If \(a_{n}\) is the greatest term in the sequence \(a_{n}=\frac{n^{3}}{n^{4}+147}, n=1,2,3 \ldots\), then \(n\) is equal to ______.

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Application of Derivatives

a_{n}

is the greatest term in the sequence

a_{n} = \frac{n ^{3}}{n ^{4} + 147}, n = 1, 2, 3 \dots

, then

n

is equal to ______.

Answer: 5

Treating the general term as a continuous function

f (x) = \frac{x ^{3}}{x ^{4} + 147}

and differentiating,

f^{'} (x) = \frac{3 x ^{2} ( x ^{4} + 147 ) - x ^{3} ( 4 x ^{3} )}{( x ^{4} + 147 ) ^{2}} = \frac{x ^{2} ( 441 - x ^{4} )}{( x ^{4} + 147 ) ^{2}} .

The sign of

f^{'} (x)

is governed by

441 - x^{4}

, which is positive for

x^{4} < 441

and negative for

x^{4} > 441

. Thus

f

increases up to

x = 44 1^{1/4} = 21 \approx 4.58

and decreases afterwards, so the greatest term occurs at the nearest integer,

n = 4

n = 5

a_{4} = \frac{64}{403} \approx 0.1588, a_{5} = \frac{125}{772} \approx 0.1619.

Since

a_{5} > a_{4}

, the greatest term of the sequence is attained at

n = 5

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Integer answer