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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 ax2+ax+a+1=0ax^2 + ax + a + 1 = 0 is less than 11 and the other is greater than 11, then the complete set of values of aa is
Aa∈(−12,0)a \in \left(-\dfrac{1}{2}, 0\right)
Ba∈(−1,0)a \in (-1, 0)
Ca∈(−13,0)a \in \left(-\dfrac{1}{3}, 0\right)correct
Da∈(0,13)a \in \left(0, \dfrac{1}{3}\right)
Solution
Step 1: For a=0a = 0: the equation becomes 1=01 = 0, no solution. So a≠0a \neq 0. Step 2: Divide by aa: f(x)=x2+x+a+1a=0f(x) = x^2 + x + \dfrac{a+1}{a} = 0. For x=1x = 1 to lie strictly between the roots of f(x)=0f(x) = 0 (with leading coefficient 1>01 > 0), require f(1)<0f(1) < 0:
f(1)=1+1+a+1a=2+a+1a=2a+a+1a=3a+1a<0f(1) = 1 + 1 + \frac{a+1}{a} = 2 + \frac{a+1}{a} = \frac{2a + a + 1}{a} = \frac{3a+1}{a} < 0
 Step 3: 3a+1a<0\dfrac{3a+1}{a} < 0. The numerator and denominator have opposite signs: a<0a < 0 and 3a+1>03a + 1 > 0 (i.e., a>−1/3a > -1/3): gives a∈(−13,0)a \in \left(-\dfrac{1}{3}, 0\right). a>0a > 0 and 3a+1<03a+1 < 0 (i.e., a<−1/3a < -1/3): impossible. Correct answer: (3)
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