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If a > 0 and b^2-4ac<0, then the graph o...

If `a > 0` and `b^2-4ac<0,` then the graph of `y= ax^2+bx+c`

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if a gt 0 and b^2 - 4ac =0 , then the curve y= ax^2 +bx +c

Statement I If a gt 0 and b^(2)- 4ac lt 0 , then the value of the integral int(dx)/(ax^(2)+bx+c) will be of the type mu tan^(-1) . (x+A)/(B)+C , where A, B, C, mu are constants. Statement II If a gt 0, b^(2)- 4ac lt 0 , then ax^(2)+bx +C can be written as sum of two squares .

Statement I If a gt 0 and b^(2)- 4ac lt 0 , then the value of the integral int(dx)/(ax^(2)+bx+c) will be of the type mu tan^(-1) . (x+A)/(B)+C , where A, B, C, mu are constants. Statement II If a gt 0, b^(2)- 4ac lt 0 , then ax^(2)+bx +C can be written as sum of two squares .

Statement I If a gt 0 and b^(2)- 4ac lt 0 , then the value of the integral int(dx)/(ax^(2)+bx+c) will be of the type mu tan^(-1) . (x+A)/(B)+C , where A, B, C, mu are constants. Statement II If a gt 0, b^(2)- 4ac lt 0 , then ax^(2)+bx +C can be written as sum of two squares .

If a gt 0 and b^(2) - 4 ac = 0 then solve ax^(3) + (a + b) x^(2) + (b + c) x + c gt 0 .

If a gt 0 and b^(2) - 4 ac = 0 then solve ax^(3) + (a + b) x^(2) + (b + c) x + c gt 0 .

If a gt 0 and b^(2) - 4 ac = 0 then solve ax^(3) + (a + b) x^(2) + (b + c) x + c gt 0 .

If f(x) = a + bx + cx^(2) where c > 0 and b^(2) - 4ac < 0 . Then the area enclosed by the coordinate axes,the line x = 2 and the curve y = f(x) is given by

If f(x) = a + bx + cx^(2) where c > 0 and b^(2) - 4ac < 0 . Then the area enclosed by the coordinate axes,the line x = 2 and the curve y = f(x) is given by

If b^2 - 4ac