BITSAT Mathematics Question (2021) — Solution

Step-by-step solution

The equation of the given curve is array l y=4 x^ 3 -2 x^ 5 \\ dy dx =12 x^ 2 -10 x^ 4 array Therefore, the slope of the tangent at point (x, y) is 12 x^ 2 -10 x^ 4 . The equation of the tangent at (x, y) is given by Y -y= (12 x^ 2 -10 x^ 4 )( 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 x^ 2 -10 x^ 4 )(-x) \\ y=12 x^ 3 -10 x^ 5 array Also, we have y=4 x^ 3 -2 x^ 5 array l 12 x^ 3 -10 x^ 5 \\ =4 x^ 3 -2 x^ 5 \\ 8 x^ 5 -8 x^ 3 =0 \\ x^ 5 -x^ 3 =0 \\ x^ 3 (x^ 2 -1 )=0 \\ x=0, 1 array When, x=0 y=4(0)^ 3 -2(0)^ 5 =0 When, x=1, y=4(1)^ 3 -2(1)^ 5 =2 When, x=-1, y=4(-1)^ 3 -2(-1)^ 5 =-2 Hence, the require points are (0,0),(1,2) and (-1,-2).