JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 5)
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?
f(2) < 1 $$-$$ loge 2
f(2) > 1 $$-$$ loge 2
g(1) > 1 $$-$$ loge 2
g(1) < 1 $$-$$ loge 2
Explanation
We have,
$$f'(x) = {e^{(f(x) - g(x))}}g'(x)\forall x \in R$$
$$ \Rightarrow f'(x) = {{{e^{f(x)}}} \over {{e^{g(x)}}}}g'(x)$$
$$ \Rightarrow {{f'(x)} \over {{e^{f(x)}}}} = {{g'(x)} \over {{e^{g(x)}}}}$$
$$ \Rightarrow {e^{ - f(x)}}f'(x) = {e^{ - g(x)}}g'(x)$$
On integrating both side, we get
$${e^{ - f(x)}}$$ = $${e^{ - g(x)}}$$ + C
At x = 1
$${e^{ - f(1)}}$$ = $${e^{ - g(1)}}$$ + C
$${e^{ - 1}} = {e^{ - g(1)}} + C$$ [$$ \because $$ f(1) = 1] ...(i)
At x = 2
$${e^{ - f(2)}}$$ = $${e^{ - g(2)}}$$ + C
$$ \Rightarrow $$ $${e^{ - f(2)}}$$ = $${e^{ - 1}}$$ + C [$$ \because $$ g(2) = 1] .... (ii)
From Eqs. (i) and (ii)
$${e^{ - f(2)}}$$ = $$2{e^{ - 1}}$$ $$-$$ $${{e^{ - g(1)}}}$$ ...(iii)
$$ \Rightarrow $$ $${e^{ - f(2)}}$$ > $$2{e^{ - 1}}$$
We know that, e$$-$$x is decreasing
$$ \therefore $$ $$-$$f(2) < loge 2$$-$$1
f(2) > 1 $$-$$ loge 2
$$ \Rightarrow $$ $${{e^{ - g(1)}}}$$ + $${e^{ - f(2)}}$$ = $$2{e^{ - 1}}$$ [from Eq. (iii)]
$$ \Rightarrow $$ $${{e^{ - g(1)}}}$$ < $$2{e^{ - 1}}$$ $$-$$g(1) < loge 2 $$-$$1
$$ \Rightarrow $$ g(1) >$$-$$loge 2
$$f'(x) = {e^{(f(x) - g(x))}}g'(x)\forall x \in R$$
$$ \Rightarrow f'(x) = {{{e^{f(x)}}} \over {{e^{g(x)}}}}g'(x)$$
$$ \Rightarrow {{f'(x)} \over {{e^{f(x)}}}} = {{g'(x)} \over {{e^{g(x)}}}}$$
$$ \Rightarrow {e^{ - f(x)}}f'(x) = {e^{ - g(x)}}g'(x)$$
On integrating both side, we get
$${e^{ - f(x)}}$$ = $${e^{ - g(x)}}$$ + C
At x = 1
$${e^{ - f(1)}}$$ = $${e^{ - g(1)}}$$ + C
$${e^{ - 1}} = {e^{ - g(1)}} + C$$ [$$ \because $$ f(1) = 1] ...(i)
At x = 2
$${e^{ - f(2)}}$$ = $${e^{ - g(2)}}$$ + C
$$ \Rightarrow $$ $${e^{ - f(2)}}$$ = $${e^{ - 1}}$$ + C [$$ \because $$ g(2) = 1] .... (ii)
From Eqs. (i) and (ii)
$${e^{ - f(2)}}$$ = $$2{e^{ - 1}}$$ $$-$$ $${{e^{ - g(1)}}}$$ ...(iii)
$$ \Rightarrow $$ $${e^{ - f(2)}}$$ > $$2{e^{ - 1}}$$
We know that, e$$-$$x is decreasing
$$ \therefore $$ $$-$$f(2) < loge 2$$-$$1
f(2) > 1 $$-$$ loge 2
$$ \Rightarrow $$ $${{e^{ - g(1)}}}$$ + $${e^{ - f(2)}}$$ = $$2{e^{ - 1}}$$ [from Eq. (iii)]
$$ \Rightarrow $$ $${{e^{ - g(1)}}}$$ < $$2{e^{ - 1}}$$ $$-$$g(1) < loge 2 $$-$$1
$$ \Rightarrow $$ g(1) >$$-$$loge 2
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