Calculus - 5

+2 votes

The point on the curve y = \sqrt x which is closest to the point (4,0) is

(A) (4,2)

(B) \left ( \frac{7}{2} , \sqrt{\frac{7}{2}} \right)

(C) \left ( \frac{1}{2} , {\frac{1}{ \sqrt2}} \right)

(D) \left ( \frac{1}{2} , {\frac{3}{ \sqrt2}} \right)

asked May 30 in Basic Maths by gbmentor (14,390 points)
reshown May 31 by gbmentor

2 Answers

0 votes
from the above options only A,B,C points are the part of curve , now find distance from (4,0) to these options A, B, C . hence correct answer is B
answered May 31 by tskaran25gupta (770 points)
+1 vote

Distance of (4,0) from any point (x,y) is \sqrt{(x-4)^2 + (y-0)^2} . Here y=\sqrt{x} so

f(x)=(x-4)^2 + x  (removing sqrt)

so f'(x)=2x - 7

putting f'(x)= 0 we get x = 7 /2

hence y=\sqrt{7/2}

So the point becomes (7/2,\sqrt{7/2})

Hence the correct answer is B

answered May 31 by (310 points)
edited Jun 3 by
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