f(x) is a polynomial of degree 2, f(0) = 4, f'(0) = 3 and f''(0) = 4, then f(- 1) is equal to
3
- 2
2
- 3
Solution of differential equation sec(x)dy - cosec(y)dx = 0 is
cos(x) + sin(y) = c
sin(x) + cos(y) = 0
sin(y) - cos(x) = c
cos(y) - sin(x) = c
The point P(9/2, 6) lies on the parabola y2 = 4ax, then parameter of the point P is
D.
We know that, coordinates of parametric point on the parabola y2 = 4ax is (at2, 2at).
On putting this value in Eq. (i), we get
On putting the value of a in Eq. (ii), we get
Parameter of the point P is
Using Trapezoidal rule and following table is equal to
x | 0 | 0 | 4 | 6 | 8 |
f(x) | 2 | 5 | 10 | 17 | 26 |
184
92
46
- 36