Question

If \(A\) and \(B\) are the minimum and maximum values of \[ \sin^6 x+\cos^6 x, \] then \(A+B=\)

• A. \(1\)
• B. \(-1\)
• C. \(\dfrac{5}{4}\)
• D. \(\dfrac{7}{4}\)

Hint:

For expressions involving \(\sin^6x+\cos^6x\), convert them using the identity \(a^3+b^3=(a+b)^3-3ab(a+b)\). Then use \(0\le\sin^2x\cos^2x\le\frac14\) to find the extrema.

Answer 1

The Correct Option is C

Concept: Use the identity \[ a^3+b^3=(a+b)^3-3ab(a+b). \] Taking \[ a=\sin^2x, \qquad b=\cos^2x, \] and using \[ \sin^2x+\cos^2x=1, \] the expression can be simplified in terms of \(\sin^2x\cos^2x\).

Step 1: Simplify the given expression. Let \[ S=\sin^6x+\cos^6x. \] Using \[ a^3+b^3=(a+b)^3-3ab(a+b), \] we get \[ S = (\sin^2x+\cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x+\cos^2x). \] Since \[ \sin^2x+\cos^2x=1, \] \[ S = 1-3\sin^2x\cos^2x. \]

Step 2: Find the maximum value of \(S\). Since \[ \sin^2x\cos^2x\ge 0, \] the maximum value of \(S\) occurs when \[ \sin^2x\cos^2x=0. \] Thus, \[ B=1. \]

Step 3: Find the minimum value of \(S\). Using \[ \sin^2x\cos^2x = \frac{\sin^22x}{4}, \] we have \[ 0\le \sin^2x\cos^2x\le \frac14. \] Hence, \[ S_{\min} = 1-3\left(\frac14\right) = \frac14. \] Therefore, \[ A=\frac14. \]

Step 4: Calculate \(A+B\). \[ A+B = \frac14+1 = \frac54. \] \[ \boxed{A+B=\frac54} \] \[ \boxed{\text{Answer = (C)}} \]