Answer :


m at (x = 1) = 2


y at (x = 1) = (1)4 – 6(1)3 + 13(1)2 – 10(1) + 5 = 3


Tangent : y – b = m(x – a)


y – 3 = 2(x – 1)


2x – y + 1 = 0


Normal :



x + 2y – 7 = 0


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