#### Question

Find a and b, so that the function f(x) defined by

f(x)=-2sin x, for -π≤ x ≤ -π/2

=a sin x+b, for -π/2≤ x ≤ π/2

=cos x, for π/2≤ x ≤ π

is continuous on [- π, π]

#### Solution

`f(x)=-2sinx, " for " -pi<=x<=-pi/2`

`=asinx+b , " for " -pi/2<x<pi/2`

`=cosx , " for " pi/2<=x<pi`

f(x) is continuous for x=-π/2

RHL,

`=lim_(x->-pi/2)asinx+b`

`=asin(-pi/2)+b`

=-a+b

`f(-pi/2)=-2sin(-pi/2)`

`therefore-a+b=2........(i) [because f(x) " is continuous for "x=-x/2]`

`f(x) " is continuous for "x=pi/2`

LHL,

`=lim_(x->pi/2)asinx+b`

`=asin(pi/2)+b`

`=a+b`

`f(pi/2)=cos(pi/2)=0`

`a+b=0` .........(ii)

Solving (i) and (ii)

a= -1 and b=1

Is there an error in this question or solution?

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Solution Find a and b, so that the function f(x) defined by Concept: Continuity - Continuity of a Function at a Point.

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