Let f:R R, be a continuous function defined by f ( x ) = 1 e^x + … - Vidyadip

MCQ

Let $f:R \to R,$ be a continuous function defined by $f\left( x \right) = \frac{1}{{{e^x} + 2{e^{ - x}}}}$

Statement $-1 :$ $f\left( c \right) = \frac{1}{3}$ for some $c\; \in R$

Statement $-2 :$$0 < f\left( x \right) < \frac{1}{{2\sqrt 2 }}\;,\forall x\; \in R$

A
Statement $-1$ is false, Statement $-2$ is true
B
Statement $-1$ is true, Statement $-2$ is false
C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
✓
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

✓

Answer

Correct option: D.

Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

d
$f\left( x \right) = \frac{1}{{{e^x} + 2{e^{ - x}}}} = \frac{{{e^x}}}{{{e^{2x}} + 2}}$

${f^1}\left( x \right) = \frac{{\left( {{e^{2x + 2}}} \right){e^x} - 2{e^{2x}}.{e^x}}}{{{{\left( {{e^{2x + 2}}} \right)}^2}}}$

${f^1}\left( x \right) = 0$

${e^{2x}} = 2$

$\Rightarrow e^{2 x}+2=2 e^{2 x}$

$\Rightarrow e^{x}=\sqrt{2}$

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

$0 < f\left( x \right) \le \frac{1}{{2\sqrt 2 }}\,\,\,\,\,\,\,\forall x \in R$

since $0 < f\left( x \right) \le \frac{1}{{2\sqrt 2 }}\,$ or $0 < \frac{1}{3} < \frac{1}{{2\sqrt 2 }}$

For sum $C \in R$

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

Statement $1$ is true

Statement $2$ is true

Statement $2$ is correct explanation of statement $1$

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