The point on the curve `y=6x-x^2` at which the tangent to the curve is inclined at `pi//4` to the line `x+y=0` is `(-3,\ -27)` (b) `(3,\ 9)` (c) `7//2,\ 35//4)` (d) `(0,\ 0)`

Explanation

`y=6x−x^2
`

⇒`y′=6−2x` as tangent is parallel to line `x+y=0` slope `=-1`

⇒`6−2x=-1`
⇒`x=7/2`
So `y=37/4`

Hence the point of tangency is `(7/2,37/4)`