Let f(x) be a polynomial function such that $f(x) + f'(x) + f''(x) = {x^5} + 64$. Then, the value of $\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$ is equal to:

<p>Given, $f(x) + f'(x) + f''(x) = {x^5} + 64$ .........(i)</p> <p>$\Rightarrow f(x)$ is a polynomial in $x$ whose degree is 5.</p> <p>Let $f(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e$</p> <p>$f'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d$</p> <p>$f''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c$</p> <p>On substituting the value of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$ in Eq. (i), we get</p> <p>$${x^5}+(a + 5){x^4} + (b + 4a + 20){x^3} + (c + 3b + 12a){x^2} + (d + 2c + 6b)x + e + d + 2c = {x^5} + 64$$</p> <p>Now, equating the coefficient, we get</p> <p>$\Rightarrow a + 5 = 0$</p> <p>$b + 4a + 20 = 0$</p> <p>$c + 3b + 12a = 0$</p> <p>$d + 2c + 6b = 0$</p> <p>$e + d + 2c = 64$</p> <p>$\therefore$ $a = - 5,\,b = 0,\,c = 60,\,d = - 120,\,e = 64$</p> <p>$\therefore$ $f(x) = {x^5} - 5{x^4} + 60{x^2} - 120x + 64$</p> <p>Now, $$\mathop {\lim }\limits_{x \to 1} {{{x^5} - 5{x^4} + 60{x^2} - 120x + 64} \over {x - 1}}$$ is (${0 \over 0}$ form)</p> <p>By L' Hospital rule</p> <p>$\mathop {\lim }\limits_{x \to 1} {{5{x^4} - 20{x^3} + 120x - 120} \over 1}$</p> <p>$= - 15$</p>