - 2,1\rbrack$
Let $g(t) = 2bt - t^{2} - 3{\text{ }b}^{2}\ t \in \lbrack 0,4\rbrack$
$$g(t) < 0\forall t \in \lbrack 0,4\rbrack$$
Case-I If $b \leq 0$, then $g(t)$ will decrease in $\lbrack 0,4\rbrack$
Maximum value of $f(t) = f(4) = \frac{1}{8\text{ }b - 16 - 3{\text{ }b}^{2}}$
Case
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