Which of the following statement is (are) correct?

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Let $f : [0, 1] \to R$ be a differentiable function with $f (0) = 0,$ and $0 \leq f^{1} (x) \leq 2 f (x),$ then $f (x)$ is identically zero $\forall X \in [0, 1] .$

If $f (x) = x^{3} + a x^{2} + b x + 5 \sin^{2} x$ is an increasing function $\forall x \in R,$ then $a^{2} - 3 b + 15 \leq 0.$

If $x^{a} \leq a^{x} \forall x \in (1, \infty)$ where $a \in (1, \infty)$ then number of value of a satisfying the inequality is exactly one.

Let f(x) be a differentiable function in $[0, \infty)$ with $f (0) = 0$ and $f^{1} (x)$ is increasing function $\forall x \geq 0,$ then $g (x) = {\begin{cases} \frac{f (x)}{x}, x > 0 \\ f^{1} (0), x = 0 \end{cases}$ is a decreasing function.

answer is B, A, C.

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Detailed Solution

(A) given $f^{1} (x) - 2 f (x) \leq 0 ..... (1)$ and $f^{1} (x) \underline{>} 0 ..... (2)$ from equation (2), f(x) is increasing function $\forall x \underline{>} 0 \Rightarrow f (x) \underline{>} f (0) \Rightarrow f (x) \underline{>} 0$ ……(3)
From equation (1), $\frac{d}{d x} (f (x) e^{- 2 x}) \leq 0$
$\Rightarrow (f (x) e^{- 2 x})$ is decreasing function $\forall x \underline{>} 0$
Hence $f (x) e^{- 2 x} \leq f (0) \leq 0$
Hence $f (x) \leq 0 .... (4)$
From (3) and (4), f(x)=0
(B) $f (x) = x^{3} + a x^{2} + b x + 5 \sin^{2} x$
$f^{1} (x) = 3 x^{2} + 2 a x + b + \sin 2 x$
Hence $3 x^{2} + 2 a x + b + 5 \leq 3 x^{2} + 2 a x + b + 5 + \sin 2 x \leq 3 x^{2} + 2 a x + b + 5$
For f(x) to be strictly increasing $f^{1} (x) > 0$
$\Rightarrow 3 x^{2} + 2 a x + b - 5 \underline{>} 0 \forall x$
$\Rightarrow D \leq 0 \Rightarrow a^{2} - 3 b + 15 \leq 0$
(C) Let $f (x) = x^{1 / x}$
f(x) is increasing in (1,e) and decreasing (e, $\infty$ ) and hence maximum value of f(x) occurs at x=e
$\Rightarrow f (x) \leq f (e) \forall x \underline{>} 1$ $\Rightarrow x^{1 / e} < e^{1 / e}$
Hence $x^{a} \leq a^{x} \forall x \in (1, \infty)$ $\Rightarrow a = e$
(D) $g^{1} (x) = \frac{x f^{1} (x) - f (x)}{x^{2}} f o r x > 0$
Now given $f^{1} (x)$ is an $↑$ function for $x > 0$
$f (x) = \int_{0}^{x} f^{1} (t) d t < \int_{0}^{x} f^{1} (x) d t$
Hence $f (x) < x f^{1} (x)$
$\Rightarrow g^{1} (x) > 0 \forall x > 0$
Hence g(x) is an $↑$ function.