Let f ( x ) be a polynomial of degree 5 . When f ( x ) is divided by ( x − 1 ) 3 , the remainder 33, and when f ( x ) is divided by ( x + 1 ) 3 , the remainder is -3. Then, equation to the tangent dra

Question

Let f ( x ) be a polynomial of degree 5 . When f ( x ) is divided by ( x − 1 ) 3 , the remainder 33, and when f ( x ) is divided by ( x + 1 ) 3 , the remainder is -3. Then, equation to the tangent drawn to y = f ( x ) at x = 0 is

TestHub · Accepted Answer

f ( 1 ) = 33 , f ( − 1 ) = − 3 , and f ′ ( x ) is divisible by ( x − 1 ) 2 ( x + 1 ) 2 , so f ′ ( x ) = a ( x 2 − 1 ) 2 ​ ⇒ f ( x ) = a ( 5 x 5 ​ − 3 2 x 3 ​ + x ) + b ; f ( 1 ) − f ( − 1 ) = 15 16 a ​ = 36 ⇒ a = 4 135 ​ , b = 15 ​ f ( x ) = 4 27 x 5 ​ − 2 45 x 3 ​ + 4 135 x ​ + 15 ; at x = 0 , y = 15 ⇒ f ′ ( 0 ) = 4 135 ​ ​ ⇒ tangent equation is y − 15 = 4 135 ​ ( x ) ⇒ 135 x − 4 y + 60 = 0 ​

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