Here, f(x)=x2 -6 logx-3=0
f(2)=4-6 log2-3=-0.806
f(3)=9-6 log3-3=3.1373
f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.
Hence, the root lies between 2 and 3
c0 =(2+3)/2=2.5
f(2.5)=6.25-6 log 2.5-3=0.8623
Now
| n | a(-ve) | b(+ve) | cn | f(cn) |
| 0 | 2 | 3 | 2.5 | 0.8623 |
| 1 | 2 | 2.5 | 2.25 | -0.050595 |
| 2 | 2.25 | 2.5 | 2.375 | 0.38664 |
| 3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |
| 4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |
| 5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |
From the table,
f(2.265625)=0.001928<10-2
Therefore, the...
Balance on a bicycle is a matter of constantly correcting against falls, and it's easier when the speed is higher because the inertia of moving forward overcomes the need for corrective actions.
Here, the given equation of parabola is y2= 8x.
The equation of tangent to the parabola y2=8x is,
y= mx + 2/m
This tangent passes through the point (-2, 3)
So, 3 = -2m + 2/m
or, 3m + 2m2 = 2
or, 2m2+3m - 2= 0
or, 2m2 + (4 - 1)m -2 = 0
or, 2m2 + 4m - m - 2 = 0
or, 2m(m + 2) - 1(m+2) = 0
or, (m + 2) (2m - 1) = 0
Either, Or,
m = -2 m = 1/2
Required angle is,
