JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q9

← JEE Main 2022 · 25 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 9 of 90

Let $f(x)$ be a polynomial function such that $f(x)+f'(x)+f''(x)=x^5+64$ . Then, the value of $\displaystyle\lim_{x\to1}\frac{f(x)}{x-1}$ is equal to :

-15

-60

Answer and worked solution

Official answer: A

Invert the operator. $\dfrac1{1+D+D^2}=\dfrac{1-D}{1-D^3}=1-D+D^3-D^4+\dots$

Apply. $f=x^5-5x^4+60x^2-120x+64$ (the constant 64 is unchanged).

Limit. $f(1)=0$ , so the limit is $f'(1)=5-20+120-120=-15$ .

(A)

Topics:Formation and applications Limits Polynomials and remainder theorem

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JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 9