UPSEE Solved Model Paper 1

Let A(x, y) be the position of airplane and man is at B(3,4)
Distance between man and airplane
AB = √(x−3)2+(y−4)2
⇒ AB = √(x−2)2+(x2+4−4)2
[Since y = x2 + 4]
⇒ (AB)2 = x

1

4

= (x−3)2+x4
w = (x−3)2+x4
Distance AB will be maximum or minimum according as 'w' is maximum or minimum
Now, w = (x−3)2+x4

dw

dx

= 2(𝑥 − 3) + 4𝑥3 ,

d2w

dx2

= 12x2+2
Critical points of 'w' are given by

dx

dx

= 0
2(𝑥 − 3) + 4𝑥3 = 0
2𝑥3 + 𝑥 − 3 = 0
⇒ (𝑥 − 1)(2𝑥2 + 2𝑥 + 3) = 0
𝑥 = 1
[2𝑥2 + 2𝑥 + 3 = 0 gives imaginary value of 𝑥]
𝑁𝑜𝑤,⋅ (

d2w

dx2

)x=1 = 12x2 + 2 = 14 > 0
So, x = 1, will give minimum value of 'w'
Hence, distance between man and airplane AB will also be minimum when x = 1
(AB)min = √(x−3)2+x4 at x = 1
= √(1−3)2+14
= √5