Let displaystyle ∫ (1 - ln x/x2)dx=f(x), for all positive x . If f(e)=(1/e), then f(2)+f(4) is equal to

Question

Let displaystyle ∫ (1 - ln x/x2)dx=f(x), for all positive x . If f(e)=(1/e), then f(2)+f(4) is equal to (A) ln 4 (B) ln 2 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

f(x)= displaystyle ∫ (1/x2)dx- displaystyle ∫ ln x⋅ (1/x2)dx =-(1/x)- ln x (- (1/x)) + displaystyle ∫ (1/x2) d x =-(1/x)+(ln x/x)+(1/x)+c=(ln â¡ x/x)+c Now, f(e)=(1/e)⇒ c=0 Hence, f(2)+f(4)=(ln 2/2)+(ln â¡ 4/4)=ln2

Q. Let $\int \frac{1 - l n x}{x ^{2}} d x = f (x),$ for all positive $x$ . If $f (e) = \frac{1}{e},$ then $f (2) + f (4)$ is equal to

$l n 4$

$l n 2$

$1$

$0$

Q. Let ∫x21−lnx​dx=f(x), for all positive x . If f(e)=e1​, then f(2)+f(4) is equal to

ln4

ln2

1

0

f(x)=∫x21​dx−∫lnx⋅x21​dx =−x1​−{lnx(−x1​)+∫x21​dx} =−x1​+xlnx​+x1​+c=xln⁡x​+c Now, f(e)=e1​⇒c=0 Hence, f(2)+f(4)=2ln2​+4ln⁡4​=ln2

Q. Let $\int \frac{1 - l n x}{x ^{2}} d x = f (x),$ for all positive $x$ . If $f (e) = \frac{1}{e},$ then $f (2) + f (4)$ is equal to

$l n 4$

$l n 2$

$1$

$0$