If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$, then $f' \left( \frac{\pi}{4} \right)$ is equal to

Let $f$ be a real-valued function defined on the interval $(0, \infty)$ by $f(x)=\ln x+\int_0^x \sqrt{1+\sin t} \, dt$. Then which of the following statement$(s)$ is (are) true?
$(A)$ $f^{\prime \prime}(x)$ exists for all $x \in(0, \infty)$
$(B)$ $f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$,but not differentiable on $(0, \infty)$
$(C)$ there exists $\alpha>1$ such that $|f^{\prime}(x)|<|f(x)|$ for all $x \in(\alpha, \infty)$
$(D)$ there exists $\beta>0$ such that $|f(x)|+|f^{\prime}(x)| \leq \beta$ for all $x \in(0, \infty)$

In the usual notation,the value of $\Delta \nabla$ is equal to

Let $h$ be a twice continuously differentiable positive function on an open interval $J.$ Let $g(x) = \ln(h(x))$ for each $x \in J$. Suppose $(h'(x))^2 > h''(x) h(x)$ for each $x \in J$. Then

Let $f$ be a differentiable function and the equation of the normal to the graph of $y = f(x)$ at $x = 3$ is $3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {3 + {{\left( {f\left( 3 \right) - x - 6} \right)}^2}} \right)}}{{{{\sin }^2}\left( {x - 1} \right)}}$,then:

Let $f$ and $g$ be real-valued functions defined on the interval $(-1, 1)$ such that $g^{\prime \prime}(x)$ is continuous,$g(0) \neq 0$,$g^{\prime}(0) = 0$,$g^{\prime \prime}(0) \neq 0$,and $f(x) = g(x) \sin x$.
$STATEMENT-1$: $\lim_{x \rightarrow 0} [g(x) \cot x - g(0) \operatorname{cosec} x] = f^{\prime \prime}(0)$.
$STATEMENT-2$: $f^{\prime}(0) = g(0)$.