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← JEE Advanced 2018 · Paper 1
MathematicsMultiple correct+4 / -2Question 41 of 54

JEE Advanced 2018 · Paper 1, Mathematics question 5

Let f:R→Rf: \mathbb{R} \to \mathbb{R} and g:R→Rg: \mathbb{R} \to \mathbb{R} be two non-constant differentiable functions. If

f′(x)=(e(f(x)−g(x)))g′(x)for all x∈R,f'(x) = \left(e^{(f(x) - g(x))}\right) g'(x) \quad \text{for all } x \in \mathbb{R},

and f(1)=g(2)=1f(1) = g(2) = 1, then which of the following statement(s) is (are) TRUE?

Choose one or more options
Answer and worked solution

Official answer: B, C

Separate.
e−ff′=e−gg′  ⟹  e−f(x)−e−g(x)=k (constant).e^{-f}f'=e^{-g}g'\implies e^{-f(x)}-e^{-g(x)}=k\ \text{(constant)}.

Use the data. At x=1x=1: k=e−1−e−g(1)k=e^{-1}-e^{-g(1)}. At x=2x=2: e−f(2)=e−1+ke^{-f(2)}=e^{-1}+k. Hence
e−f(2)+e−g(1)=2e.e^{-f(2)}+e^{-g(1)}=\frac2e.

Bounds. Both terms are positive, so each is less than 2e\frac2e:
e−f(2)<2e  ⟹  f(2)>1−ln⁡2,e−g(1)<2e  ⟹  g(1)>1−ln⁡2.e^{-f(2)}<\frac2e\implies f(2)>1-\ln2,\qquad e^{-g(1)}<\frac2e\implies g(1)>1-\ln2.

(B), (C)

Topics:Variable separableMonotonicity
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