Question:
If \(f(x)=ax^{3}+bx^{2}+cx+1\) attains an extreme value \(2\) at \(x=1\) and another extreme value at \(x=\frac{2}{3}\), then \(2b+3c\) is equal to
If \(f(x)=ax^{3}+bx^{2}+cx+1\) attains an extreme value \(2\) at \(x=1\) and another extreme value at \(x=\frac{2}{3}\), then \(2b+3c\) is equal to
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If the roots of \(f'(x)\) are known, write the derivative directly in factorized form and compare coefficients.
Updated On: Jun 9, 2026
- \(a\)
- \(2a\)
- \(3a\)
- \(4a\)
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The Correct Option is A
Solution and Explanation
Concept:
At points of maxima or minima, the first derivative vanishes.
\[
f'(x)=0
\]
Hence the extremum points become roots of the derivative.
Step 1: Form the derivative.
\[ f'(x)=3ax^2+2bx+c \] Given extrema occur at \[ x=1,\qquad x=\frac23 \] Therefore \[ f'(x)=3a(x-1)\left(x-\frac23\right) \] \[ =3ax^2-5ax+2a \]
Step 2: Compare coefficients.
Comparing with \[ 3ax^2+2bx+c \] gives \[ 2b=-5a \] \[ b=-\frac{5a}{2} \] and \[ c=2a \]
Step 3: Find \(2b+3c\).
\[ 2b+3c = 2\left(-\frac{5a}{2}\right)+3(2a) \] \[ =-5a+6a \] \[ =a \] center minipage0.35
\(2b+3c=a\) minipage center
Step 1: Form the derivative.
\[ f'(x)=3ax^2+2bx+c \] Given extrema occur at \[ x=1,\qquad x=\frac23 \] Therefore \[ f'(x)=3a(x-1)\left(x-\frac23\right) \] \[ =3ax^2-5ax+2a \]
Step 2: Compare coefficients.
Comparing with \[ 3ax^2+2bx+c \] gives \[ 2b=-5a \] \[ b=-\frac{5a}{2} \] and \[ c=2a \]
Step 3: Find \(2b+3c\).
\[ 2b+3c = 2\left(-\frac{5a}{2}\right)+3(2a) \] \[ =-5a+6a \] \[ =a \] center minipage0.35
\(2b+3c=a\) minipage center
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