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

• A. \(a\)
• B. \(2a\)
• C. \(3a\)
• D. \(4a\)

Hint:

If the roots of \(f'(x)\) are known, write the derivative directly in factorized form and compare coefficients.

Answer 1

The Correct Option is A

Step 1: Differentiate the cubic.
For $f(x)=ax^3+bx^2+cx+1$, the derivative is $f'(x)=3ax^2+2bx+c$.
Step 2: Use that extrema are roots of $f'$.
Extreme values occur at $x=1$ and $x=\tfrac23$, so these are the roots of $f'(x)=0$.
Step 3: Write $f'$ in factored form.
\[ f'(x)=3a(x-1)\left(x-\frac23\right)=3a\left(x^2-\frac53x+\frac23\right)=3ax^2-5ax+2a. \]
Step 4: Match coefficients.
Comparing with $3ax^2+2bx+c$: the $x$-term gives $2b=-5a$, and the constant gives $c=2a$.
Step 5: Form the required combination.
\[ 2b+3c=(-5a)+3(2a). \]
Step 6: Simplify.
\[ 2b+3c=-5a+6a=a. \]
\[ \boxed{a} \]