Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma3abe: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma3abe Solution: Let f : R R and g : R R be two non-constant differentiable…

Let f : R →\to R and g : R →\to R be two non-constant differentiable functions. If f'(x) = (e(f(x) −- g(x))) g'(x) for all x ∈\in R and f(1) = g(2) = 1, then which of the following statement(s) is (are) TRUE?

Multiple Choice Options

  • A f(2) < 1 −- loge 2
  • B f(2) > 1 −- loge 2
  • C g(1) > 1 −- loge 2
  • D g(1) < 1 −- loge 2

Step-by-Step Verified Solution

We have,

f′(x)=e(f(x)−g(x))g′(x)∀x∈Rf'(x) = {e^{(f(x) - g(x))}}g'(x)\forall x \in R

⇒f′(x)=ef(x)eg(x)g′(x)\Rightarrow f'(x) = {{{e^{f(x)}}} \over {{e^{g(x)}}}}g'(x)

⇒f′(x)ef(x)=g′(x)eg(x)\Rightarrow {{f'(x)} \over {{e^{f(x)}}}} = {{g'(x)} \over {{e^{g(x)}}}}

⇒e−f(x)f′(x)=e−g(x)g′(x)\Rightarrow {e^{ - f(x)}}f'(x) = {e^{ - g(x)}}g'(x)

On integrating both side, we get

e−f(x){e^{ - f(x)}} = e−g(x){e^{ - g(x)}} + C

At x = 1

e−f(1){e^{ - f(1)}} = e−g(1){e^{ - g(1)}} + C

e−1=e−g(1)+C{e^{ - 1}} = {e^{ - g(1)}} + C [ ∵\because f(1) = 1] ...(i)

At x = 2

e−f(2){e^{ - f(2)}} = e−g(2){e^{ - g(2)}} + C

⇒\Rightarrow e−f(2){e^{ - f(2)}} = e−1{e^{ - 1}} + C [ ∵\because g(2) = 1] .... (ii)

From Eqs. (i) and (ii)

e−f(2){e^{ - f(2)}} = 2e−12{e^{ - 1}} −- e−g(1){{e^{ - g(1)}}} ...(iii)

⇒\Rightarrow e−f(2){e^{ - f(2)}} > 2e−12{e^{ - 1}}

We know that, e −- x is decreasing

∴\therefore −- f(2) < loge 2 −- 1

f(2) > 1 −- loge 2

⇒\Rightarrow e−g(1){{e^{ - g(1)}}} + e−f(2){e^{ - f(2)}} = 2e−12{e^{ - 1}} [from Eq. (iii)]

⇒\Rightarrow e−g(1){{e^{ - g(1)}}} < 2e−12{e^{ - 1}} −- g(1) < loge 2 −- 1

⇒\Rightarrow g(1) > −- loge 2

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