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If f(x)={(x-1)/(2x^2-7x+5), when x !=1 -...

If `f(x)={(x-1)/(2x^2-7x+5),` when `x !=1 -1/3,` when `x=1,` prove that `f prime(1)=-2/9.`

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Verified by Experts

The correct Answer is:
`(-2/9)`
Given that , `f(x) ={{:((x-1)/(2x^(2) -7x+5)", when " x ne 1),(-1/3" , when " x = 1):}`
RHD ` = underset( h to 0) lim (f(1+h)-f(1))/h `
` = underset( h to 0) lim [(1+h-1)/(2(1+h)^(2) -7(1+h)+5)-(-1/3)]/h`
`=underset( h to 0) lim [(3h+2(1+h)^(2)-7(1+h)+5)/(3h{2(1+h)^(2)-7 (1+h)+5})]`
`=underset( h to 0) lim ((2h^(2))/(3h(-3h+2h^(2)))) =- 2/9`
LHD = ` underset( h to 0) lim (f(1-h) - f(1))/(-h) `
` = underset( h to 0) lim ([(1-h-1)/(2(1-h)^(2)-7(1-h)+5)-(-1/3)])/(-h)`
`=underset( h to 0) lim (-3h+2(1+h^(2)-2h)-7(1-h)+5)/(-3h[2(1-h)^(2) -7 (1-h)+5])`
`= underset( h to 0) lim (2h^(2))/(-3h(2h^(2)+3h)) =-2/9:. "LHD = RHD"`
Hence, required value of `f'(1) =- 2/9`.
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