Question

Let f be a real valued function defined on the interval $\left( {0,\infty } \right)$ by $f\left( x \right) = \ln x + \int_0^x {\sqrt {1 + \sin t} dt} $, then which of the following statement(s) is(are) true?
$\left( a \right)f''\left( x \right)$ exists for all $x \in \left( {0,\infty } \right)$
$\left( b \right)$ f (x) exists for all $x \in \left( {0,\infty } \right)$ and f’ is continuous on $\left( {0,\infty } \right)$, but not differentiable on $\left( {0,\infty } \right)$
$\left( c \right)$ There exists a > 1 such that $\left| {f'\left( x \right)} \right| < \left| {f\left( x \right)} \right|$ for all $x \in \left( {a,\infty } \right)$
$\left( d \right)$ There exists $\beta > 0$ such that $\left| {f'\left( x \right)} \right| + \left| {f\left( x \right)} \right| \leqslant \beta $ for all $x \in \left( {0,\infty } \right)$

Answer 1

Hint: In this particular question use the concept that if the function is defined in the given interval than it is continuous in that interval and use the concept that if the left-hand derivative is not equal to the right-hand derivative than function is not differentiable in the given interval and use that $ - 1 \leqslant \sin x \leqslant 1$ so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
$f\left( x \right)$ be a real valued function defined on the interval $\left( {0,\infty } \right)$ by $f\left( x \right) = \ln x + \int_0^x {\sqrt {1 + \sin t} dt} $
Option (b)
So, f (x) should be continuous in the interval $\left( {0,\infty } \right)$.
Now as we know that $\dfrac{d}{{dx}}\left( {\int\limits_a^b {g\left( x \right)dx} } \right) = {\left( {g\left( x \right)} \right)_{x = b}} - {\left( {g\left( x \right)} \right)_{x = a}}$ so according to this property differentiate the given equation we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {\ln x + \int_0^x {\sqrt {1 + \sin t} dt} } \right)$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{1}{x} + {\left( {\sqrt {1 + \sin t} } \right)_{t = x}} - {\left( {\sqrt {1 + \sin t} } \right)_{t = 0}}$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{1}{x} + \left( {\sqrt {1 + \sin x} } \right) - \left( {\sqrt {1 + \sin 0} } \right)$
$ \Rightarrow f'\left( x \right) = \dfrac{1}{x} + \sqrt {1 + \sin x} - 1$.............. (1)
Now as we all know that $\sqrt x = \pm \sqrt x $
So, $\sqrt {1 + \sin x} = \pm \sqrt {1 + \sin x} $
So if we take left hand derivative, $\sqrt {1 + \sin x} = - \sqrt {1 + \sin x} $ in the interval $\left( {0,\infty } \right)$ and if we take right hand derivative, $\sqrt {1 + \sin x} = + \sqrt {1 + \sin x} $ in the interval $\left( {0,\infty } \right)$
So LHD $ \ne $ RHD so f (x) is not differentiable in the interval $\left( {0,\infty } \right)$.
Hence option (b) is correct.
Option (a)
Now again differentiate equation (1) we have,
$ \Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{x} + \sqrt {1 + \sin x} - 1} \right)$
Now as we know that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}\sqrt {g\left( x \right)} = \dfrac{1}{{2\sqrt {g\left( x \right)} }}\dfrac{d}{{dx}}g\left( x \right),\dfrac{d}{{dx}}\sin x = \cos x$ so we have,
$ \Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = \left( {\dfrac{{ - 1}}{{{x^2}}} + \dfrac{1}{{2\sqrt {1 + \sin x} }}\dfrac{d}{{dx}}\left( {1 + \sin x} \right) - 0} \right)$
$ \Rightarrow f''\left( x \right) = \left( {\dfrac{{ - 1}}{{{x^2}}} + \dfrac{{\cos x}}{{2\sqrt {1 + \sin x} }}} \right)$
Now in the interval $\left( {0,\infty } \right)$ at a particular value the value of sin x = -1, so $\sqrt {1 + \sin x} = 0$
$ \Rightarrow \dfrac{1}{{\sqrt {1 + \sin x} }}$ is not defined in the interval $\left( {0,\infty } \right)$.
So f”(x) is not differentiable in the interval $\left( {0,\infty } \right)$.
Hence option (a) is false.
Option (c) There exists a > 1 such that $\left| {f'\left( x \right)} \right| < \left| {f\left( x \right)} \right|$ for all $x \in \left( {a,\infty } \right)$
$ \Rightarrow \left| {f\left( x \right)} \right| - \left| {f'\left( x \right)} \right| > 0$
Now substitute the values we have,
$ \Rightarrow \left| {\ln x + \int_0^x {\sqrt {1 + \sin t} dt} } \right| - \left| {\dfrac{1}{x} + \sqrt {1 + \sin x} - 1} \right| > 0$
Now as we know that $ - 1 \leqslant \sin x \leqslant 1$ so the maximum value of $\left| {\dfrac{1}{x} + \sqrt {1 + \sin x} - 1} \right| = \sqrt 2 $
So in the interval there exists a value a > 1, such that $\left| {f\left( x \right)} \right| - \left| {f'\left( x \right)} \right| > 0$
Therefore, $\left| {f'\left( x \right)} \right| < \left| {f\left( x \right)} \right|$ for all $x \in \left( {a,\infty } \right)$
So option (c) is also correct.
Option (d) There exists $\beta > 0$ such that $\left| {f'\left( x \right)} \right| + \left| {f\left( x \right)} \right| \leqslant \beta $ for all $x \in \left( {0,\infty } \right)$
Now as we see that f (x) is an increasing function in the interval $x \in \left( {0,\infty } \right)$ whereas f’(x) is a bounded curve (I.e. not increasing function) in the interval $x \in \left( {0,\infty } \right)$, so after a certain value $\beta > 0$, f (x) dominating f’(x) in the interval $x \in \left( {0,\infty } \right)$.
So, $\left| {f'\left( x \right)} \right| + \left| {f\left( x \right)} \right| \geqslant \beta $, for $\beta > 0$ and $x \in \left( {0,\infty } \right)$.
Hence option (d) is also incorrect.
Hence options (b) and (c) are the correct answers.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic property that how to differentiate the definite integral which is given as $\dfrac{d}{{dx}}\left( {\int\limits_a^b {g\left( x \right)dx} } \right) = {\left( {g\left( x \right)} \right)_{x = b}} - {\left( {g\left( x \right)} \right)_{x = a}}$ so simply apply this property as above applied.