Evaluating the value of first derivative at $x=1$ for a polynomial $f$ satisfying $f(x)+f'(x)+f''(x)=x^5+64$ Let $f(x)$ be a polynomial function such that $f(x)+f'(x)+f''(x)=x^5+64$. Then, the value of $\lim_{x \to 1}\frac{f(x)}{x-1}$ is
$\boxed{A) \; -15}$
$B) \; -60$
$C) \; 60$
$D) \; 15$
I have solved it by considering a $5$ degree general polynomial of the form $ax^5+bx^4+cx^3+dx^2+ex+f$ and then algebraically solving to obtain the required result.
Is there any other method which doesn't include assumption of a function of any kind?

Source: JEE Mains 2022 25th June Shift-1
 A: $$f(x) + f'(x) + f''(x) = x^5 + 64$$
Take the derivative.
$$f'(x) + f''(x) + f'''(x) = 5x^4$$
Subtract the two equations:
$$f(x) - f'''(x) = x^5 - 5x^4 + 64$$
Differentiate some more.
$$f'(x) - f''''(x) = 5x^4 - 20x^3$$
$$f''(x) - f'''''(x) = 20x^3 - 60x^2$$
$$f'''(x) - f''''''(x) = 60x^2 - 120x$$
Assuming that $f$ is a quintic, its sixth derivative is everywhere zero.  So the last equation is just $f'''(x) = 60x^2 - 120x$.  Substitute into a previous equation.
$$f(x) - (60x^2 - 120x) = x^5 - 5x^4 + 64$$
$$f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$$
So now we have an explicit formula for $f$.  To double-check, differentiate it twice.
$$f'(x) = 5x^4 - 20x^3 + 120x - 120$$
$$f''(x) = 20x^3 - 60x^2 + 120$$
And confirm that all the middle coefficients in $f + f' + f''$ cancel out, leaving you with $x^5 + 64$.
Now, $f(1) = 0$, so evaluating $\frac{f(x)}{x-1}$ directly at $x=1$ gives you the indeterminate form $\frac{0}{0}$.  L'Hôpital's Rule applies.
$$\lim_{x \rightarrow 1} \frac{f(x)}{x-1} = \lim_{x \rightarrow 1} \frac{f'(x)}{1} = f'(1) = 5 - 20 + 120 - 120 = -15$$
QED
A: If you were not told that $f(x)$ is a polynomial, then you just need to solve the differential equation
$$f(x)+f'(x)+f''(x)=x^5+64$$ With the usual method, the solution is
$$f(x)=e^{-x/2} \left(c_1 \sin \left(\frac{\sqrt{3} x}{2}\right)+c_2 \cos
   \left(\frac{\sqrt{3} x}{2}\right)\right)+ \left(x^5-5 x^4+60 x^2-120 x+64\right)$$ Differentiate and make $x=1$ to obtain
$$f'(1)=-15+\frac{\left(c_1\sqrt{3} -c_2\right) \cos
   \left(\frac{\sqrt{3}}{2}\right)-\left(c_1+c_2\sqrt{3} \right) \sin
   \left(\frac{\sqrt{3}}{2}\right)}{2 \sqrt{e}}$$ and the result depends on the boundary conditions which fix the values of $c_1$ and $c_2$.
A: (This answer originates from this page given by @DatBoi in the comments under the question.)
$$
\lim_{x\rightarrow 1}{f(x)\over x-1}=f'(1) 
$$
(and $f(1)=0$)
$$
f(x)+f'(x)+f''(x)=x^5+64\\
f'(x)+f''(x)+f'''(x)=5x^4\\
f''(x)+f'''(x)+f^{iv}(x)=20x^3\\
f'''(x)+f^{iv}(x)+f^{v}(x)=60x^2\\
$$
so $$f^v(x)-f''(x)=60x^2-20x^3,$$ and
since $f(x)=x^5+\text{lower order terms}$, $$f^v(x)={d^5\over dx^5}x^5=5!=120$$ we have $f^v(1)=120$, so this becomes
$$120-f''(1)=40,$$ thus $f''(1)=80$. Since $$f(1)+f'(1)+f''(1)=65,$$ $f'(1)=-15$.
