Question

If \[ u=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta} + \sqrt{a^2\sin^2\theta+b^2\cos^2\theta}, \] then the difference between the maximum and minimum values of \(u^2\) is

• A. \((a+b)^2\)
• B. \((a-b)^2\)
• C. \(2\sqrt{a^2+b^2}\)
• D. \(2\sqrt{a^2-b^2}\)

Hint:

For expressions involving \(\sqrt A+\sqrt B\), square first. Then use identities such as \(\sin^2\theta\cos^2\theta=\frac14\sin^22\theta\) to determine extrema.

Answer 1

The Correct Option is B

Concept: To find the maximum and minimum values of \(u^2\), first expand \(u^2\) and then use the bounds of \(\sin^2 2\theta\).

Step 1: Let \[ A=a^2\cos^2\theta+b^2\sin^2\theta, \] \[ B=a^2\sin^2\theta+b^2\cos^2\theta. \] Then \[ u=\sqrt A+\sqrt B. \] Hence, \[ u^2=A+B+2\sqrt{AB}. \]

Step 2: Simplify \(A+B\). \[ A+B = a^2(\cos^2\theta+\sin^2\theta) + b^2(\sin^2\theta+\cos^2\theta). \] \[ A+B=a^2+b^2. \] Therefore, \[ u^2=a^2+b^2+2\sqrt{AB}. \]

Step 3: Find \(AB\). \[ AB = (a^2\cos^2\theta+b^2\sin^2\theta) (a^2\sin^2\theta+b^2\cos^2\theta). \] Expanding, \[ AB = a^2b^2 + (a^2-b^2)^2\sin^2\theta\cos^2\theta. \] Using \[ \sin^2\theta\cos^2\theta = \frac14\sin^22\theta, \] we get \[ AB = a^2b^2 + \frac{(a^2-b^2)^2}{4}\sin^22\theta. \]

Step 4: Find the maximum value of \(u^2\). Since \[ 0\le \sin^22\theta \le 1, \] the maximum value of \(AB\) occurs when \[ \sin^22\theta=1. \] Then \[ AB_{\max} = a^2b^2+\frac{(a^2-b^2)^2}{4} = \frac{(a^2+b^2)^2}{4}. \] Hence, \[ \sqrt{AB_{\max}} = \frac{a^2+b^2}{2}. \] Therefore, \[ u^2_{\max} = a^2+b^2+2\left(\frac{a^2+b^2}{2}\right) = 2(a^2+b^2). \]

Step 5: Find the minimum value of \(u^2\). The minimum value of \(AB\) occurs when \[ \sin^22\theta=0. \] Thus, \[ AB_{\min}=a^2b^2. \] Hence, \[ u^2_{\min} = a^2+b^2+2ab = (a+b)^2. \]

Step 6: Find the required difference. \[ u^2_{\max}-u^2_{\min} = 2(a^2+b^2)-(a+b)^2. \] \[ = a^2+b^2-2ab. \] \[ =(a-b)^2. \] Therefore, \[ \boxed{(a-b)^2} \] \[ \boxed{\text{Answer = (B)}} \]