TestHub
TestHub

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaHard2 minPYQ_2022
MathematicsHardsingle choice

If the absolute maximum value of the functionfx=x2-2x+7e4x3-12x2-180x+31in the interval-3,0isfα, then

Options:

Answer:
B
Solution:

Given,

fx=x2-2x+7e4x3-12x2-180x+31

Now differentiating the function to find function is increasing or decreasing,

f'x=e4x3-12x2-180x+3112x2-2x-15+2x-1

f'x=e4x3-12x2-180x+3112x+3x-5+2x-1

f'x<0 for x-3,0

So, fxis decreasing function on -3,0

The absolute maximum value of the function fx is at x=-3

α=-3

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima
2mℹ️ Source: PYQ_2022

Doubts & Discussion

Loading discussions...