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Question: If the function f(x) = \(\frac { a x + b } { ( x - 1 ) ( x - 4 ) }\)has a local maxima at (2, -1) t...

If the function f(x) = ax+b(x−1)(x−4)\frac { a x + b } { ( x - 1 ) ( x - 4 ) }has a local maxima at

(2, -1) then

A

b = 1, a = 0

B

a = 1, b = 0

C

b = -1, a = 0

D

a = −1, b = 0

Answer

a = 1, b = 0

Explanation

Solution

Clearly f(2) =-1

⇒ −1 = 2a+b(2−1)(2−4)\frac { 2 a + b } { ( 2 - 1 ) ( 2 - 4 ) } ⇒ 2a + b = 2

Now f '(x) = 4a+5b−2bx−ax2(2−1)(2−4)\frac { 4 a + 5 b - 2 b x - a x ^ { 2 } } { ( 2 - 1 ) ( 2 - 4 ) }, f '(2) = 0

⇒ a = 1 ⇒ b = 0

⇒ f'(x) = −(x−2)(x+2)(x−1)(x−4)2\frac { - ( x - 2 ) ( x + 2 ) } { ( x - 1 ) ( x - 4 ) ^ { 2 } }

Clearly for x > 2, f '(x) < 0 and for x < 2, f '(x) > 0.

Thus x = 2 is indeed the point of local maxima for y = f(x).