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Question

# Find the values of a and b so that the function f(x)=⎧⎪⎨⎪⎩x+a√2sinx,0≤x<π/42xcotx+b,π/4≤x≤π/2acos2x−bsinx,π/2<x≤π is continuous for 0≤x≤π.

A
a=π6
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B
b=π12.
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C
b=π6.
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D
a=π12
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Solution

## The correct options are A a=π6 D b=−π12.We apply the test of continuity atx=π/4 and x=π/2 to get the values of a and b.At x=π/4, we have f(π4)=2,π4cotπ4+b=π2+b f(π4−0)=limh→0[π4−h+a√2sin(π4−h)] =π4+a√2.1√(2)=a+π4 and f(π4+0)=limh→0[2(π4+h)cot(π4+h)+b] =π2.1+b=π2+b. ∴ For continuity at x=π/4, we have π2+b=a+π4ora−b=π4....(1) At x=π/2, we have f(π2)=2.π2cotπ2+b=b f(π2+0)=limh→0[2(π2+h)cot(π2−h)+b]=b f(π2+0)=limh→0[acos2(π2+h)−bsin(π2+h)] =−a−b. ∴ For continulity at x=π/2, we have b=−a−bora+2b=0...(2) Solving (1)and(2),we get a=π6,b=−π12.Ans: A,B

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