Warmup Test(3)!

Calculus Level 2

Let f : R → R f :R\rightarrow R and g : R → R g:R \rightarrow R be two non-constant differentiable functions . If f ′ ( x ) = ( e f ( x ) − g ( x ) ) g ′ ( x ) f'(x) = (e^{f(x) - g(x)}) g'(x) for all real values of x x and f ( 1 ) = g ( 2 ) = 1 , f(1) = g(2) = 1, then which of the following statements is(are) TRUE?

1 ) f ( 2 ) < 1 − ln ⁡ ( 2 ) 1) f(2) < 1 - \ln(2)

2 ) f ( 2 ) > 1 − ln ⁡ ( 2 ) 2) f(2) > 1 - \ln(2)

3 ) g ( 1 ) > 1 − ln ⁡ ( 2 ) 3) g(1) > 1 - \ln(2)

4 ) g ( 1 ) < 1 − ln ⁡ ( 2 ) 4) g(1) < 1 - \ln(2)

Suppose your answer is(are) option numbers a a and b b where ( a < b ) (a<b) numerically . Then enter your answer as ∣ 4 a − b ∣ . |4a -b|.


The answer is 5.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

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

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

Integrating we get, e − f ( x ) = e − g ( x ) + C e^{-f(x)} = e^{-g(x)} + C where C C is a abritrary constant.

Now, we know f ( 1 ) = g ( 2 ) = 1 f(1) = g(2) = 1

Put x = 1 , x=1, e − 1 = e − g ( 1 ) + C e^{-1} = e^{-g(1)} + C

Put x = 2 , x=2, e − f ( 2 ) = e − 1 + C e^{-f(2)} = e^{-1} + C

1 e − 1 e g ( 1 ) = 1 e f ( 2 ) − 1 e \dfrac 1e - \dfrac {1}{e^{g(1)}} = \dfrac {1}{e^{f(2)}} - \dfrac 1e

e g ( 1 ) = e f ( 2 ) + 1 2 e f ( 2 ) − e > 0 ⇒ e f ( 2 ) > e 2 ⇒ f ( 2 ) > 1 − ln ⁡ ( 2 ) e^{g(1)} = \dfrac {e^{f(2)+1}}{2e^{f(2)} - e} >0 \Rightarrow e^{f(2)} > \dfrac e2 \Rightarrow\boxed{ f(2) > 1 - \ln(2)}

Also, e f ( 2 ) = 2 e g ( 1 ) − 1 e g ( 1 ) + 1 > 0 ⇒ e g ( 1 ) > e 2 ⇒ g ( 1 ) > 1 − ln ⁡ ( 2 ) e^{f(2)} = \dfrac {2e^{g(1)} - 1}{e^{g(1) +1}} >0 \Rightarrow e^{g(1)} > \dfrac e2 \Rightarrow \boxed {g(1)>1 - \ln(2)}

Therefore , A N S W E R = 2 + 3 = 5 ANSWER = 2 + 3 = \boxed{5}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Just a suggestion, but if you wanna keep multi correct options, I suggest keeping the answer as the concatenation of the serial numbers of options, instead of the sums. For example, someone getting a 1) and a 3) would be equivalent to someone getting just a 4). This won't be a problem if we use concatenation, i.e. if someone gets 1,3 they can input the answer as 13.

Parth Sankhe - 2 years, 6 months ago

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Nice point, I will make the changes, Thanks :)

A Former Brilliant Member - 2 years, 6 months ago

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