Home
Class 12
MATHS
The least positive integral value of 'k'...

The least positive integral value of `'k'` for which there exists at least one line that the tangent to the graph of the curve `y=x^(3)-kx` at one point and normal to the graph at another point is

A

`2`

B

`4`

C

`3`

D

`1`

Text Solution

Verified by Experts

The correct Answer is:
A
Tangent at `(t,t^(3)-kt)` is `y=(3t^(2)-k)x-2t^(3)`
This line cuts the curve again at `x=-2t` where slope of tangent is `12t^(2)-k`
`implies(12t^(2)-k)(3t^(2)-k)=-1`
for positive root `Dge0implies9(3k-4)(3k+4)ge0`
`kepsilon(4/2,oo)implies` Ans `2`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

Find the least positive value of k for which the equation x^(2)+kx+4=0 has real roots.

Find the equation of tangent and normal to the curve y =3x^(2) -x +1 at the point (1,3) on it.

Knowledge Check

  • The complete set of value of 'a' for which there exists at least one line that is tangent to the graph of the curve y=x^(3)-ax at one point and normal to the graph at another point is given by

    A
    `(-oo, -4//3] `
    B
    `[-4//3,oo) `
    C
    `[4//3,oo) `
    D
    `(-oo,4//3]`

  • Equation of the tangent and normal to the curve y=x^(3)-3x^(2)-x+5 at the point x=3 on it are respectively

    A
    `x-8y=19, 8x+y=22`
    B
    `8x+y=19, x-8y=22`
    C
    `8x-y=22, x+8y=19`
    D
    `x+3y=19, 3x-y=22`

  • The set of real values of 'a' for which at least one tangent to y^(2)=4ax becomes normal to the circle x^(2)+y^(2)-2ax-4ay+3a^(2)-0, is

    A
    `[1, 2]`
    B
    `[sqrt2, 3]`
    C
    `R`
    D
    none of these

    • Similar Questions

    Explore conceptually related problems

    Does there exists line/lines which is/are tangent to the curve y=sin x at (x_(1),y_(1)) and normal to the curve at (x_(2),y_(2))?

    Find the number of positive integral values of k for which kx ^(2) +(k-3)x+1 lt 0 for atleast one positive x.

    If the tangent at the point (p,q) to the curve x^(3)+y^(2)=k meets the curve again at the point (a,b), then

    Find the value of k for which the point (1,-2) lies on the graph of the linear equation x-2y+k=0

    The tangents to the curve y=(x-2)^(2)-1 at its points of intersection with the line x-y=3, intersect at the point:

    Home
    Profile