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The minimum ..............

Answer»


Solution :Equation of normal to `y=tan X` at `(alpha, tan alpha)`
`y-tan alpha= (-1)/(sec^(2) alpha) (x-alpha)`
`tan alpha. (sec^(2) alpha)=(2+pi/4-alpha)`
`tan alpha (1+tan^(2) alpha)= (2+pi/4-alpha)`
`alpha=pi/4`
`implies` minimum DISTANCE = distance between `(pi/4, 1)` and `(2+pi/4, 0)` - RADIUS `=SQRT(5)-1`


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