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

Question

Let ∫ (1- ln x/x2) d x=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)=∫ (1/x2) d x-∫ ln x ⋅ (1/x2) d x =-(1/x)- ln x(-(1/x))+∫ (1/x2) d x =-(1/x)+( ln x/x)+(1/x)+c=( ln x/x)+c Now, f(e)=(1/e) ⇒ c=0 Hence, f(2)+f(4)=( ln 2/2)+( ln 4/4)= ln 2

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

$ln 4$

$ln 2$

$1$

$0$

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=xlnx​+c Now, f(e)=e1​⇒c=0 Hence, f(2)+f(4)=2ln2​+4ln4​=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

$ln 4$

$ln 2$

$1$

$0$