Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minQB

MathematicsMediuminteger

$f (x) = \frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 1∣}$ . Let $x_{1}, x_{2}$ are points where $f (x)$ attains local minimum and global maximum respectively. Let $k = f (x_{1}) + f (x_{2})$ then $6 k - 9$

Answer:

Solution:

Local minimum $= f (\frac{1}{2}) = \frac{4}{3}$ Global maximum

$= f (0) = f (1) = \frac{3}{2} k = \frac{4}{3} + \frac{3}{2} = \frac{17}{6}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: QB

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