Question

# If the solution of y=xdydx+dydx−(dydx)2;d2ydx2≠0 is y=f(x) and it is defined as f:R→R then f(x) is

A
one-one
B
into
C
even
D
onto

Solution

## The correct option is B intoy=xdydx+dydx−(dydx)2 Differentiating w.r.t. x, y′=xy′′+y′+y′′−2y′y′′⇒0=(x+1−2y′)y′′⇒y′=x+12 Putting it in the given differential equation we get, y=(x+1)×(x+1)2−(x+1)24⇒y=(x+1)24 So, f(x)=(x+1)24 The function is many one. As f(x)≠f(−x) so, it is not even. The range of function is [0,∞) Therefore, the function is into.Mathematics

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