JEE MAIN - Mathematics (2010 - No. 13)
Let $$f:R \to R$$ be a continuous function defined by
$$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$$
Statement - 1 : $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$.
Statement - 2 : $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$
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 false.
Statement - 1 is false, Statement - 2 is true.
Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1.
Explanation
$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$
$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$
$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$
$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$
maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$
$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$
Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$
$$f\left( c \right) = {1 \over 3}$$
$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$
$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$
$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$
maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$
$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$
Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$
$$f\left( c \right) = {1 \over 3}$$
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