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Question

Let f:RR be a differentiable function satisfying f(x+y3)=2+f(x)+f(y)3x,yR and f(2)=2, then answer the following questions:
If g(x)=x4f(|x|)26; then the value of g(k1), where k denotes the solution as well as number of solutions of equation g(x)=0 is

A
9
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B
8
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C
55
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D
None of these
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Solution

The correct option is C 9
Substitute x=3x,y=0 in f(x+y2)=2+f(x)+f(y)3, we get
f(3x+03)=2+f(3x)+f(0)3f(x)=2+f(3x)+f(0)3 ...(1)
Substitute x=y=0
f(0)=2+f(0)+f(0)33f(0)=2+2f(0)f(0)=2 ....(2)
From (1)
f(x)=2+f(3x)+233f(x)=f(3x)+4 ...(3)
Now f(x)=limh0f(x+h)f(x)h=limh0f(3x+3h3)f(x)h
=limh0⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪2+f(3x)+f(3h)3f(x)h⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪=limh0[2+f(3x)+f(3h)3f(x)3h]
=limh0[2+3f(x)4+f(3h)3f(x)3h]
f(x)=limh0[f(3h)23h]
Since f(x) is differentiable x;limh0f(3h)=2
f(0)=2f(x)=limh03f(3h)3=f(0)=k (say)
f(x)=kf(x)=kx+c
f(x)=kf(2)=k
But f(2)=2k=2
f(x)=2x+c
Also f(0)=2c=2
f(x)=2x+2
g(x)=x4f(|x|)26=x4(2|x|2+2)6=x42|x|28
=|x|42|x|28=(|x|24)(|x|2+2)
g(x)=0
|x|2=4 or |x|2=2
|x|2=4(|x|0)
|x|=±2(|x|2)
|x|=2x=±2
Solution of equation g(x)=0 are 2 and 2
there are two solutions of equation g(x)=0
k is the value of solutions as well as number of solutions
k=2
g(k1)=g(21)=g(1)
g(k1)=g(1)=(1)42|1|28=128=9

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