If u=√(a^2)^2θ+b^2^2θ+√(a^2)^2θ+b^2^2θ, then the difference between the maximum and minimum values of u^2 is given by - Mathematics MCQ AIEEE 2004 2004 | ExamTest.live

Solution & Step-by-step Explanation

Square the expression for

u

u^{2} = (a^{2} cos^{2} θ + b^{2} sin^{2} θ) + (a^{2} sin^{2} θ + b^{2} cos^{2} θ) + 2 (a^{2} cos^{2} θ + b^{2} sin^{2} θ) (a^{2} sin^{2} θ + b^{2} cos^{2} θ)

u^{2} = a^{2} + b^{2} + 2 a^{2} b^{2} (cos^{4} θ + sin^{4} θ) + (a^{4} + b^{4}) sin^{2} θ cos^{2} θ

.
Let

x = sin^{2} θ cos^{2} θ = \frac{1}{4} sin^{2} 2 θ .

Note that

0 \leq x \leq 1/4.

Using

cos^{4} θ + sin^{4} θ = 1 - 2 sin^{2} θ cos^{2} θ = 1 - 2 x

u^{2} = a^{2} + b^{2} + 2 a^{2} b^{2} (1 - 2 x) + (a^{4} + b^{4}) x

u^{2} = a^{2} + b^{2} + 2 a^{2} b^{2} + (a^{2} - b^{2})^{2} x .

Minimum value occurs when

x = 0

(

θ = 0

u_{min}^{2} = a^{2} + b^{2} + 2 a^{2} b^{2} = a^{2} + b^{2} + 2 ab = (a + b)^{2}

.Maximum value occurs when

x = 1/4

(

θ = π /4

u_{ma x}^{2} = a^{2} + b^{2} + 2 a^{2} b^{2} + \frac{( a ^{2} - b ^{2} ) ^{2}}{4} = a^{2} + b^{2} + 2 \frac{4 a ^{2} b ^{2} + a ^{4} + b ^{4} - 2 a ^{2} b ^{2}}{4}

u_{ma x}^{2} = a^{2} + b^{2} + 2 \frac{( a ^{2} + b ^{2} ) ^{2}}{2} = 2 (a^{2} + b^{2})

.
Difference

= u_{ma x}^{2} - u_{min}^{2} = 2 (a^{2} + b^{2}) - (a + b)^{2} = 2 a^{2} + 2 b^{2} - (a^{2} + b^{2} + 2 ab) = a^{2} + b^{2} - 2 ab = (a - b)^{2}

If $u = a^{2} cos^{2} θ + b^{2} sin^{2} θ + a^{2} sin^{2} θ + b^{2} cos^{2} θ,$ then the difference between the maximum and minimum values of $u^{2}$ is given by