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Question

Let f:RR be a differentiable function satisfying f(x+y2)=f(x)+f(y)2 for all x,yR. If f(0)=1,f(0)=1 and g(x)=2x22x+1, then which of the following is (are) CORRECT?


Correct Answer
A
f(|x|) is not differentiable at only one point
Your Answer
B
Number of solutions of the equation f(x)=f1(x) is exactly one
Correct Answer
C
10r=0(f(r))2=286
Correct Answer
D
Area bounded between the curves y=f(x) and y=g(x) is 124 sq. units

Solution

The correct options are
A f(|x|) is not differentiable at only one point
C 10r=0(f(r))2=286
D Area bounded between the curves y=f(x) and y=g(x) is 124 sq. units
Given, f is differentiable and
f(x+y2)=f(x)+f(y)2
f(x)=limh0f(x+h)f(x)h 
         =limh0f(2x+2h2)f(2x+2×02)h
         =limh0f(2x)+f(2h)f(2x)f(0)2h
         =limh0f(2h)f(0)2h     [00]form
Applying L'Hopitals' rule, we get
f(x)=limh02f(2h)2
f(x)=f(0)=1
So, f(x)=x+C
But f(0)=1C=1
f(x)=x+1

f(|x|)=1|x| which is not differentiable at x=0

f(x)=1x
f1(x)=1x
f(x)=f1(x) for all xR
f(x)=f1(x) will have infinitely many solutions.

10r=0(f(r))2
=(f(0))2+(f(1))2+(f(2))2++(f(10))2
=1+0+(1)2+(2)2++(9)2
=1+9×10×196
=1+285=286 

Area bounded between y=f(x) and y=g(x)
Finding intersection point :
1x=2x22x+12x2x=0x=0,12

Area =1/20(f(x)g(x))dx
=1/20(2x2+x)dx=[23x3+x22]1/20 
=124




 

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