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प्रश्न
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]
उत्तर
`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
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