JEE Mains Model Paper 4

Let ABC be a right angle triangle in which side BC = x(say) And hypotenuse AC = y(say)
Given x + y = k(const) = y = k - x
Now the area of the triangle ABC is given by
A =

1

2

BC . AB =

1

2

x√(y2−x2) =

1

2

x√[(k−x)2−x2]
For max. or min. of u ,

du

dx

= 0 = x =

k

3

(∵ x ∊ 0)
When x =

k

3

,

d2u

dx2

=

1

2

k (k - 2k) = −

1

2

k2 < 0
u , i.e. A , is max. when x =

k

3

and when y = k - x = 2k/3
Now cosθ = BC/AC = x/y = 1/2 = θ = π/3
Hence the required angle is π / 3