1.

The point on the curve `y=sqrt(x-1)`, where the tangent is perpendicular to the line 2x+y-5=0 isA. `(2,-1)`B. `(10,3)`C. `(2,1)`D. `(5,-2)`

Answer» Correct Answer - C
Let slope of the curve `y = sqrt(x-1)` is `m_(1)` and slope of the line `2x + y - 5 = 0` is `m_(2)`
Now, `m_(1) = (dy)/(dx) = d/(dx) sqrt(x-1)`
and `m_(2) = (dy)/(dx) = (d)/(dx) (5-2x)`
`rArr m_(1) = (1)/(2sqrt(x-1))` and `m_(2) = -2`
We know that, lines are perpendicular it and only if `m_(1)m_(2) = - 1`
`:. (1)/(2sqrt(x - 1)) = 1 rArr sqrt(x-1) = 1`
On squaring both sides, we get
`x - 1= 1`
`rArr x = 2`
On substituting `x = 2` in `y = sqrt(x - 1)`, we get
Hence. coordination of the point on the curve, `y = sqrt(x-1)`, where tangent is perpendicular to the line `2x+y = 5` is `(2,1)`.


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