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 = $\sqrt{(x-3)^2+(y-4)^2}$ ⇒ AB = $\sqrt{(x-2)^2+(x^2+4-4)^2}$ [Since y = $x^2$ + 4] ⇒ $(AB)^2$ = $x^{1/4}$ = $(x-3)^2+x^4$ w = $(x-3)^2+x^4$ Distance AB will be maximum or minimum according as 'w' is maximum or minimum Now, w = $(x-3)^2 + x^4$ $\frac{dw}{dx}$ = 2(𝑥 − 3) + $4x^3$ , $\frac{d^2w}{dx^2}$ = $12x^2+2$ Critical points of 'w' are given by $\frac{dx}{dx}$ = 0 2(𝑥 − 3) + $4x^3$ = 0 $2x^3$ + 𝑥 − 3 = 0 ⇒ (𝑥 − 1)($2x^2$ + 2𝑥 + 3) = 0 𝑥 = 1 [$2x^2$ + 2𝑥 + 3 = 0 gives imaginary value of 𝑥] 𝑁𝑜𝑤,⋅ $\left(\frac{d^2w}{dx^2}\right)_{x=1}$ = $12x^2$ + 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 $(AB)_{\min}$ = $\sqrt{(x-3)^2+x^4}$ at x = 1 = $\sqrt{(1-3)^2+1^4}$ = $\sqrt{5}$