Question

If $f\left(x\right)$ is a polynomial of degree $4$ such that $\underset{x\to -1}{lim}\frac{f\left(x\right)}{(x+1{)}^3}=1$ and $f^{\prime \prime \prime }\left(0\right)=-12$, then the maximum value of $f\left(x\right)$ is
​​​​​​​(correct answer + 1, wrong answer - 0.25)

• A. $1$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $2$

Answer 1

The correct option is B $\frac{1}{4}$

Given : $\underset{x\to -1}{lim}\frac{f\left(x\right)}{(x+1{)}^3}=1$
By observation, we get
$f\left(x\right)=(x+1{)}^3[ax+b]$
Now,
$\underset{x\to -1}{lim}\frac{f\left(x\right)}{(x+1{)}^3}=1\Rightarrow b-a=1\Rightarrow f\left(x\right)=(x+1{)}^3[ax+1+a]\Rightarrow f(x)=a(x+1{)}^4+(x+1{)}^3$
Also,
$f^{\prime \prime \prime }\left(0\right)=-12\Rightarrow 24a+6=-12\Rightarrow a=-\frac{3}{4}\therefore f\left(x\right)=(x+1{)}^3-\frac{3}{4}(x+1{)}^4$

Now, for maxima and minima
$f^{\prime }\left(x\right)=0\Rightarrow 3(x+1{)}^2-3(x+1{)}^3=0\Rightarrow (x+1{)}^2[1-x-1]=0\Rightarrow x=-1,0$
$f(-1)=0,f\left(0\right)=\frac{1}{4}$

Hence, the maximum value of the function is $\frac{1}{4}$