If f ( x ) + f ( y ) = f ( x ) f ( y ) + f ( x y ) , f ( 1 ) = 0 , f ′ ( 1 ) = − 2 then, equation to the tangent, drawn to the curve y = f ( x ) at x = 2 is,

Question

If f ( x ) + f ( y ) = f ( x ) f ( y ) + f ( x y ) , f ( 1 ) = 0 , f ′ ( 1 ) = − 2 then, equation to the tangent, drawn to the curve y = f ( x ) at x = 2 ​ is,

TestHub · Accepted Answer

1 − f ( x y ) = ( 1 − f ( x )) ( 1 − f ( y )) ⇒ 1 − f ( x ) = x c . As f ′ ( 1 ) = − 2 , c = 2 Clearly f ( x ) = 1 − x 2 ; at x = 2 ​ , y = − 1 , f ′ ( 2 ​ ) = − 2 2 ​ ⇒ tangent at ( 2 ​ , − 1 ) is y + 1 = − 2 2 ​ ( x − 2 ​ ) , i.e. 2 2 ​ x + y − 3 = 0

Mathematics - Application of Derivative Question with Solution | TestHub

Options:

Doubts & Discussion