Home
Class 12
MATHS
The curve y=ax^(3)+bx^(2)+cx is inclined...

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

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`
Promotional Banner

Topper's Solved these Questions

  • APPLICATIONS OF DERIVATIVES

    CENGAGE|Exercise Comprehension Type|5 Videos

  • APPLICATIONS OF DERIVATIVES

    CENGAGE|Exercise Subjective Type|2 Videos

  • APPLICATIONS OF DERIVATIVES

    CENGAGE|Exercise Subjective Type|2 Videos

  • APPLICATION OF INTEGRALS

    CENGAGE|Exercise Solved Examples And Exercises|17 Videos

  • AREA

    CENGAGE|Exercise Comprehension Type|2 Videos

Similar Questions

Explore conceptually related problems

The curve y=ax^(4)+bx^(2)" with "abgt0

The curve y=ax^3 + bx^2 + cx + d has a point of inflexion at x=1 , then :

The curve y=a x^3+b x^2+c x+5 touches the x-axis at P(-2,0) and cuts the y-axis at the point Q where its gradient is 3. Find the equation of the curve completely.

A curve with equation of the form y=a x^4+b x^3+c x+d has zero gradient at the point (0,1) and also touches the x- axis at the point (-1,0) then the value of x for which the curve has a negative gradient are:

Points on the curve f(x)=x/(1-x^2) where the tangent is inclined at an angle of pi/4 to the x-axis are (a)(0,0) (b) (sqrt(3),-(sqrt(3))/2) (-2,2/3) (d) (-sqrt(3),(sqrt(3))/2)

A vector vecn is inclined to x -axis at 45^(@) , to y-axis at 60^(@) and at an angle to z-axis. If vecn is a normal to the plane passing through the point (sqrt(2),-1,1) , then the equation of plane is

If the tangent to the curve x y+a x+b y=0 at (1,1) is inclined at an angle tan^(-1)2 with x-axis, then find aa n db ?

The curve ay^2=x^2(3a-x) cuts the y - axis at .........

In the curve y=x^(3)+ax and y=bx^(2)+c pass through the point (-1,0) and have a common tangent line at this point then the value of a+b+c is