BITSAT2021MathematicsApplication of DerivativesActual
The points at which the tangent passes through the origin for the curve y=4 x³-2 x⁵ are
Options
- A(0,0),(2,1) and (-1,-2)
- B(0,0),(2,1) and (-2,-1)
- C(2,0),(2,1) and (-3,1)
- D(0,0),(1,2) and (-1,-2)
Correct answer
D. (0,0),(1,2) and (-1,-2)
Step-by-step solution
The equation of the given curve is array l y=4 x³-2 x⁵ dy dx =12 x²-10 x⁴ array Therefore, the slope of the tangent at point (x, y) is 12 x²-10 x⁴ . The equation of the tangent at (x, y) is given by Y -y= (12 x²-10 x⁴ )( X -x) (i) When, the tangent passes through the origin (0,0) , then X = Y =0 Therefore, eq. (i) reduce to array l -y= (12 x²-10 x⁴ )(-x) y=12 x³-10 x⁵ array Also, we have y=4 x³-2 x⁵ array l 12 x³-10 x⁵ =4 x³-2 x⁵ 8 x⁵-8 x³=0 x⁵-x³=0 x³ (x²-1 )=0 x=0, 1 array When, x=0y=4(0)³-2(0)⁵=0 When, x=1 , y=4