The range of absinx+b1−a2⋅cosx+c, (|a| 0) respectively are

c±(ab)2+(b1−a2)2=c±a2b2+b2(1−a2)=c±b2(a2+1−a2)=c±b2(1)=c±bRange ={c−b, c+b}.

The range of absinx+b1−a2⋅cosx+c, (|a| 0) respectively are

c±(ab)2+(b1−a2)2=c±a2b2+b2(1−a2)=c±b2(a2+1−a2)=c±b2(1)=c±bRange ={c−b, c+b}.

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The range of $a b \sin x + b \sqrt{1 - a^{2}} \cdot \cos x + c$ , $(| a | < 1, b > 0)$ respectively are

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${c - b, b + c}$

${c + b, c - b}$

${b + c, b - c}$

${b - c, b + c}$

answer is C.

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Detailed Solution

$c \pm \sqrt{{(a b)}^{2} + {(b \sqrt{1 - a^{2}})}^{2}}$

$= c \pm \sqrt{a^{2} b^{2} + b^{2} (1 - a^{2})}$

$= c \pm \sqrt{b^{2} (a^{2} + 1 - a^{2})}$

$= c \pm \sqrt{b^{2} (1)}$

$= c \pm b$

Range $= {c - b, c + b}$ .

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