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Q.

 Column – I Column – II
A)If  f(x)=limn(r=1nrcosrx) where xR{nπ,nI} ,then the value of [limx0((1cosx)2f(x))1cosx1]  is (where [ . ] is greatest integer function)P)2
B)If the function  f(x)  is given by f(x)=x822x56x3+3x2+55  then the number of integral solutions of the equation f(x)=0  is Q)0
C)The number of local maxima of function f(x)=x2+4cosx+5 is k and g(x)=2|x|3+3x212|x|+1  ,where x[1,2]  then greatest value of  g(x)  is m then m-k isR)4
D)If  f:R{1}R and fx (which is not a constant function and fxx,x0,2  )is a differentiable function satisfies f(x+f(y)+xf(y))=y+f(x)+yf(x),x,yR{1} then the value of 
2023 {1+f(2022)}  is
S)1

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a

AP:BQ:CS:DR

b

AP:BP:CS:DR

c

AP:BQ:CR:DS

d

AP:BP:CP:DR

answer is C.

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Detailed Solution

detailed_solution_thumbnail

A)  f(x)=cosx(11cosx+cosx(1cosx)2)=cosx(1cosx)2
  limx0[((1cosx)2f(x))1cosx1]=limx0[(cosx)1cosx1]=[e1cosx1(cosx1)]=2
B) No integral value of x such that  f(x)=0
C)  f'=2x4sinx
x=0  only point of maximum 
 g'(x)=6(|x|+2)(|x|1)
maximum value of  g(x)=5
D) x=0=yf(0)=0  also differentiate w.r.t x and y and simplifying  
f'(x)1+f(x)=1x+1f(x)=xx+1

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