InterviewSolution
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InterviewSolution
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The equation of the curve satisfying thedifferential equation `y_2(x^2+1)=2x y_1`passingthrough the point (0,1) and having slope of tangent at `x=0`as 3 (where `y_2`and `y_1`represent 2ndand 1st order derivative), then(a)`( b ) (c) y=f(( d ) x (e))( f )`(g) is astrictly increasing function(h)`( i ) (j) y=f(( k ) x (l))( m )`(n) is anon-monotonic function(o)`( p ) (q) y=f(( r ) x (s))( t )`(u) has a threedistinct real roots(v)`( w ) (x) y=f(( y ) x (z))( a a )`(bb)has only one negative root.A. `y=f(x)` is a strictly increasing functionB. `y=f(x)` is a non-monoatomic functionC. `y=f(x)` has three distinct real rootD. `y=f(x)` has only one negative root |
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Answer» Correct Answer - A::D 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. |
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