Q. 265.0( 1 Vote )

A jet of enemy is

Answer :

Given: The curve is y = x2 + 2


To find: minimum distance between the soldier and the jet


Proof: Let jet be at point (x, y) = (x, x2 + 2)


Let d be the distance between jet and soldier at (3, 2)





We need to find the minimum distance .i.e. minimum value of d


Let f(x) = d2 [to make calculation easy]



f(x) = (x – 3)2 + x4


When f(x) is minimum, d is minimum


Finding f’(x)



f’(x) = 2 × (x – 3) × 1 + 4 × x3


= 2x – 6 + 4x3


= 4x3 + 2x – 6


Factorizing f’(x)


f’(1) = 4(1)3 + 2(1) – 6


= 4 + 2 – 6


= 6 – 6


= 0



Hence, (x – 1) is a factor of 4x3 + 2x – 6


Thus,


f’(x) = (x – 1)(4x2 + 4x + 6)


Hence, f’(x) = 0 gives


either x – 1 = 0


x = 1


or 4x2 + 4x + 6 = 0


2(2x2 + 2x +3) = 0


2x2 + 2x + 3 = 0


Now, we find the value of x



Here, a = 2, b = 2 and c = 3





This is not possible as there are no real roots


Hence, there is only one point x = 1


This is either maxima or minima


Hence, we find f’’(x)



f’’(x) = 3 × 4x2 + 2


f’’(x) = 12x2 + 2


Finding value at x = 1


f’’(1) = 12(1)2 + 2


= 12 + 2


= 14 > 0


f’’(x) > 0


x = 1 is the minima


The value of f(1) is


f(1) = (1 – 3)2 + (1)4


= (-2)2 + 1


= 4 + 1


= 5


Hence, minimum distance between soldier and jet




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