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Question

For what value of is the function defined by continuous at x = 0? What about continuity at x = 1?

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Solution

The given function is, f( x )={ λ( x 2 −2x )   ,  x≤0 4x+1           ,  x>0 The left hand limit of the function is, LHL= lim x→ 0 − f( x ) = lim x→ 0 − λ( x 2 −2x ) =λ[ 0 2 −2( 0 ) ] =0 The right hand limit of the function is, RHL= lim x→ 0 + f( x ) = lim x→ 0 + ( 4x+1 ) =4(0)+1 =1 As the function is continuous at x=0, so LHL=RHL. lim x→ 0 − f( x )= lim x→ 0 + f( x ) 0=1 This is not possible. Therefore, there is no real value of λ for which the given function is continuous at x=0. When x=1, the function becomes, f( 1 )=4( 1 )+1 =5 The limit of the function is, lim x→1 f( x )= lim x→1 4( 1 )+1 =5 It can be observed that, lim x→1 f( x )=f( 1 ). Therefore, the function is continuous for all real values of λ at x=1.

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