Let f(x) be differentiable on the interval (0, ) such that f(1) =1 and for each x > 0. Then, f(x) is equal to
Let . Then, the value of f'(x) at x = (2n + 1) , n (the set of integers) is equal to
(- 1)n
(- 1)n + 1
3
9
The number of points, where f(x) = [sin(x) + cos(x)] (where [] denotes the greatest integerfunction) and x (0, 2) is not continuous, is
3
4
5
6
According to Simpson's rule, the value of is
1.358
1.958
1.625
1.458
B.
1.958
Then, we have the following table
x | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 | 5 | 5.5 | 6 | 6.5 | 7 |
1 | 0.67 | 0.5 | 0.4 | 0.33 | 0.29 | 0.25 | 0.22 | 0.2 | 0.18 | 0.17 | 0.15 | 0.14 | |
y0 | y1 | y2 | y3 | y4 | y5 | y6 | y7 | y8 | y9 | y10 | y11 | y12 |
Now, by Simpson's rule,
Equation of a plane passing through (- 1, 1, 1) and (1, - 1, 1) and perpendicular to x + 2y + 2z = 5 is
2x + 3y - 3z + 3 = 0
x + y + 3z - 5 = 0
2x+ 2y - 3z + 3 = 0
x + y + z - 3 = 0