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If f ( x ) is continuous at x = 3, then the value of P is
when x is not equal to 1, 2, 3, 4, write f(x)= P + [2x(x-4) + 3(x-1)]/[(x-1)(x-3)(x-4)]= P + (2x^2 – 5x – 3)/[(x-1)(x-3)(x-4)]= P + [(x-3)(2x+1)]/[(x-1)(x-3)(x-4)]= P + (2x+1)/[(x-1)(x-4)] (since x-3 cannot be 0)Now, lim x tends to 3 f(x) = f(3) for it to be continuous at x=3.or lim x tends to 3 P + (2x+1)/[(x-1)(x-4)] = 2or P – 7/2= 2or P= 11/2KINDLY APPROVE :))
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