Theory of EquationsmediumFree

Theory of Equations: Roots Less Other Greater Complete Set Values

JEE Maths question with a full step-by-step solution.

Question

If one of the roots of

ax^2 + ax + a + 1 = 0

is less than

1

and the other is greater than

1

, then the complete set of values of

a

a \in \left(-\dfrac{1}{2}, 0\right)

a \in (-1, 0)

a \in \left(-\dfrac{1}{3}, 0\right)

correct

a \in \left(0, \dfrac{1}{3}\right)

Solution

Step 1: For

a = 0

: the equation becomes

1 = 0

, no solution. So

a \neq 0

. Step 2: Divide by

a

f(x) = x^2 + x + \dfrac{a+1}{a} = 0

. For

x = 1

to lie strictly between the roots of

f(x) = 0

(with leading coefficient

1 > 0

), require

f(1) < 0

f(1) = 1 + 1 + \frac{a+1}{a} = 2 + \frac{a+1}{a} = \frac{2a + a + 1}{a} = \frac{3a+1}{a} < 0

Step 3:

\dfrac{3a+1}{a} < 0

. The numerator and denominator have opposite signs:

a < 0

and

3a + 1 > 0

(i.e.,

a > -1/3

): gives

a \in \left(-\dfrac{1}{3}, 0\right)

a > 0

and

3a+1 < 0

(i.e.,

a < -1/3

): impossible. Correct answer: (3)

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