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f(x)= x^2+3x+b (x>1,X=1)
2ax+3 (x<1)
f(x) is continous and differenstiable then find value of a and b
at x=1 f(1) = 1+3+b=b+4
at x=1-h whr h-->0 , f(1-h)=2a(1-h)+3=2a+3
at x=1+h whr h-->0 , f(1+h)=(1+h)^2+3(1+h)+b=1+3+b
as all 3 eqns shld be equal for function to be continuous, it means : 2a+3 = b+4
f(x) is differentiable
f'(x) = 2x+3 at x=1 its f'(1)=5
f'(x+h)= [ f(x+h)-f(x)] / h = [ f(1+h)-f(1)] / h = [ (1+h)^2+3(1+h)+b - b-4 ] / h = h+5 = 5
f'(x-h)=[ f(x-h)-f(x) ]/ -h = [ f(1-h)-f(1)] / -h = [ 2a -2ah +3 - b-4 ] / h
to cancel on both numerator and denominator, (b+1-2a) shld be 0
= 2a = 5
a=5/2 .. on putting above we have
b=4
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regards
Ramesh
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