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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minPYQ_2021
MathematicsMediumnumerical

Let and be defined by f_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t, \quad x > 0 $ and f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450, \quad x > 0 na_{1}, a_{2}, \ldots, a_{n}, \prod_{i=1}^{n} a_{i}a_{1}, a_{2}, \ldots, a_{n} .m_{i}n_{i}f_{i}, i=1,2(0, \infty)$. The value of6m2+4n2+8m2n2is ____.

Answer:
6.00
Solution:

f2x=98x-150-600x-149+2450,  x>0

f2'x=98×50x149600×49x148

=4900x148x16

=4900x148x7

So extrema is at x=7 only, which is minima

m2=1,n2=0

Hence 6m2+4n2+8m2n2=6

Stream:JEE_ADVSubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima
2mℹ️ Source: PYQ_2021

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