Let f:[0,1]→R be a twice differentiable function in (0,1) such that f(0)=3 and f(1)=5. If the line y=2x+3 intersects the graph of f at only two distinct points in (0,1), then the least number of points x∈(0,1), at which f′′(x)=0, is
Now plotting the diagram of given data we have,

Given f(x) cuts y=2x+3 at two distinct point betweenx∈(0,1),
So, f′(a)=f′(b)=f′(c)=2 {as slope of given line y=2x+3 is 2}
So we can conclude that f′′(x) is zero for atleast {x}_{1}\in (a,b)&{x}_{2}\in (b,c)