If the slope of the tangent to the curve is maximum at x = 1 and curve has minim value 1  at x = 0, then the curve which also satisfies the equation  $\frac{d^{3}y}{d{x}^3}=4x-3$ is
 

detailed solution

Correct option is B

we have  d3ydx3=4x−3 Integrating, we get   d2ydx2=2x2−3x+c Given slope of tangent to the curve is maximum at x = 1,slope is dydx, to find slope max equate d2ydx2 to zero Then  d2ydx2x=1=0     ⇒2−3+c=0  ⇒  c=1d2ydx2=2x2−3x+1   Integrating we get dydx=2x33−3x22+x+D  ∴curve has a minimum value 1 at x = 0to find y=f(x) minimum equate f'(x)=0 or dydx=0∴dydxx=0=0         ⇒     D=0  ∴dydx=2x33−3x22+x   Again integrating, we get   y=x46−x32+x22+E Given minimum value of curve is 1 at x = 0 Then 1 = 0 + E  ⇒  E = 1 Hence the curve is   y=1+x22−x32+x46