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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Differentiation

Let

f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3, x \in R

. Then the natural number

n

for which

lim_{x \to 1} \frac{x ^{n} f ( 1 ) - f ( x )}{x - 1} = 44

is ______.

Answer: 7

Solution

With

f (1) = 1 + 2 + 1 + 2 + 3 = 9

, the numerator vanishes at

x = 1

, so the limit has the form

\frac{0}{0}

. Applying L'Hopital's rule,

x \to 1 lim \frac{n x ^{n - 1} f ( 1 ) - f ^{'} ( x )}{1} = 9 n - f^{'} (1) .

Since

f^{'} (x) = 6 x^{5} + 8 x^{3} + 3 x^{2} + 2

,

f^{'} (1) = 6 + 8 + 3 + 2 = 19

. Setting

9 n - 19 = 44

gives

9 n = 63

, so

n = 7

.

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Integer answer

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