IIT-JEE 2007 Paper 1 Offline | Quadratic Equation and Inequalities Question 26 | Mathematics

IIT-JEE 2007 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$\alpha,\beta$$ be the roots of the equation $$x^2-px+r=0$$ and $$\frac{\alpha}{2},2\beta$$ be the roots of the equation $$x^2-qx+r=0$$. Then the value of r is

$$\frac{2}{9}(p-q)(2q-p)$$

$$\frac{2}{9}(q-p)(2p-q)$$

$$\frac{2}{9}(q-2p)(2q-p)$$

$$\frac{2}{9}(2p-q)(2q-p)$$

IIT-JEE 2006

MCQ (Single Correct Answer)

-1

Let $$a, b, c$$ be the sides of a triangle. No two of them are equal and $$\lambda \in R$$. If the roots of the equation $$x^{2}+2(a+b+c) x+3 \lambda(a b+b c+c a)=0$$ are real, then,

$$\lambda<\frac{4}{3}$$

$$\lambda>\frac{5}{3}$$

$$\lambda \in\left(\frac{1}{3}, \frac{5}{3}\right)$$

$$\lambda \in\left(\frac{4}{3}, \frac{5}{3}\right)$$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

If one root is square of the other root of the equation $${x^2} + px + q = 0$$, then the realation between $$p$$ and $$q$$ is

$${p^3} - q\left( {3p - 1} \right) + {q^2} = 0$$

$${p^3} - q\left( {3p + 1} \right) + {q^2} = 0$$

$${p^3} + q\left( {3p - 1} \right) + {q^2} = 0$$

$${p^3} + q\left( {3p + 1} \right) + {q^2} = 0$$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

For all $$'x',{x^2} + 2ax + 10 - 3a > 0,$$ then the interval in which '$$a$$' lies is

$$a < - 5$$

$$ - 5 < a < 2$$

$$a > 5$$

$$2 < a < 5$$

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