The equation of the curve satisfying the differential equation `y_2(x^2+1)=2x y_1` passing through the point (0,1) and having slope of tangent at `x=0` as 3 (where `y_2` and `y_1` represent 2nd and 1st order derivative), then (a) `( b ) (c) y=f(( d ) x (e))( f )` (g) is a strictly increasing function (h) `( i ) (j) y=f(( k ) x (l))( m )` (n) is a non-monotonic function (o) `( p ) (q) y=f(( r ) x (s))( t )` (u) has a three distinct real roots (v) `( w ) (x) y=f(( y ) x (z))( a a )` (bb) has only one negative root.

Options

A. `y=f(x)` is a strictly increasing function
B. `y=f(x)` is a non-monoatomic function
C. `y=f(x)` has three distinct real root
D. `y=f(x)` has only one negative root

Answer

A, D

Explanation

The given differential equation is
`y_(2)(x^(2)-1)=2xy_(1)`or `y_(2)/y_(1)=(2x)/(x^(2)+1)`
Integrating both sides, we get
`logy_(1)=log(x^(2)+1)+logC`………(1)
It is given that `y_(1)=3` at x=0
Putting `x=0, y_(1)=3` at `x=0`
Substituting the value of C in (1), we obtain
`y_(1)=3(x^(2)+1)` ..............(2)
Integrating both sides w.r.t to x, we get
`y=x^(3)+3x+C_(2)`
This passes through the point (0,1). Therefore, 1`=C_(2)`
Hence, the required equation of the curve is `y=x^(3)+3x+1`
Obviously, is it strictly increasing from equation (2).
Also, `f(0)=1 gt0`. Then the only root is negative.