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Question: Let \(f:R\to R\) and \(g:R\to R\) be two non-constant differentiable function. If \(f'(x)={{e}^{(f(x...

Let f:R→Rf:R\to R and g:R→Rg:R\to R be two non-constant differentiable function. If f′(x)=e(f(x)−g(x))g′(x)f'(x)={{e}^{(f(x)-g(x))}}g'(x) for all x∈Rx\in R , and f ( 1 ) = g ( 2 ) = 1, then which of the following statement(s) is are true?
( a ) f(2)<1−log⁡e2f(2)<1-{{\log }_{e}}2
( b ) f(2)>1−log⁡e2f(2)>1-{{\log }_{e}}2
( c ) g(1)>1−log⁡e2g(1)>1-{{\log }_{e}}2
( d ) g(1)<1−log⁡e2g(1)<1-{{\log }_{e}}2

Explanation

Solution

To solve this question, first we will separate the question by variable separation method, then we will integrate the function. After that, we will use some properties of positive real numbers and also of exponential and logarithmic function and hence we will obtain the two relations.

Complete step-by-step answer:
Now, it is given that f′(x)=e(f(x)−g(x))g′(x)f'(x)={{e}^{(f(x)-g(x))}}g'(x).
So, we can write f′(x)=e(f(x)−g(x))g′(x)f'(x)={{e}^{(f(x)-g(x))}}g'(x) as
f′(x)=ef(x)e−g(x)g′(x)f'(x)={{e}^{f(x)}}{{e}^{-g(x)}}g'(x)
Or, f′(x)ef(x)=e−g(x)g′(x)\dfrac{f'(x)}{{{e}^{f(x)}}}={{e}^{-g(x)}}g'(x)
Or, e−f(x)f′(x)=e−g(x)g′(x){{e}^{-f(x)}}f'(x)={{e}^{-g(x)}}g'(x)
Or, e−f(x)f′(x)−e−g(x)g′(x)=0{{e}^{-f(x)}}f'(x)-{{e}^{-g(x)}}g'(x)=0
Integrating both side, we get
∫(e−f(x)f′(x)−e−g(x)g′(x))=∫0dx\int{({{e}^{-f(x)}}f'(x)-{{e}^{-g(x)}}g'(x))}=\int{0}dx
Or, ∫e−f(x)f′(x)−∫e−g(x)g′(x)=c1\int{{{e}^{-f(x)}}f'(x)-\int{{{e}^{-g(x)}}g'(x})}={{c}_{1}}, as integration of 0 is constant
Let, f ( x ) = t, then f’( x )dx = dt
And, g ( x ) = u, then g’( x )dx = du
∫e−tdt−∫e−udu=c1\int{{{e}^{-t}}dt-\int{{{e}^{-u}}du}}={{c}_{1}}
We know that ∫eaxdx=eaxa+c\int{{{e}^{ax}}dx=\dfrac{{{e}^{ax}}}{a}+c} ,
So, −e−t+c2−(−e−u)+c3=c1-{{e}^{-t}}+{{c}_{2}}-(-{{e}^{-u}})+{{c}_{3}}={{c}_{1}}
On simplifying, we get
−e−t+e−u=c1−c2−c3-{{e}^{-t}}+{{e}^{-u}}={{c}_{1}}-{{c}_{2}}-{{c}_{3}}
On replacing t and u, we get
−e−f(x)+e−g(x)=c-{{e}^{-f(x)}}+{{e}^{-g(x)}}=c, where c1−c2−c3=c{{c}_{1}}-{{c}_{2}}-{{c}_{3}}=c
So, −e−f(x)+e−g(x)=c-{{e}^{-f(x)}}+{{e}^{-g(x)}}=cis constant function.
Now at x = 1 and x = 2, we get
−e−f(1)+e−g(1)=c-{{e}^{-f(1)}}+{{e}^{-g(1)}}=c and −e−f(2)+e−g(2)=c-{{e}^{-f(2)}}+{{e}^{-g(2)}}=c,
So, −e−f(1)+e−g(1)=−e−f(2)+e−g(2)-{{e}^{-f(1)}}+{{e}^{-g(1)}}=-{{e}^{-f(2)}}+{{e}^{-g(2)}}
Also it is given in question that f ( 1 ) = g ( 2 ) = 1, we get
−e−1+e−g(1)=−e−f(2)+e−1-{{e}^{-1}}+{{e}^{-g(1)}}=-{{e}^{-f(2)}}+{{e}^{-1}}
On simplifying, we get
−1e+e−g(1)=−e−f(2)+1e-\dfrac{1}{e}+{{e}^{-g(1)}}=-{{e}^{-f(2)}}+\dfrac{1}{e}
Or, e−g(1)+e−f(2)=1e+1e{{e}^{-g(1)}}+{{e}^{-f(2)}}=\dfrac{1}{e}+\dfrac{1}{e}
Again, on simplification
e−g(1)+e−f(2)=2e{{e}^{-g(1)}}+{{e}^{-f(2)}}=\dfrac{2}{e}
Now, as exponential function is always positive for any real x,
So, if we have two positive quantities say, a and b and if on adding we get c which will also be positive,
That is, a + b = c, then a < c and b < c
So, for e−g(1)+e−f(2)=2e{{e}^{-g(1)}}+{{e}^{-f(2)}}=\dfrac{2}{e}, we can say that
e−g(1)<2e{{e}^{-g(1)}}<\dfrac{2}{e} and e−f(2)<2e{{e}^{-f(2)}}<\dfrac{2}{e}
Taking log on both side, we get
ln⁡(e−g(1))<ln⁡(2e)\ln ({{e}^{-g(1)}})<\ln \left( \dfrac{2}{e} \right)and ln⁡(e−f(2))<ln⁡(2e)\ln ({{e}^{-f(2)}})<\ln \left( \dfrac{2}{e} \right)
As we know that ln⁡a(af(x))=f(x){{\ln }_{a}}({{a}^{f(x)}})=f(x) ,
So, −g(1)<ln⁡(2e)-g(1)<\ln \left( \dfrac{2}{e} \right) and −f(2)<ln⁡(2e)-f(2)<\ln \left( \dfrac{2}{e} \right),
We know that, ln⁡(ab)=ln⁡a−ln⁡b\ln \left( \dfrac{a}{b} \right)=\ln a-\ln band ln( e ) = 1, then we get
f(2)>1−log⁡e2f(2)>1-{{\log }_{e}}2 and g(1)>1−log⁡e2g(1)>1-{{\log }_{e}}2

So, the correct answers are “Option (b) and (c)”.

Note: To solve such question one must know all functions properties properly, and always some formulas and values of log such that ln⁡(ab)=ln⁡a−ln⁡b\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b, ln⁡a(af(x))=f(x){{\ln }_{a}}({{a}^{f(x)}})=f(x) and ln( e ) = 1. Try to avoid silly calculation errors in the solution of the question.