Question:

If \(f(x)=ax^2+bx+c\) satisfies \[ f(1)+2f(2)=0 \] and \[ 2f(1)+f(2)=0, \] then \(3a+b=\)

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When two linear equations in \(f(1)\) and \(f(2)\) are given, first solve for \(f(1)\) and \(f(2)\), then substitute the polynomial values.
Updated On: Jun 25, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Write the given conditions.
We are given \[ f(1)+2f(2)=0 \] and \[ 2f(1)+f(2)=0 \] Let \[ f(1)=p \] and \[ f(2)=q \] Then the equations become \[ p+2q=0 \] and \[ 2p+q=0 \]

Step 2: Solve for \(p\) and \(q\).
From \[ p+2q=0, \] we get \[ p=-2q \] Substitute this in \[ 2p+q=0 \] Then, \[ 2(-2q)+q=0 \] \[ -4q+q=0 \] \[ -3q=0 \] \[ q=0 \] Therefore, \[ p=0 \] Hence, \[ f(1)=0 \] and \[ f(2)=0 \]

Step 3: Use the quadratic polynomial.
Given, \[ f(x)=ax^2+bx+c \] Now, \[ f(1)=a+b+c=0 \] Also, \[ f(2)=4a+2b+c=0 \]

Step 4: Subtract the equations.
Subtract \[ a+b+c=0 \] from \[ 4a+2b+c=0 \] We get \[ (4a+2b+c)-(a+b+c)=0 \] \[ 3a+b=0 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{0} \]
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