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Question

# If the function f(x)=⎧⎪ ⎪⎨⎪ ⎪⎩x+a2√2sin x,0≤x<π4x cot x+b,π4≤x<π2b sin 2x−a cos 2x,π2≤x≤π is continuous in the interval [0,π] , then the values of (a,b) are

A
(–1, –1)
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B
(0,0)
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C
(–1, 1)
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D
(1,-1)
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Solution

## The correct option is B (0,0) since f is continuous at x = π4 ∴f(π4)−fh→0(π4+h)−fh→0(π4−h) ⇒π4cotπ4+b−fh→0(π4+h)+a2√2sin(π4+h) ⇒π4(1)+b−(π4+0)+a2√2 sin(π4+0) ⇒π4+b−π4+a2√2 sinπ4 ⇒b−a2√21√2⇒b−a2 Also as f is continuous at x - π4 ∴f(π2)=limx→π2−0f(x)=limx→−0f(π2−h) ⇒b sin 2π2−a cos 2π2=limh→−0[(π2−h)cot(π2−h)+b] ⇒b.0−a(−1)=0+b⇒a=b Hence (0,0) satisfy the above relations.

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