Let f: R arrow R be a continuous function defined by f(x)= 1 e^x+2 e^ -x . Statement-1: f(c)=(1)/(3) , for some c ∈ R . Statement-2: 0 < f(x) ≤ 1 2 2 , for all x ∈ R

Correct answer: D. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 f(x)= 1 e^x+2 e^ -x = e^x e^ 2 x +2 f^ (x)= (e^ 2 x +2) e^x-2 e^ 2 x · e^x (e^ 2 x+2 )^2 f^ (x)=0 e^ 2 x +2=2 e^ 2 x e^ 2 x =2 e^ x = 2 maximum f(x)= 2 4 = 1 2 2 0 < f(x) ≤ 1 2 2 x ∈ R Since 0 < (1)/(3) < 1 2 2 for some c ∈ R f(c)=(1)/(3)

JEE Main 2010

Exam: JEE Main
Year: 2010
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2010 · Mathematics · Continuity and Differentiability

Let f: R arrow R be a continuous function defined by f(x)= 1 e^x+2...

Let

f: R \rightarrow R

be a continuous function defined by

f(x)=\frac{1}{e^x+2 e^{-x}}

. Statement-1:

f(c)=\frac{1}{3}

, for some

c \in R

. Statement-2:

0 < f(x) \leq \frac{1}{2 \sqrt{2}}

, for all

x \in R

Options

AStatement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
BStatement-1 is true, Statement-2 is false
CStatement-1 is false, Statement-2 is true
DStatement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Correct answer

D. Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Step-by-step solution

f(x)=\frac{1}{e^x+2 e^{-x}}=\frac{e^x}{e^{2 x}+2}

f^{\prime}(x)=\frac{\left(e^{2 x}+2\right) e^x-2 e^{2 x} \cdot e^x}

{

\left(e^{2 x+2}\right)^2}

f^{\prime}(x)=0 \quad \Rightarrow e^{2 x}+2=2 e^{2 x}

\mathrm{e}^{2 \mathrm{x}}=2 \quad \Rightarrow \mathrm{e}^{\mathrm{x}}=\sqrt{2}

maximum

f(x)=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}}

0 < f(x) \leq \frac{1}{2 \sqrt{2}} \quad \forall x \in R

Since

0 < \frac{1}{3} < \frac{1}{2 \sqrt{2}} \quad \Rightarrow

for some

c \in R

f(c)=\frac{1}{3}

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