  Question

# Let f:R→R be a differentiable function satisfying f(x+y2)=f(x)+f(y)2 for all x,y∈R. If f(0)=1,f′(0)=−1 and g(x)=2x2−2x+1, then which of the following is (are) CORRECT?

A
f(|x|) is not differentiable at only one point
B
Number of solutions of the equation f(x)=f1(x) is exactly one
C
10r=0(f(r))2=286
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 10∑r=0(f(r))2=286 D Area bounded between the curves y=f(x) and y=g(x) is 124 sq. unitsGiven, f is differentiable and f(x+y2)=f(x)+f(y)2 f′(x)=limh→0f(x+h)−f(x)h           =limh→0f(2x+2h2)−f(2x+2×02)h          =limh→0f(2x)+f(2h)−f(2x)−f(0)2h          =limh→0f(2h)−f(0)2h     form Applying L'Hopitals' rule, we get f′(x)=limh→02f′(2h)2 ⇒f′(x)=f′(0)=−1 So, f(x)=−x+C But f(0)=1⇒C=1 ∴f(x)=−x+1 f(|x|)=1−|x| which is not differentiable at x=0 f(x)=1−x ∴f−1(x)=1−x f(x)=f−1(x) for all x∈R ⇒f(x)=f−1(x) will have infinitely many solutions. 10∑r=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 : 1−x=2x2−2x+1⇒2x2−x=0⇒x=0,12 Area =1/2∫0(f(x)−g(x))dx =1/2∫0(−2x2+x)dx=[−23x3+x22]1/20  =124  Suggest corrections   