\(\frac{\sqrt17+3}{2}\)
\(\frac{\sqrt{17}+5}{2}\)
5
\(\frac{9-\sqrt{17}}{2}\)
The correct answer is (A) : \(\frac{\sqrt17+3}{2}\)
ƒ(x) = |x2– 3x – 2| – x
\(∀x∈[−1,2]\)
\(f(x) = \begin{cases} x^2 - 4x - 2 & \text{if } -1 \leq x < \frac{3 - \sqrt{17}}{2} \\ -x^2 + 2x + 2 & \text{if } \frac{3 - \sqrt{17}}{2} \leq x \leq 2 \end{cases}\)

\(ƒ(x)_{max} = 3\)
\(ƒ(x)_{min}=ƒ(\frac{3−\sqrt{17}}{2})\)
\(=\frac{\sqrt{17}-3}{2}\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
The extrema of a function are very well known as Maxima and minima. Maxima is the maximum and minima is the minimum value of a function within the given set of ranges.

There are two types of maxima and minima that exist in a function, such as: