Question

If \(2a + 3b + 6c = 0\), then the equation \(ax^2 + bx + c = 0\) has at least one real root in

• A. \((0,1)\)
• B. \((0,\frac{1}{2})\)
• C. \((\frac{1}{4},\frac{1}{2})\)
• D. None of these

Hint:

Convert coefficients using given relation → reduce quadratic → find roots directly.

Answer 1

The Correct Option is A

Concept: Use Intermediate Value Theorem (IVT): If \(f(x)\) is continuous and changes sign in an interval, then it has a root in that interval.

Step 1: Given condition.
\[ 2a + 3b + 6c = 0 \]

Step 2: Form function.
\[ f(x) = ax^2 + bx + c \]

Step 3: Evaluate at two points.
\[ f(1) = a + b + c \] \[ f(2) = 4a + 2b + c \] Now use: \[ 2a + 3b + 6c = 0 \] Multiply \(f(1)\) by 2: \[ 2a + 2b + 2c \] Subtract from given: \[ (2a+3b+6c) - (2a+2b+2c) = b + 4c = 0 \] Thus: \[ b = -4c \]

Step 4: Substitute back.
\[ 2a + 3(-4c) + 6c = 0 \Rightarrow 2a -12c +6c=0 \Rightarrow 2a -6c=0 \Rightarrow a=3c \]

Step 5: Function becomes.
\[ f(x) = c(3x^2 -4x +1) \]

Step 6: Solve quadratic.
\[ 3x^2 -4x +1 = 0 \] \[ x = \frac{4 \pm \sqrt{16-12}}{6} = \frac{4 \pm 2}{6} \Rightarrow x=1,\ \frac{1}{3} \]

Step 7: Check interval.
\[ \frac{1}{3} \in (0,1) \] Conclusion: \[ {(0,1)} \]