JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 16)
Let $$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$$, x $$\in$$ R. Then the natural number n for which $$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$ is __________.
Answer
7
Explanation
$$f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3$$
$$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$
$$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$$
$$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^2} + 2)} \over 1} = 44$$
$$\Rightarrow$$ 9n $$-$$ (19) = 44
$$\Rightarrow$$ 9n = 63
$$\Rightarrow$$ n = 7
$$\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44$$
$$\mathop {\lim }\limits_{x \to 1} {{9{x^n} - ({x^6} + 2{x^4} + {x^3} + 2x + 3)} \over {x - 1}} = 44$$
$$\mathop {\lim }\limits_{x \to 1} {{9n{x^{n - 1}} - (6{x^5} + 8{x^3} + 3{x^2} + 2)} \over 1} = 44$$
$$\Rightarrow$$ 9n $$-$$ (19) = 44
$$\Rightarrow$$ 9n = 63
$$\Rightarrow$$ n = 7
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