Home
Class 12
MATHS
If a+b+c=0, then the equation 3ax^(2)+2b...

If `a+b+c=0`, then the equation `3ax^(2)+2bx+c=0` has :

A

at least one real root in (0, 1)

B

one root is (-1, 0) and other in (1, 2)

C

both imaginary roots

D

two coincident roots.

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • APPLICATION OF DERIVATIVES

    MODERN PUBLICATION|Exercise Latest Questions from AIEEE/JEE Examinations|13 Videos
  • APPLICATION OF DERIVATIVES

    MODERN PUBLICATION|Exercise Recent Competitive Questions (Questions from Karnataka CET & COMED)|25 Videos
  • APPLICATION OF DERIVATIVES

    MODERN PUBLICATION|Exercise Recent Competitive Questions (Questions from Karnataka CET & COMED)|25 Videos
  • AREA UNDER CURVES

    MODERN PUBLICATION|Exercise Recent Competitive Questions (Question from karnataka CET & COMED)|9 Videos

Similar Questions

Explore conceptually related problems

If 2 a+3 b+6 c=0, a, b, c in R then the equation a x^(2)+b x+c=0 has a root in

Let a, b,c in R and a ne 0. If alpha is a root of a^(2) x^(2) + bx + c = 0 , beta is a root of a^(2) x^(2) - bx - x = 0 and 0 lt alpha lt beta . Then the equation a^(2) x^(2) + 2 bx + 2c = 0 has a root gamma that always satisfies :

If a + b + c = 0 , then the equation equation : 3ax^(2) + 2bx + c = 0 has :

If a,b,c are in G.P.L, then the equations ax^(2) + 2bx + c = 0 0 and dx^(2) + 2ex+ f = 0 have a common root if ( d)/( a) , ( e )/( b ) , ( f )/( c ) are in :

If a,b, c epsilon R(a!=0) and a+2b+4c=0 then equatio ax^(2)+bx+c=0 has

If 2a+3b+6c = 0, then show that the equation a x^2 + bx + c = 0 has atleast one real root between 0 to 1.

If the equation ax^(2) + 2bx - 3c = 0 has non-real roots and ((3c)/(4)) lt (a + b) , then c is always :

The discriminant of the quadratic equations ax^(2) + bx+ c =0 is :

In a triangle PQR, /_R = pi//2 , If tan(P//2). and tan(Q//2) are the roots of the equation : ax^(2) + bx + c = 0 a != 0 , then :

a, b, c in R, a!= 0 and the quadratic equation ax^2+bx+c=0 has no real roots, then

MODERN PUBLICATION-APPLICATION OF DERIVATIVES -Multiple Choice Questions LEVEL-II
  1. A cannon ball is fired at an angle theta, 0 lt theta lt pi//2, with th...

    Text Solution

    |

  2. The area of the triangle formed by the co-ordinate axes and a tangent ...

    Text Solution

    |

  3. If a+b+c=0, then the equation 3ax^(2)+2bx+c=0 has :

    Text Solution

    |

  4. Let f(x) be twice differentiable on [1, 3], and let f(1)=f(3). Further...

    Text Solution

    |

  5. Let f be a real valued function defined on (0, 1) cup (2, 4), such tha...

    Text Solution

    |

  6. If x in (0, pi//2), then :

    Text Solution

    |

  7. The curve y=ax^(3)+bx^(2)+cx is inclined at 45^(@) to X-axis at (0, 0)...

    Text Solution

    |

  8. If log(0.3) (x-1) lt log(0.09) (x-1), then x lies in the interval :

    Text Solution

    |

  9. The angle of intersection of the two curves xy=a^(2) and x^(2)-y^(2)=2...

    Text Solution

    |

  10. The length of subtangent to the curve x^(2)+xy+y^(2)=7 at (1, -3) is ...

    Text Solution

    |

  11. The equation 3x^(2)+4ax+b=0 has atleast one root in (0, 1) if :

    Text Solution

    |

  12. The curve y-e^(xy)+x=0 has a vertical tangent at the point :

    Text Solution

    |

  13. If the normal to the curve y = f(x) at the point (3, 4) makes an angle...

    Text Solution

    |

  14. If x+y=k is normal to y^(2)=12x, then k is :

    Text Solution

    |

  15. The two curves x^(3)-3xy^(2)+2=0 and 3x^(2)y-y^(3)-2=0:

    Text Solution

    |

  16. The greatest distance of the point P(10, 7) form the circle x^(2)+y^(...

    Text Solution

    |

  17. The greatest value of f(x)=(x+1)^(1//3)-(x-1)^(1//3) on [0, 1] is :

    Text Solution

    |

  18. The point(s) on the curve y^(3)+3x^(2)=12y, where the tangent is verti...

    Text Solution

    |

  19. If the function f(x)=2x^(3)-9ax^(2)+12a^(2)x+1, where a gt0 attains it...

    Text Solution

    |

  20. If minimum value of f(x)=x^(2)+2bx+2c^(2) is greater than maximum valu...

    Text Solution

    |