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UPSC 2019 Maths Optional Paper 1 Q4c-ii — Step-by-Step Solution 8 marks · Section A
Jacobian · Calculus · asked 4× in 13 yrs · Read the full method →
Question
Using the Jacobian method, show that if f ′ ( x ) = 1 1 + x 2 f'(x)=\dfrac{1}{1+x^2} f ′ ( x ) = 1 + x 2 1 f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 f ( x ) + f ( y ) = f ( x + y 1 − x y ) . f(x)+f(y)=f\!\left(\dfrac{x+y}{1-xy}\right). f ( x ) + f ( y ) = f ( 1 − x y x + y ) .
Technique
Jacobian test for functional dependence: ∂ ( u , w ) / ∂ ( x , y ) = 0 ⇒ u = Φ ( w ) \partial(u,w)/\partial(x,y)=0\Rightarrow u=\Phi(w) ∂ ( u , w ) / ∂ ( x , y ) = 0 ⇒ u = Φ ( w ) Φ \Phi Φ y = 0 y=0 y = 0 f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0
Solution
Step 1 — Set up two functions and their Jacobian
Define
u ( x , y ) = f ( x ) + f ( y ) , w ( x , y ) = x + y 1 − x y . u(x,y)=f(x)+f(y),\qquad w(x,y)=\frac{x+y}{1-xy}. u ( x , y ) = f ( x ) + f ( y ) , w ( x , y ) = 1 − x y x + y . The Jacobian method : if the Jacobian ∂ ( u , w ) ∂ ( x , y ) ≡ 0 \dfrac{\partial(u,w)}{\partial(x,y)}\equiv0 ∂ ( x , y ) ∂ ( u , w ) ≡ 0 u u u w w w functionally dependent , i.e. there is a relation u = Φ ( w ) u=\Phi(w) u = Φ ( w ) Φ = f \Phi=f Φ = f
Step 2 — Partial derivatives
Using f ′ ( t ) = 1 1 + t 2 f'(t)=\dfrac{1}{1+t^2} f ′ ( t ) = 1 + t 2 1
u x = 1 1 + x 2 , u y = 1 1 + y 2 . u_x=\frac{1}{1+x^2},\qquad u_y=\frac{1}{1+y^2}. u x = 1 + x 2 1 , u y = 1 + y 2 1 . For w = x + y 1 − x y w=\dfrac{x+y}{1-xy} w = 1 − x y x + y
w x = ( 1 − x y ) ⋅ 1 − ( x + y ) ( − y ) ( 1 − x y ) 2 = 1 − x y + x y + y 2 ( 1 − x y ) 2 = 1 + y 2 ( 1 − x y ) 2 , w_x=\frac{(1-xy)\cdot1-(x+y)(-y)}{(1-xy)^2}=\frac{1-xy+xy+y^2}{(1-xy)^2}=\frac{1+y^2}{(1-xy)^2}, w x = ( 1 − x y ) 2 ( 1 − x y ) ⋅ 1 − ( x + y ) ( − y ) = ( 1 − x y ) 2 1 − x y + x y + y 2 = ( 1 − x y ) 2 1 + y 2 , w y = 1 + x 2 ( 1 − x y ) 2 ( by symmetry ) . w_y=\frac{1+x^2}{(1-xy)^2}\quad(\text{by symmetry}). w y = ( 1 − x y ) 2 1 + x 2 ( by symmetry ) . Step 3 — The Jacobian vanishes
∂ ( u , w ) ∂ ( x , y ) = ∣ u x u y w x w y ∣ = u x w y − u y w x = 1 1 + x 2 ⋅ 1 + x 2 ( 1 − x y ) 2 − 1 1 + y 2 ⋅ 1 + y 2 ( 1 − x y ) 2 . \frac{\partial(u,w)}{\partial(x,y)}=\begin{vmatrix}u_x&u_y\\w_x&w_y\end{vmatrix}=u_xw_y-u_yw_x=\frac{1}{1+x^2}\cdot\frac{1+x^2}{(1-xy)^2}-\frac{1}{1+y^2}\cdot\frac{1+y^2}{(1-xy)^2}. ∂ ( x , y ) ∂ ( u , w ) = u x w x u y w y = u x w y − u y w x = 1 + x 2 1 ⋅ ( 1 − x y ) 2 1 + x 2 − 1 + y 2 1 ⋅ ( 1 − x y ) 2 1 + y 2 . = 1 ( 1 − x y ) 2 − 1 ( 1 − x y ) 2 = 0. =\frac{1}{(1-xy)^2}-\frac{1}{(1-xy)^2}=0. = ( 1 − x y ) 2 1 − ( 1 − x y ) 2 1 = 0. Since the Jacobian is identically zero, u u u w w w f ( x ) + f ( y ) = Φ ( x + y 1 − x y ) f(x)+f(y)=\Phi\!\left(\dfrac{x+y}{1-xy}\right) f ( x ) + f ( y ) = Φ ( 1 − x y x + y ) Φ \Phi Φ
Step 4 — Identify Φ = f \Phi=f Φ = f
To find Φ \Phi Φ y = 0 y=0 y = 0 w = x + 0 1 − 0 = x w=\dfrac{x+0}{1-0}=x w = 1 − 0 x + 0 = x
u ( x , 0 ) = f ( x ) + f ( 0 ) = f ( x ) + 0 = f ( x ) . u(x,0)=f(x)+f(0)=f(x)+0=f(x). u ( x , 0 ) = f ( x ) + f ( 0 ) = f ( x ) + 0 = f ( x ) . Hence Φ ( x ) = u ( x , 0 ) = f ( x ) \Phi(x)=u(x,0)=f(x) Φ ( x ) = u ( x , 0 ) = f ( x ) Φ = f \Phi=f Φ = f
Answer
f ( x ) + f ( y ) = f ( x + y 1 − x y ) . \boxed{\;f(x)+f(y)=f\!\left(\frac{x+y}{1-xy}\right).\;} f ( x ) + f ( y ) = f ( 1 − x y x + y ) .