Question

Let g : (0, ∞) ${\displaystyle \to }$ R be a differentiable function such that ${\displaystyle \int (\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)(e^x+1-x{e}^x)}{{(e^x+1)}^2})dx=\frac{xg\left(x\right)}{e^x+1}+c}$, for all x > 0, where c is an arbitrary constant. Then ______.

Options

• g is decreasing in ${\displaystyle (0,\frac{\pi }{4})}$

• g’ is increasing in ${\displaystyle (0,\frac{\pi }{4})}$

• g + g’ is increasing in ${\displaystyle (0,\frac{\pi }{2})}$

• g – g’ is increasing in ${\displaystyle (0,\frac{\pi }{2})}$

Answer 1

Let g : (0, ∞) ${\displaystyle \to }$ R be a differentiable function such that ${\displaystyle \int (\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)(e^x+1-x{e}^x)}{{(e^x+1)}^2})dx=\frac{xg\left(x\right)}{e^x+1}+c}$, for all x > 0, where c is an arbitrary constant. Then ${\displaystyle \underline{g-g^{’} \text{is increasing in}(0,\frac{\pi }{2})}}$.

Explanation:

Given integral is

${\displaystyle \int (\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)(e^x+1-x{e}^x)}{{(e^x+1)}^2})dx=\frac{xg\left(x\right)}{e^x+1}+c}$

On differentiating both sides w.r.t. x, we get

${\displaystyle (\frac{x(\cos x-\sin x)}{e^x+1}+\frac{g\left(x\right)(e^x+1-x{e}^x)}{{(e^x+1)}^2})}$

= ${\displaystyle \frac{(e^x+1)(g\left(x\right)+x{g}^{\prime }\left(x\right))-e^x.x.g\left(x\right)}{{(e^x+1)}^2}}$

(e^x + 1) x (cos x – sin x) + g(x) (e^x + 1 – xe^x)

= (e^x + 1) (g(x) + xg'(x)) – e^x. x g(x)

${\displaystyle \Rightarrow }$ g'(x) = cos x – sin x  ...(i)

Take integral both sides,

${\displaystyle \Rightarrow }$ g(x) = sin x + cos x + C

Take g(x) = 0; then x = ${\displaystyle \frac{\overline{x}}{4}}$

So, g(x) is increasing in ${\displaystyle (0,\frac{\pi }{4})}$

Again, differentiate w.r.t. x in equation (i),

g''(x) = – sin x – cos x < 0

${\displaystyle \Rightarrow }$ g'(x) is decreasing function

Let r (x) = g(x) + g'(x) = 2 cos x + C

${\displaystyle \Rightarrow }$ r'(x) = g'(x) + g''(x) = –2 sin x < 0

${\displaystyle \Rightarrow }$ r is decreasing

Let l(x) = g(x) – g'(x) = 2 sin x + C

Differentiate w.r.t. x

${\displaystyle \Rightarrow }$ l'(x) = g'(x) – g''(x) = 2 cos x > 0

Take l"(x) = 0; cos x = 0; x = ${\displaystyle \frac{\pi }{2}}$

${\displaystyle \Rightarrow }$ l is increasing

Therefore, l(x) is increasing at ${\displaystyle (0,\frac{\pi }{2})}$