COMPASSLearning

Talk to us Register Login

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Limits XII

If

lim_{t \to 0} (\int_{0}^{1} (3 x + 5)^{t} d x)^{\frac{1}{t}} = \frac{α}{5 e} (\frac{8}{5})^{\frac{2}{3}}

, then

α

is equal to ____.

Answer: 64

Solution

The limit has the

1^{\infty}

form. Writing

I (t) = \int_{0}^{1} (3 x + 5)^{t} d x = \frac{8 ^{t + 1} - 5 ^{t + 1}}{3 ( t + 1 )}

, note

I (0) = 1

, so

ln L = t \to 0 lim \frac{ln I ( t )}{t} = \frac{I ^{'} ( 0 )}{I ( 0 )} = I^{'} (0)

Differentiating

I (t)

and evaluating at

t = 0

:

I^{'} (0) = \frac{( 8 ln 8 - 5 ln 5 ) \cdot 3 - 3 \cdot 3}{9} = \frac{8 ln 8 - 5 ln 5 - 3}{3}

So

L = e^{\frac{8 l n 8 - 5 l n 5 - 3}{3}} = \frac{1}{e} (\frac{8 ^{8}}{5 ^{5}})^{1/3} = \frac{1}{e} \cdot \frac{2 ^{8}}{5 ^{5/3}} = \frac{256}{5 e} \cdot 5^{- 2/3}

Writing

5^{- 2/3} = (\frac{8}{5})^{2/3} \cdot \frac{1}{4}

(since

(\frac{8}{5})^{2/3} = \frac{4}{5 ^{2/3}}

):

L = \frac{64}{5 e} (\frac{8}{5})^{2/3}

Comparing with

\frac{α}{5 e} (\frac{8}{5})^{2/3}

gives

α = 64

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs by year and chapter →

More JEE Main questions on Limits XII

Let $[t]$ be the greatest integer less than or equal to $t$ . Then the least value of $p \in N$ for which …
JEE Main 2025View solution →
Let $f : R \to R$ be a continuous function satisfying $f (0) = 1$ and $f (2 x) - f (x) = x$ for all $x \in R$ . If $lim_{n \to \infty} {f (x) - f (\frac{x}{2 ^{n}})} = G (x)$ , then …
JEE Main 2025View solution →
Let $f : R \to R$ be a positive increasing function with $lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )} = 1$ . Then $lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} =$
JEE Main 2010View solution →