If left| a sin^2 theta + b sin theta cos thet | vedclass

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(A) Let $f(	heta) = a \sin^2 	heta + b \sin 	heta \cos 	heta + c \cos^2 	heta - \frac{1}{2}(a + c)$.Using the identities $\sin^2 	heta = \frac{1

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$b^2 + (a - c)^2$

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(A) Let $f(	heta) = a \sin^2 	heta + b \sin 	heta \cos 	heta + c \cos^2 	heta - \frac{1}{2}(a + c)$.Using the identities $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ and $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$f(	heta) = a \left( \frac{1 - \cos 2	heta}{2} ight) + b \left( \frac{\sin 2	heta}{2} ight) + c \left( \frac{1 + \cos 2	heta}{2} ight) - \frac{1}{2}(a + c)$$f(	heta) = \frac{1}{2} [a - a \cos 2	heta + b \sin 2	heta + c + c \cos 2	heta - a - c]$$f(	heta) = \frac{1}{2} [b \sin 2	heta - (a - c) \cos 2	heta]$We know that for any expression of the form $A \sin x + B \cos x$,the range is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.Thus,$|b \sin 2	heta - (a - c) \cos 2	heta| \le \sqrt{b^2 + (a - c)^2}$.Therefore,$|f(	heta)| = \frac{1}{2} |b \sin 2	heta - (a - c) \cos 2	heta| \le \frac{1}{2} \sqrt{b^2 + (a - c)^2}$.Comparing this with the given inequality $\left| f(	heta) ight| \le \frac{1}{2}k$,we have $k = \sqrt{b^2 + (a - c)^2}$.Squaring both sides,$k^2 = b^2 + (a - c)^2$.

If $\left| a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta - \frac{1}{2}(a + c) \right| \le \frac{1}{2}k,$ then $k^2$ is equal to

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