  Question

# Let y=f(x) be a non-negative function defined on [0,2] satisfying the differential equation y3y′′+1=0. If f′(1)=0 and f(1)=1, then the maximum value of f(x) is 1.the minimum value of f(x) is 0. the solution y=f(x) of the given differential equation represents a semi-circle with centre (1,0).the area bounded by the curve y=f(x) and x−axis is π4 sq. units.

Solution

## The correct options are A the maximum value of f(x) is 1. B the minimum value of f(x) is 0.  C the solution y=f(x) of the given differential equation represents a semi-circle with centre (1,0).y3y′′=−1 ⇒y′y′′=−y′y3 Integrating both the sides, (y′)22=12y2+C f′(1)=0, f(1)=1⇒C=−12 ⇒(y′)22=12y2−12 ⇒y′=±√1−y2y ⇒y√1−y2y′=±1 Integrating both the sides, ∫y√1−y2y′ dx=±∫dx Put 1−y2=z2⇒−2yy′dx=2zdz ⇒−∫z dzz=±x+c ⇒−z=±x+c ⇒−√1−y2=±x+c y(1)=1⇒c=∓1 ⇒−√1−y2=±x∓1 ⇒−√1−y2=x−1 or −x+1 Squaring both the sides, we get 1−y2=x2−2x+1  ⇒y2=2x−x2     ...(1) ⇒y=√2x−x2=f(x) From eqn (1), x2+y2−2x=0  ⇒(x−1)2+y2=1,  x∈[0,2] Clearly, x=1 is the point of maxima. Maximum value of f(x)=√2−1=1  Minimum value of f(x)=0 at x=0,2 Area bounded by the curve y=f(x) and x−axis is : π×122=π2 sq. units.  Suggest corrections   