Application of Derivatives 4 Question 40

42. If f(x) is twice differentiable function such that f(a)=0, f(b)=2,f(c)=1,f(d)=2,f(e)=0, where a<b<c<d<e, then the minimum number of zeroes of g(x)=f(x)2+f(x)f(x) in the interval [a,e] is

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Answer:

Correct Answer: 42. P0=12,54 and Q0=54,12

Solution:

  1. Let g(x)=ddx[f(x)f(x)]

To get the zero of g(x), we take function

h(x)=f(x)f(x)

between any two roots of h(x), there lies atleast one root of h(x)=0. g(x)=0h(x)=0

f(x)=0 or f(x)=0

If f(x)=0 has 4 minimum solutions.

f(x)=0 has 3 minimum solutions.

h(x)=0 has 7 minimum solutions.

h(x)=g(x)=0 has 6 minimum solutions.



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