The value of the function f(x)=1+x+∫ limits1x( ln 2 t+2 ln t) d t, where f prime(x) vanishes is:

Question

The value of the function f(x)=1+x+∫ limits1x( ln 2 t+2 ln t) d t, where f prime(x) vanishes is: (A) e-1 (B) 0 (C) 2 e -1 (D) 1+2 e-1

Tardigrade Admin · Accepted Answer

f(x)=1+x+∫ limits1x(ℓ n2 t+2 ℓ n t) d t Differentiate both sides w.r.t. x by using Leibinitz theorem f prime( x )=1+ℓ n 2 x +2 ln x =0 (1+ℓ nx )2=0 ∴ ℓ nx =-1 ∴ x =(1/ e )

Q. The value of the function $f (x) = 1 + x + 1 \int x (ln^{2} t + 2 ln t) d t$ , where $f^{'} (x)$ vanishes is:

$e^{- 1}$

0

$2 e^{- 1}$

$1 + 2 e^{- 1}$

$f (x) = 1 + x + 1 \int x (ℓ n^{2} t + 2 ℓ n t) d t$
Differentiate both sides w.r.t. $x$
by using Leibinitz theorem
$f^{'} (x) = 1 + ℓ n^{2} x + 2 ln x = 0$
$(1 + ℓ n x)^{2} = 0 ∴ ℓ n x = - 1$
$∴ x = \frac{1}{e}$

Q. The value of the function f(x)=1+x+1∫x​(ln2t+2lnt)dt, where f′(x) vanishes is:

e−1

0

2e−1

1+2e−1

f(x)=1+x+1∫x​(ℓn2t+2ℓnt)dt Differentiate both sides w.r.t. x by using Leibinitz theorem f′(x)=1+ℓn2x+2lnx=0 (1+ℓnx)2=0∴ℓnx=−1 ∴x=e1​

Q. The value of the function $f (x) = 1 + x + 1 \int x (ln^{2} t + 2 ln t) d t$ , where $f^{'} (x)$ vanishes is:

$e^{- 1}$

$2 e^{- 1}$

$1 + 2 e^{- 1}$

$f (x) = 1 + x + 1 \int x (ℓ n^{2} t + 2 ℓ n t) d t$
Differentiate both sides w.r.t. $x$
by using Leibinitz theorem
$f^{'} (x) = 1 + ℓ n^{2} x + 2 ln x = 0$
$(1 + ℓ n x)^{2} = 0 ∴ ℓ n x = - 1$
$∴ x = \frac{1}{e}$