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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:

JEE · Math · previous-year question

  1. A.$$-$$15correct
  2. B.$$-$$60
  3. C.60
  4. D.15

Answer

A. $$-$$15

Explanation

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

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