Home
Class 12
MATHS
Points on the curve f(x)=x/(1-x^2) where...

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

`(0,0)`

B

`(sqrt(3),-(sqrt3)/(2))`

C

`(-2,(2)/(3))`

D

`(sqrt3,-(sqrt3)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • APPLICATION OF DERIVATIVES

    CENGAGE|Exercise Exercise (Numerical)|12 Videos

  • APPLICATION OF DERIVATIVES

    CENGAGE|Exercise JEE Previous Year|9 Videos

  • APPLICATION OF DERIVATIVES

    CENGAGE|Exercise Exercise (Single)|48 Videos

  • 3D COORDINATION SYSTEM

    CENGAGE|Exercise DPP 3.1|11 Videos

  • APPLICATION OF INTEGRALS

    CENGAGE|Exercise Solved Examples And Exercises|17 Videos

Similar Questions

Explore conceptually related problems

If at each point of the curve y=x^3-a x^2+x+1, the tangent is inclined at an acute angle with the positive direction of the x-axis, then a >0 (b) a<-sqrt(3) -sqrt(3)<=a<= sqrt(3) (d) non eoft h e s e

The abscissas of point Pa n dQ on the curve y=e^x+e^(-x) such that tangents at Pa n dQ make 60^0 with the x-axis are. 1n((sqrt(3)+sqrt(7))/7)a n d1n((sqrt(3)+sqrt(5))/2) 1n((sqrt(3)+sqrt(7))/2) (c) 1n((sqrt(7)-sqrt(3))/2) +-1n((sqrt(3)+sqrt(7))/2)

If tan^(-1)x+2cot^(-1)x=(2pi)/3, then x , is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3) (d) sqrt(2)

Solve (sqrt(3) + sqrt(2))^(x) +(sqrt(3)-sqrt(2))^(x)=10

The incenter of the triangle with vertices (1,sqrt(3)),(0,0), and (2,0) is (a) (1,(sqrt(3))/2) (b) (2/3,1/(sqrt(3))) (c) (2/3,(sqrt(3))/2) (d) (1,1/(sqrt(3)))

The point(s) on the curve y^3+\ 3x^2=12 y where the tangent is vertical, is(are) ?? (+-4/(sqrt(3)),\ -2) (b) (+-\ sqrt((11)/3,\ )\ 1) (0,\ 0) (d) (+-4/(sqrt(3)),\ 2)

Solve : (sqrt(3)+sqrt(2))^(x)+(sqrt(3)-sqrt(2))^(x)=10

The values of parameter a for which the point of minimum of the function f(x)=1+a^2x-x^3 satisfies the inequality (x^2+x+2)/(x^2+5x+6)<0a r e (2sqrt(3),3sqrt(3)) (b) -3sqrt(3),-2sqrt(3)) (-2sqrt(3),3sqrt(3)) (d) (-2sqrt(2),2sqrt(3))

If 0lt=xlt=pi/3 then range of f(x)=sec(pi/6-x)+sec(pi/6+x) is (a) (4/(sqrt(3)),oo) (b) (4/(sqrt(3)),oo) (c) (0,4/(sqrt(3))) (d) (0,4/(sqrt(3)))

A beam of light is sent along the line x-y=1 , which after refracting from the x-axis enters the opposite side by turning through 30^0 towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is (a) (2-sqrt(3))x-y=2+sqrt(3) (b) (2+sqrt(3))x-y=2+sqrt(3) (c) (2-sqrt(3))x+y=(2+sqrt(3)) (d) y=(2-sqrt(3))(x-1)

Home
Profile