Let `y=f(x)` be a parabola, having its axis parallel to the y-axis, which is touched by the line `y=x` at `x=1.` Then, `2f(0)=1-f^(prime)(0)` (b) `f(0)+f^(prime)(0)+f^(0)=1` `f^(prime)(1)=1` (d) `f^(prime)(0)=f^(prime)(1)`

Options

A. `f'(0)=f'(1)`
B. `f'(1)=-1`
C. `f(0)+f'(0)+f''(0)=1 `
D. `2f(0)=1-f'(0) `

Answer

D

Explanation

Let `y=ax^(2) +bx +c ` be the given parabola. Then,
`f(x)=ax^(2)+bx+c `
Clearly, `(dy)/(dx)=2ax+b `
It is given that `y=x` touches the parabola at `x=1.`
`therefore ((dy)/(dx))_(x=1) =` (Slope of the line `y=x`)
` rArr 2a+b=1 " "...(i)`
Putting `x=1 ` in ` y=x,` we get `y=1. `
So, the line `y=x ` touches the parabola ` y=ax^(2)+bx+c ` at (1,1).
` a+b+c=1 " "...(ii) `
Now,
`f(x)=ax^(2)+bx +c rArr f'(x) = 2ax +b " and " f''(x)=2a `
` therefore f(0)=c,f'(0)=b, f''(0)=2a " and " f'(1) =2a+b. `
Form (ii), we have
`a+b+c=1 `
` rArr 2a+2b+2c =2 `
` rArr 2a+b+(b+2c)=2 `
` rArr 1+(b+2c)=2 " "[ because 2a+b=1 " from (i)" ] `
` rArr b+2c=1 rArr 2c=1-b rArr 2f(0)=1-f'(0) `
Also, `f'(1)=2a+b=1 " "[" Using " (i) ] `