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Question

# Let f(x)=⎧⎨⎩−2sinx, x≤−mAsinx+B,−m<x<mcosx, x≥m be a continuous function, where m is the principal solution of the equation sin3x+sinxcosx+cos3x=1. If m>0, then

A
A=1, B=1
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B
m=π
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C
m=π2
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D
A=1, B=1
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Solution

## The correct options are A A=−1, B=1 C m=π2sin3x+sinxcosx+cos3x=1 ⇒(sin3x+cos3x)−(1−sinxcosx)=0 ⇒(sinx+cosx)(1−sinxcosx)−(1−sinxcosx)=0 ⇒(1−sinxcosx)(sinx+cosx−1)=0⇒sinxcosx=1 or sinx+cosx=1⇒sin2x=2 or sinx+cosx=1 sin2x=2 is not possible. sinx+cosx=1 ⇒sinxsinπ4+cosxcosπ4=1√2 ⇒cos(x−π4)=cosπ4 ⇒x=0,π2,2π But m is a principal solution with m>0. So, m=π2 Now, f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩−2sinx, x≤−π2Asinx+B,−π2<x<π2cosx, x≥π2 Continuity at x=−π2 limx→−π/2−f(x)=limx→−π/2+f(x) ⇒2=−A+B ⋯(1) Continuity at x=π2 limx→π/2−f(x)=limx→π/2+f(x) ⇒A+B=0 ⋯(2) Solving equation (1) and (2), we get B=1, A=−1

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