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Please solve the following problem by using Mean Valu theorem only For any Two real Numbers a and b | cos a - cos b | <= | a - b | .
Please solve the following problem by using Mean Valu theorem only
For any Two real Numbers a and b | cos a - cos b | <= | a - b | .
I am providing you the step wise solution for your problem: Step 1. Function cos x is continuous and differentiable for all real numbers. Use the mean value theorem, using 2 real numbers a and b to write (cos x) ' = [cos a - cos b] / [a - b] step 2. Take the absolute value of both sides | (cos x) ' | = | [cos a - cos b] / [a - b] | (cos x)' = - sin x, hence. | (cos x) ' | < = 1 step 3. Which gives | [cos a - cos b] / [a - b] | <= 1 step 4. But | [cos a - cos b] / [a - b] | = |cos a - cos b| / |a - b| step 5. When combined with the above gives |cos a - cos b| / |a - b| <= 1 step 6. Multiply both sides by |a - b| to obtain |cos a - cos b| <= |a - b| Thanks
I am providing you the step wise solution for your problem:
Step 1. Function cos x is continuous and differentiable for all real numbers. Use the mean value theorem, using 2 real numbers a and b to write (cos x) ' = [cos a - cos b] / [a - b]
step 2. Take the absolute value of both sides | (cos x) ' | = | [cos a - cos b] / [a - b] |
| (cos x) ' | < = 1
step 3. Which gives | [cos a - cos b] / [a - b] | <= 1
step 4. But | [cos a - cos b] / [a - b] | = |cos a - cos b| / |a - b|
step 5. When combined with the above gives |cos a - cos b| / |a - b| <= 1
step 6. Multiply both sides by |a - b| to obtain |cos a - cos b| <= |a - b|
Thanks
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