Question

Let $f:R \to R$ be a continuous function defined by$f\left( x \right) = \dfrac{1}{{{e^x} + 2{e^{ - x}}}}$.
Statement-1: $f\left( c \right) = \dfrac{1}{3},$ for some $c \in R$
Statement-2: $0 < f\left( x \right) \leqslant \dfrac{1}{{2\sqrt 2 }}$, for all $x \in R$
(1) Statement-1 is True, Statement-2 is True; Statement-2 is not the correct explanation for Statement-1
(2) Statement-1 is True, Statement-2 is false
(3) Statement-1 is False, Statement-2 is true
(4) Statement-1 is True, Statement-2 is True; Statement-2 is the correct explanation for Statement-1

Answer 1

Hint: Try to find the range of the function $f\left( x \right)$, i.e. interval where its value will lie. Do that by finding the maximum value of the function. Use the first derivative to find the maximum point then the maximum value. Check for both the statements.

Complete step-by-step answer:
We are given with a function $f\left( x \right)$ mapped from real number to real number, defined as $f\left( x \right) = \dfrac{1}{{{e^x} + 2{e^{ - x}}}}$ which is continuous.
Let’s find the maximum value for the function $f\left( x \right)$. In a smoothly changing function, a maximum or minimum is always where the function flattens out (except for a saddle point). When a function's slope is zero at x, and the second derivative at x is: less than 0, it is a local maximum and greater than 0, it is a local minimum.
So, we have $f\left( x \right) = \dfrac{1}{{{e^x} + 2{e^{ - x}}}} = \dfrac{{{e^x}}}{{{e^{2x}} + 2}}$
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{e^x}}}{{{e^{2x}} + 2}}} \right) = \dfrac{{\left( {{e^{2x}} + 2} \right){e^x} - 2{e^{2x}} \times {e^x}}}{{{{\left( {{e^{2x}} + 2} \right)}^2}}}$
So, for finding the maximum, we need to put $f'\left( x \right) = 0$
$ \Rightarrow \dfrac{{\left( {{e^{2x}} + 2} \right){e^x} - 2{e^{2x}} \times {e^x}}}{{{{\left( {{e^{2x}} + 2} \right)}^2}}} = 0$
Now, since the left-hand side is in fraction, so we can make the numerator as zero:
$ \Rightarrow \left( {{e^{2x}} + 2} \right){e^x} - 2{e^{2x}} \times {e^x} = 0$
$ \Rightarrow \left( {{e^{2x}} + 2} \right){e^x} = 2{e^{2x}} \times {e^x}$
We can remove ${e^x}$ from both sides since it can’t be zero:
$ \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$
$ \Rightarrow {e^{2x}} = 2$
Therefore, we get the maxima as: ${e^x} = \sqrt 2 $
Let’s put this back into the$f\left( x \right)$, we get that:
Maximum $f\left( x \right) = \dfrac{{\sqrt 2 }}{{{{\left( {\sqrt 2 } \right)}^2} + 2}} = \dfrac{{\sqrt 2 }}{4} = \dfrac{1}{{2\sqrt 2 }}$
And since the exponential function can never be zero, we can say that $0 < f\left( x \right) < \dfrac{1}{{2\sqrt 2 }}$ for all real values of$x$.
Now, let’s look into Statement-1
We can say: $0 < \dfrac{1}{3} < \dfrac{1}{{2\sqrt 2 }}$
Therefore, $\dfrac{1}{3}$ can be the value of $f\left( x \right)$ for some $c$ belonging to the real number.
Hence, Statement-1 is true.
Thus, we can say Statement-1 is True, Statement-2 is True; Statement-2 is the correct explanation for Statement-1.

So, the option (4) is the correct answer.

Note: Try to go step by step with the procedure. For finding the maxima or minima of a function, the function must be differentiable. The function must also be continuous, but any differentiable function is also continuous, so no need to worry about that. For differentiation of a fraction we used a differentiation formula:$\dfrac{d}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)df\left( x \right) - f\left( x \right)dg\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$