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The curve y=ax^(3)+bx^(2)+cx is incline...

The curve `y=ax^(3)+bx^(2)+cx` is inclined at `45^(@)` to x-axis at (0,0) but it touches x-axis at (1,0) , then a+b+c+10 is

A

f'(1) = 0

B

f''(1) = 2

C

f'''(2) = 12

D

f(2) = 2

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
`(dy)/(dx)=3ax^(2)+2bx+c=tan45^(@)=1" at "x=0`
`therefore" "c=1`
`(dy)/(dx)" at "(1,0)=3a+2b+c=0" as x-axis is tangent."`
`therefore" "3a+2b+1=0 because c=1`
(1, 0) lies on curve,
`therefore" "a+b+1=0`
Solving we get `a=1, b=-2`
`therefore" "f(x)=x^(3)-2x^(2)+x`
and `f'(x)=3x^(2)-4x+1`
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