InterviewSolution
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InterviewSolution
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Using problems are of the Inequality Type. Examples of this type are as follows: |
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Answer» Solution :Let `P(n): tan n alpha gt n tan alpha` Step I For `n=2, tan 2 alpha gt 2 tan alpha ` `rArr (2 tan alpha)/(1-tan^2alpha)-2tan alpha gt 0` `rArr 2 tan alpha ((1-(1-tan^2alpha)/(1-tan^2alpha))gt0` `rArr tan^2alpha.tan 2alpha gt 0 [because 0 lt alpha lt (pi)/(4) "for"n=2]` `rArr tan 2 alpha gt 0 [because 0 lt 2 alpha lt (pi)/(2)]` Which is true (`because` in first quadrant , `tan2alpha ` is POSITIVE) Therefore , P(2)is true. Step II ASSUME that P(k) is true , then `P(k): tan k alpha gt tan alpha` Step III For `n=k+1`, we shall prove that `tan(k+1)alpha gt (k+1)tan alpha` `becausetan (k+1)alpha=(tan k alpha +tan alpha)/(1-tank alpha tan alpha )` .......(i) When `0 lt alpha lt (pi)/(4k)or 0 lt kalpha lt (pi)/(4)` i.e., `0 lt tan k alpha lt1`, also `0 lt tan alpha lt 1` `therefore tan k alpha tan alpha lt1` `1-tan k alpha tan alpha gt 0 and 1-tan k alpha tan alpha lt 1`......(ii) From Eqs. (i) and (ii) , we GET `tan(k+1alpha gt (tan kalpha +tanalpha)/(1) gt tan kalpha +tanalpha gt k tan alpha +tan alpha ` [ by assumption step ] `therefore tan (k+1)alpha gt (k+1)tan alpha` Therefore , `P(k+1)` is true , Hence by the PRINCIPLE of mathematical induction P(n) is true for all `n in N`. |
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