Question

Let g : (0, ∞) `rightarrow` R be a differentiable function such that `int((x(cosx - sinx))/(e^x + 1) + (g(x)(e^x + 1 - xe^x))/(e^x + 1)^2)dx = (xg(x))/(e^x + 1) + c`, for all x > 0, where c is an arbitrary constant. Then ______.

विकल्प

• g is decreasing in `(0, π/4)`

• g’ is increasing in `(0, π/4)`

• g + g’ is increasing in `(0, π/2)`

• g – g’ is increasing in `(0, π/2)`

Answer 1

Let g : (0, ∞) `rightarrow` R be a differentiable function such that `int((x(cosx - sinx))/(e^x + 1) + (g(x)(e^x + 1 - xe^x))/(e^x + 1)^2)dx = (xg(x))/(e^x + 1) + c`, for all x > 0, where c is an arbitrary constant. Then `underlinebb(g - g^’  "is increasing in" (0, π/2)`.

Explanation:

Given integral is

`int((x(cosx - sinx))/(e^x + 1) + (g(x)(e^x + 1 - xe^x))/(e^x + 1)^2)dx = (xg(x))/(e^x + 1) + c`

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

`((x(cosx - sinx))/(e^x + 1) + (g(x)(e^x + 1 - xe^x))/(e^x + 1)^2)`

= `((e^x + 1)(g(x) + xg^'(x)) - e^x.x.g(x))/(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)

`\implies` g'(x) = cos x – sin x  ...(i)

Take integral both sides,

`\implies` g(x) = sin x + cos x + C

Take g(x) = 0; then x = `overlinex/4`

So, g(x) is increasing in `(0, π/4)`

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

g''(x) = – sin x – cos x < 0

`\implies` g'(x) is decreasing function

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

`\implies` r'(x) = g'(x) + g''(x) = –2 sin x < 0

`\implies` r is decreasing

Let l(x) = g(x) – g'(x) = 2 sin x + C

Differentiate w.r.t. x

`\implies` l'(x) = g'(x) – g''(x) = 2 cos x > 0

Take l"(x) = 0; cos x = 0; x = `π/2`

`\implies` l is increasing

Therefore, l(x) is increasing at `(0, π/2)`