# Let g : (0, ∞) → R be a differentiable function such that ∫(x(cosx-sinx)ex+1+g(x)(ex+1-xex)(ex+1)2)dx=xg(x)ex+1+c, for all x > 0, where c is an arbitrary constant. Then ______. - Mathematics (JEE Main)

MCQ
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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 ______.

#### Options

• 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)

#### Solution

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

(ex + 1) x (cos x – sin x) + g(x) (ex + 1 – xex)

= (ex + 1) (g(x) + xg'(x)) – ex. 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)

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